Comprehensive Guide to Circular Motion, Centripetal Force, and Gravitational Dynamics

Introduction to Circular Motion and Forces

Understanding how objects move in circular paths involves analyzing the interplay of velocity and force vectors. When force and velocity vectors align, objects speed up; when they oppose, objects slow down. Crucially, if these vectors are perpendicular, objects change direction but maintain speed, resulting in circular motion driven by centripetal force, a center-seeking force. For more on fundamental concepts of motion, see Understanding Motion: A Comprehensive Guide.

Sources of Centripetal Force

Tension: E.g., a ball swung on a rope has tension providing the centripetal force.
Gravity: For planetary bodies and satellites, gravitation acts as the centripetal force facilitating orbital motion.
Friction: Static friction between tires and road enables cars to navigate turns.
Normal Force: For objects following curved paths (like a ball on a curved road), normal force components provide centripetal force.
Magnetic Force: Moving charges in magnetic fields experience centripetal force from magnetic interactions.

Centripetal Force and Acceleration Equations

Centripetal force: (F_c = \frac{mv^2}{r})
Centripetal acceleration: (a_c = \frac{v^2}{r})
Frequency ((f)) and period ((T)) describe revolution rates and times.
Speed in circular motion relates to radius and period via (v = \frac{2\pi r}{T}).
Alternate formula for centripetal acceleration: (a_c = \frac{4\pi^2 r}{T^2}).

Effects of Changing Parameters on Centripetal Force

Doubling mass doubles (F_c).
Doubling velocity quadruples (F_c) (since velocity squared).
Increasing radius decreases (F_c) inversely.

Practical Applications and Example Problems

Tension in ropes: At high speeds, tension approximates centripetal force; at low speeds, tension includes vertical components to counteract weight.
Vertical circle motion: Tension varies with position, maximum at bottom and minimum at top due to weight components assisting or opposing centripetal force.
Banked curves: Calculations of friction and normal force components determine maximum safe speeds and angles for turns without slipping.
Acceleration in circular tracks: Combination of tangential (speed-up) and radial (direction change) accelerations requires vector addition for net acceleration.
Normal force on cars: Varies depending on vertical position along curved roads, combining weight and centripetal force effects.
Elevator physics: Scale readings reflect combined effects of gravitational and acceleration forces, altering perceived weight.

Gravity and Orbital Mechanics

Newton's law of gravitation quantifies force: (F = G \frac{m_1 m_2}{r^2}). For a detailed explanation, see Understanding Gravitational Forces and Their Applications in Physics.
Effects of mass and distance changes on gravitational force are proportional and inversely proportional to the square of the distance, respectively.
Calculations of gravitational acceleration on Earth, Moon, and varying altitudes demonstrate dependence on mass and radius.
Satellite speed and orbital period derive from balancing gravitational and centripetal forces, enabling orbital stability. For further insights, revisit Understanding Orbital Motion: The Role of Gravity and Speed.
Kepler's third law relates orbital periods and radii, used to predict planetary motions such as Mars' and Mercury's orbits.
Estimations of planetary masses and satellite geosynchronous heights illustrate application of gravitational and circular motion principles.

Summary

This overview provides foundational physics concepts and applied calculations for circular motion and gravitational systems. By mastering the forces responsible for circular paths, tension, friction, and gravitational attraction, one can solve complex real-world problems involving cars on curves, satellites in orbit, and planetary dynamics. Understanding these concepts is crucial for fields ranging from mechanical engineering to astrophysics. To expand your foundational knowledge further, consider reviewing the Complete Guide to Motion: Distance, Velocity, Acceleration & Projectile Physics.

in this video we're going to go over circular motion we're going to cover topics like centripetal force tension

satellite motion Newton's law of gravitation so gravity and things like

that so let's begin now let's say if you have a force Vector direct to the right and also a

velocity Vector what's going to happen if if an object is moving to the right and

there's a net force acting on it towards the right whenever the force and velocity

vectors are in the same direction the object speeds up the velocity will continue to

increase now what's going to happen if the force and velocity vectors if they're

anti-parallel or if they're in opposite directions in this case the object is going to slow down the speed is going to

decrease now what happens if the velocity vector and if the force

Vector are perpendicular to each other what is the object going to do will it speed up will it slow down or

will will the speed remain the same if the force is perpendicular to the

velocity it has no effect on the speed the magnitude of the Velocity Remains the Same however the direction

changes the object will begin to move in a circle whenever the force is

perpendicular to the velocity it acts as a cental force it causes the object to turn whenever you have a force that

directs an object to turn into a circle it's called a centripetal force or a center seeking

Force now at this point the velocity is directed in the upward Direction but the force is going to be

directed towards the center so the object is going to turn again it's going to turn in a direction

of the force so it's going to turn towards the left now at this point the velocity

Vector is directed West but the centripetal force still points towards the

center so it's going to turn towards the South and now the velocity Vector is directed towards the South and the

centripetal force is directed towards the center and it's going to turn towards

the east so it's going to move in a circle so the centripetal force causes an object to move in a

circle now there's many things that can provide the centri force the cental force is caused by something else for

example let's say if you have a rope and a ball attached to the end of the rope and if you swing it in a horizontal

Circle what provides the centripetal force on the ball in this case the centripetal force

is provided by tension tension is the force that acts through a rope and so tension is the force that

keeps the ball moving in a circle now what about the Earth and the moon what keeps the moon in orbit around

the Earth in this case it's gravity gravity provides the centripetal force gravity

is a force of attraction the Earth fills a gravitational force from the Moon and

the moon exerts a gravitational force on the earth they both exert a force on each other they're equal but opposite in

direction so the gravitational force caused by the Earth on the moon keeps the moon moving around the

earth so at this point the moon has a velocity that wants to send it out of space but gravity

keeps it in the circle so now when the moon is over here it has a tangental velocity that is directed towards the

West without gravity the moon would basically fly out of orbit but gravity is a force that's going to cause it to

turn and so it's going to turn towards the South now when the moon is over here it's going to be moving towards the

South but the gravitational force that the Earth exerts on the moon will cause it to turn East so now the Moon is over

here moving East and then the gravitational force is going to cause it to move towards the

North and so this cycle is going to keep repeating itself and so the moon remains in

orbit so in the case of planetary motion gravity provides the Cent tribal Force so gravity caused the moon to

orbit the earth and it caused the earth to orbit the sun now let say if we have a road and

let's say there's a car on a road I'm just going to draw a box to represent the

car so as the car turns we know that a c tributal Force anytime an object turns in a circle

there's always a cental force so what force provides the cental force whenever you have a car on a road

the friction between the road and the tires it provides the frictional the centripetal force it allows the car to

make that turn so static friction provides the srial force in this situation the reason why you want to

know what provides the centri force because when you're solving a problem sometimes you can set the cental force

equal to static friction sometimes you can set it equal to the gravitation force or sometimes you could set it

equal to tension so you want to know what provides the cental force because it's going to help you to come up with

the right equations to solve a problem as you'll see later in this video now let's say if you

have a road and then it curves like this so you're not turning let's say the object

is moving straight but it's not turning in a circle it's actually going down a

circle in this case what provides the centripetal force that causes it to turn so the ball in this case or the

object is not turning on its own let's say if you're driving a car you can turn the wheel towards the left in that case

static friction provides decent tribal Force but if you have an OP object that simply follows the curvature of the road

what provides these INRI force in this case I'll give you a hint what force does the road or the

surface exert on a ball the road exerts a normal force on the

ball so in this case as the ball follows the curvature of the road the road exerts a normal

force the normal force that's exerted on the road is not the same normal force here the magnitude is

different in this case the normal force is equal to the weight but in the case of this section

over here the normal force not only has to support the weight of the ball but it also has to provide an additional Force

to cause the object to turn so the normal force in this situation is greater than the wave Force

not only does it have to support the weight of the ball but it must also provide the cental force to turn the

ball into a circle or to cause it to change its direction so just understand that the

the value of the normal force is greater as the ball changes Direction in this region compared to this section over

here where it only has to support the weight of the ball now another case of a force

providing the centripetal force is the magnetic force now you may not need to know this for this chapter but later on

if you take uh a second course of physics or if you're in college if you're taking physics 2 you may need to

know this it has to do with the magnetic force on the moving

charge a magnetic field will exert a magnetic force on a charge if it's moving if the charge is not moving the

magnetic field exerts no force on it so let's say if you have a positive charge and it has a velocity Vector moving

towards the right and the magnetic field is directed into the

page so in this case the positive charge will fill a force that is directed upward so it's

going to cause this charge to move into a circle so in this case the magnet IC

Force provides the cental force on this moving charge so what is the equation for

cental force cental force is equal to mv^2 / R this is the equation that you need to

know cental force is proportional to the mass is proportional to the square of the

velocity and it's inversely related to the radius of the circle so if you increase the mass what's going

to happen to the magnitude of the cental force will it increase or decrease because the mass is on top

they're proportional so if the mass goes up the centripetal force goes up but by what factor if you double the mass the

cental force will double if you triple the mass the cental force will triple and the way you can figure this out is

by plugging the numbers in into the equation whatever doesn't change plug in a one if it changes plug in a number

that corresponds to that change so if we double the mass plug in two for M and if everything else stays the same plug in

one so then the force will equal two so that means that the force will double in magnitude if we triple the mass plug in

three for M and everything that stays the same keep it one so the force will triple

so using that same technique what happens if we increase the speed or the velocity of the object what's going to

happen to the centripetal force the centripetal force will increase so if you double the speed what effect will

that have on the cbal force so let's plug in two 2^2 is equal to 4 so the cpal force

will quadruple now what if we triple the speed so if we plug in three 3^2 is

9 so the centripetal force will increase by a factor of 9 if we quadruple the speed if we multiply by a factor of four

the cental force will increase by 4^ S or 16 now what about the

radius if we increase the radius what's going to happen to the cental force so because the radius is on the

bottom of the equation it's inversely related to the Cal Force meaning that if you increase the radius the cental force

will decrease so if we increase the radius by a factor of two how much will the

centripetal force decrease by so if we plug in one for everything that stays the same and plug in two for R the

cental force will be 12 of its original value which means that it decreases by a factor of

two now what if we triple the radius so in this casee the cental force will be 1/3 of its original value which means

that it decreases by a factor of three now what if we cut the Distance by a

half so if we make the radius smaller what effect will that have on the centripetal

force so this is 1 over2 1 / a half is equal to two so the cental force will increase by factor of

two now what if we double the mass triple the radius or better yet let's say if we

double the mass triple the speed and if we decrease the radius by a factor of four what's going to happen to

the centripetal force so let's plug in two for m 3 for V and 1/4 for R since we decrease the

radius by a factor four so it's 1/4 of its original value 3^2 is 9 * 2 that's 18 so this is

18 / 1/4 18 / 1/4 is equal to 18 * 4 so the cental force is 72 times as great so it

increases by a factor of 72 and so that's how you can answer these types of problems you can take that approach to

solve it now whenever there's a force there is also an

acceleration and according to Newton's Second Law net force is equal to mass time

acceleration so the cental force is the mass times the cental acceleration so if we divide both sides

by m we could see that the cental acceleration is v^2 / r like the cental

force the centripetal acceleration always points inward towards the center of the

circle sometimes you might see that as a test question so make sure that you are aware of that

fact now let's say if you have a circle and an object moving in that ccle Circle what is the frequency and the

period as it relates to circular motion the frequency is the number of revolutions that this object makes in 1

second so let's say if it makes five revolutions per second then the frequency is five

the period is the time it takes to make one revolution so the period is 1/

f so frequency is the number of revolutions divided by the time and so that's going to give you the revolutions

per 1 second the period is the

time divided by the number of evolutions now how can we calculate the

speed as an object moves in a circle how can we figure out the speed given the radius of the

circle now if the object is moving under constant speed this equation applies the distance

traveled is equal to the speed multip by the time so the speed is distance divided by time if you divide both sides

by T now the distance around a circle is the

circumference and the time that it takes to make one revolution around a circle is the

period so the speed of the object is basically equal to 2 pi r / T so that's one equation that you want to know

so sometimes you may need to calculate the centripetal acceleration using period centripetal acceleration is v^2 /

R which is the same as V ^2 1 * 1 R and we know that V is equal to 2 pi r / the period so what we could do is replace V

with 2i r / t^ 2 * 1 / r 2^ 2 is 4 so we have 4 pi^ 2 r^ 2/ T ^2 * 1 R so one of the RS

cancel so therefore we have the formula for cbal acceleration so the centripetal

acceleration is also equal to 4 pi^ 2 * R / t^2 so if you know the radius of the

circle and the time it takes to make one revolution you can use that to calculate the centripetal

acceleration the units for cental acceleration is me/ second squared so if you want to get uh these units make sure

that the radius is in meters and the period is in seconds so here's a practice problem

that covers all of the equations or at least most of them that we've covered at this point so so feel

free to pause the video and work out this example so we have this 2 kg ball and it

moves in a circle of radius 10 it makes 20 revolutions every 4 seconds what is the frequency period

cental acceleration centripetal force and the speed so let's say this is the

ball and it moves in a circle and the radius of that circle is 10 m so let's find the frequency first so the

frequency is the number of revolutions divided by the time is the revolutions per 1 second so

there's 20 Revolutions in 4 seconds if we divide it we'll get five revolutions per 1 second so that's the

frequency that's how many cycles of seconds I mean how many cycles per second is aurr

so the ball moves five times around a circle in 1 second now what is the

period the period is one over frequency or we can say it's the time divided by the number of

revolutions so the time is 4 seconds and there's 20 revolutions 4 over 20 is the same as one

over five so it's 1/ seconds per Revolution so the period is 1 over5 or2

seconds typically the unit for period is simply seconds but it's important to understand what it really means it's 2

seconds per one revolution now that we have the period what is the centripetal

acceleration so we can can use this formula to find the centripetal acceleration it's 4 pi^ 2 * R /

t^2 so let's plug in the data that we have the radius is 10 m and The Period is. 2 seconds and let's not forget to

square so 4 * < ^ 2 that's 39 478 and then let's multiply that by 10

which is 39478 / 2^2 so the centripetal acceleration is

very large it's 9,869

point6 m/ second squared so this ball is moving pretty fast

five Revolutions in 1 second that in a radius of 10 met that's pretty

fast now even though this problem is not practical I just want you to be familiar with the equations and how to use

them now let's calculate the centripetal force if you have the centripetal acceleration the centripetal force is

simply the mass time the Cent tribal acceleration so it's 2 kg time 98696 so this is going to be

19,731 Newtons now what about the speed how can we calculate the

speed now that we have the cental acceleration we could find the speed using this equation

or we could find the speed using this equation too so let's use this one since we

already have the radius and the period the radius is 10 m and the period we said was2

seconds 2 / 2 is 10times another 10 so this is equal to 100 Pi which is about

31416 m/s so that's how fast it's moving around a circle no wonder the cental

acceleration was so high so here's another problem that you can

try so we have a plane moving at 400 m/s and it makes a circular turn of radius 3 km what is the centripetal acceleration

in G's so let's say if this is the plane and it makes a

turn and the radius of that turn is 3 kilm and it's moving at 400 m/s what is the cental acceleration so

we can use this equation AC is equal to v^2 divid R so the speed is 400 m/s now now the radius is in kilm we

need to convert that to meters it turns out that 1 kilm is 1,000 M so 3 km is 3,000

M the units have to match if the speed is in me/ second then you want the radius to be in meters as well so 400 s

/ 3,000 that's equal to an acceleration of 53.3 m/s squared

so that is the centripetal acceleration now what is the answer in G's G which represents gravitational

acceleration that's 9.8 m/s so 1 G is 9.8 what is the value of 2G so 2 G's is 2 *

9.8 that represents uh 19.6 m/s squared three G's is equal to 3 times 9.8 which is

29.4 so how many G's correspond to 53.3 to find out Simply take the acceleration and divide it by

9.8 so the answer is 5.44 GS now if you want to understand why it works that way

you could use the principles of unit conversion so we have an acceleration of 53.3

m/s and we know that 1G is equal to 9.8 m/s squ so notice that these units cancel so by dividing you're going to

get the number of G's and so that's why it works out this way so it's 5.44 GS let's try this problem a 2 kg ball

attached to a rope moves in a horizontal circle of radius one calculate the tension

force so we have a rope a ball attached to it and it's going to move in a horizontal

Circle the radius is one and remember the tension force provides de centripetal force because

the tension force in the Rope caused the ball to move in a circle the tension

force is approximately equal to the cental force it's not exactly equal to it but for the most part it's

approximately equal to it if the speed is relatively High the tension force is going to be

equal to mv^2 of R so it's going to be 2 kilg times the speed of 20 ID the radius which is 1 20 2 is 400

400 * 2 is 800 so the tension force is approximately 800 Newtons now in

actuality the ball has a weight force and the tension force is actually at an angle it may be at a small angle but

it's not purely horizontal so the tension force has to support the weight force of the ball

otherwise it's going to fall down so the tension force really has an X component and it has a y

component so Ty Y is equal to mg so to find the actual tension force it's the square root of tx^ 2 plus

ty^ 2 based on the Pythagorean theorem so let's calculate Ty first so Ty which is mg that's going to be the mass which

is 2 * 9.8 so Ty is about 19.6 Newtons TX notice that it's directed

towards the center of the circle so therefore TX is equal to the cental force which is mv^2

R which we already got as 800 Newtons so now we can use this equation to calculate the tension

force so it's going to be the square root of 800 S Plus 19.6 s if you want to calculate the exact

answer so the actual tension force is about 80024 which is approximately equal to

800 now let's think about what this means when the speed is very high the centripetal force the 800 is

significantly larger than the weight Force so therefore the overall tension force is approximately equal to the

centripetal force when the speed is high but now let's see what the tension force will be when the speed is relatively low

so let's calculate Ty first so Ty is still mg which is 2 * 9.8 and so that's still going to be 19.6

so the weight force is not going to change TX is going to be different TX is equal to the centripetal force which is

mv^2 r and the mass is still two but the speed is now five and the radius is

still one 5^ 2 is 25 * 2 is 50 so the cbal forces is 50 now T is going to be the sare root of tx2 + Ty y^2 so that's

50 2 plus 19.6 s so this is equal to 53.7 which is significant relative to

50 so at high speeds the centripetal force is approximately equal to the tension force in a rope but at low

speeds the tension force is a little bit greater than the cental force because you have to take into account the weight

Force now let's think about this conceptually imagine if you have a ball attached to a rope in your hand and if

you swing it horizontally at a very high speed the ball will appear to look like this it's going to move in a horizontal

Circle and the tension or the Rope is going to appear very straight the the Rope is going to be almost horizontal

now if you spin it at a very slow rate the ball's going to be at an angle relative to the

horizontal so it's going to move like this so as you can see at low speed the vertical component of the

tension force is significant relative to the horizontal component at high speeds

Ty is very small compared to TX so T is approximately equal to TX but that's what you want to take from

this so if the speed is very high the tension force is approximately equal to the centripetal

force if it's low then you got to find you have to incorporate Ty when calculating the

tension force now how can we calculate the tension force of a

ball when it's moving in a vertical Circle as opposed to a horizontal Circle how can we calculate

it so let's call this point a point B Point C and point D now think about it conceptually let's

say if you have a rope and a Ball's attached to it and you're swinging it in a vertical Circle at what point do you

think the ball will feel the heaviest will it feel heavier at point a point B C or Point

D the ball should feel heavy at the bottom because at the bottom you have to apply more Force to lift it up to get to

the top but at the top it's going to feel uh very light because gravity is going to help you to bring it back down

but at Point a if you want to lift it from point A to point B not only do you have to support the

weight of the ball but you must also provide the centripetal force to turn it in the upward direction as you move from

A to B so at Point a the ball is it's going to fill the heaviest at Point

a so therefore at Point a the tension force is the sum of the centripetal force and the weight Force at Point C

it's the lightest the tension force doesn't have to be that great because gravity also

assists in providing the centripetal force gravity is going to help you it's going

to help the ball to turn it in a downward Direction right now at Point C the ball wants to go towards the left

gravity is applying a downward Force which is going to assist the ball to go in this direction as it moves from C to

D it's eventually going to turn towards the downward Direction so the tension force or the tension in the Rope doesn't

have to be that large the tension is still going to provide a significant portion of the cental Force at Point C

but the weight force is going to help it out so it feels lighter at Point C so at Point C the tension force is the

difference between the centripetal force and the weight Force at Point D and point B the tension

force is approximately equal to the cental force which is mv^2 r a 2 kg ball moves in a vertical circle

of radius 50 cm at a speed of 20 m/s calculate the tension force in the Rope attached to the ball at the top

bottom and in the middle of the circle so let's start with the top so when the ball's at the highest

position in the circle the tension force is the difference between a centripetal force

which is mv^2 r and the weight Force mg so now that we have the equation we

just simply have to plug in the numbers so the mass is 2 kg the speed is 20 m/s and the radius is in centimeters but

we need to convert that to met there's 100 cm in 1 M so simply divide 50 by 100 and you should get5

M minus the weight Force which is 2 * 9.8 20

2 is 400 * 2 that's 800 ID .5 so that's uh 1600 - 2 * 9.8 which is about

19.6 so in this case the tension force is about 158.4 at the top

now what about at the bottom of the circle at the bottom the tension force is the sum of the centripetal force and

the weight Force we know the centripetal force is, 1600 the weight force is 19.6 so at the bottom the tension force

is going to be a little bit heavier it's going to be 16 19.6

Newtons now what about at the side let's say if it's moving in this direction what's the tangent

force that acts on a rope in this case the tangent force is approximately equal to the centripetal force at the side so

it's just going to be about 1,600 Newtons and so that's all you need to know for the problem where it moves in a

vertical Circle A 2 kg ball attached to a 1.5 M string moves in the vertical Circle what

is the minimum speed that the ball must have at the top so that it continues to move in a

circle so at the Top If the ball is moving fast enough it's going to move in a nice Circle and the Rope is going to

be fully extended because there's going to be a tension force now if it's moving too

slow what's going to happen is it's going to fall down and the Rope it won't be fully extended because there's no

tension force and so it won't be straight we want to find the minimum speed such that It Moves In A Perfect

Circle and it has to be a tension force for it to move in a perfect circle so the minimum speed the

threshold occurs when the tension force is just above zero so we're going to set the tension force equal to zero at the

top the tension force is the difference between the cental force and the weight Force mg so if we set it equal to zero

and add mg to both sides we can see that the centripetal force must balance the weight force it must be just above the

weight Force for the Rope to remain straight and so that the ball can move in a perfect vertical Circle at the top

now we don't need the mass so we can divide by m so V ^2 R is equal to G and if we

multiply both sides by R we'll see that v^2 is equal to RG so therefore the minimum speed that

the ball requires so that it continues to move in a circle at the top is the square root of the radius times the

gravitational acceleration so it's the square root of 1.5 * 9.8 and the answer is

3.83 m/ second go ahead and try this problem a0 25 kg tether ball is tied to a pole with

a 75 M string calculate the tension force in the string if the ball makes 10 horizontal Revolutions in 5

seconds so let's draw the pole this is going to be the string and here's the

ball so it moves in a horizontal Circle so this is the radius of the circle let's call it r

this is the length of the string we'll call it l and the time it takes to make one

revolution is the period which we'll call Capital T so let's call the tension force ft

let's not confuse it with T so the tension force has a vertical component let's

draw the free body diagram so this is the tension force this is the vertical component of

the tension force we'll call it f Ty and it has a horizontal component

FTX and there's also the weight force acting on the ball so first let's find a relation

between the length of the string and the radius of the circle ccle let's call this angle

beta now based on sooa we can see that sin bet is equal to the opposite side

opposite to Beta is the radius divided by the hypotenuse which is the length of the

string so solving for R the radius is equal to the length of the string Time s

now we can draw a second triangle notice that the tension force is related to

FTX using beta as well so sin beta is equal to the opposite side which

is FTX divided the hypotenuse which is ft so if we multiply both sides by ft we

can see that ft sin beta is equal to

FTX now FTX points towards the center of the circle so therefore FTX provides the centripetal

force so let's make some space so we can say FTX is equal to

FC and FTX is equal to the tension force time sin beta and the cental force is mass time

the cental acceleration and the cental acceleration we can make it equal to v^2 R but we

don't want to use that equation we want to relate the cental acceleration to the period because we

can calculate the period from the number of revolutions per second the other equation for the cental

acceleration is 4 Pi squar time the radius of the circleid the period squar this is the

one we want to use so now we can replace the radius with L sin beta

now the last thing that we need to do is we need to divide both Signs by side beta so these two will

cancel so now we have the tension force in terms of the period so it's the mass times

4K * the length / the period squ so let's calculate the period the period is the time it takes

to make one revolution so if we take the time and divide it by the number of revolutions

we can get the period so if 10 revolutions occurs in 5 seconds

then 0. five revolutions I mean 0 five one revolution occurs in 05 seconds so it's 0.5 seconds per one revolution so

that's the period That's the time it takes to make one revolution so now that we have t let's

calculate the tension force so it's going to be 4 pi^ 2 time the mass which is 25

kilg times the length of the string which is uh 75 M divid by the period which is .5 second

squar so you should get a tension force of 29.6 new

so this is the answer all right let's try this problem a car makes a circular turn of radius 20

M the coefficient of static friction between the tires and the road is 60 what is the maximum speed at which the

car can safely make the turn without sliding so let's draw a picture so here's the

road and I'm going to use a box to represent the car and the car is

turning as it turns you need to realize that static friction provides the centripetal force

static friction between the tires and the road allows the car to turn and any force that causes an object to turn in a

circular Direction provides the cental force now if the car is moving too

fast static friction won't be able to make the complete turn it's not strong enough to uh cause the car to

turn if the car is moving too fast the faster the car is moving or if it's moving at a higher

speed you need a greater frictional force to keep it or to cause it to turn in a circle and static friction has its

limits so if the car is moving too fast instead of turning it could slide off the road

to solve this problem since static friction provides the centripetal force you want to set these two forces equal

to each other static friction is equal to M which is the coefficient of static

friction times the normal force the centripetal force is mv^2 over R and in this problem we don't have a

bank curve it's a level Road the car is simply making a turn on a flat surface the normal force for a flat surface is

mg so notice that we can cancel the mass so we don't need the mass so mu G is equal to V ^2

R so what we want to do at this point is multiply both sides by R so these units cancel

so now we have MS * G * R and this is equal to v^2 so now we need to take the square

root of both sides so the maximum speed at which the car can make the turn is equal to the

square root of the coefficient of static friction times gravitational acceleration times the radius of the

turn in this case the radius is about 20

M so now we just got to plug in everything into the equation and we'll get the

answer so it's going to be 60 * 9.8 times the radius of 20 and then we'll take the square root

of that result so for this problem the maximum speed is about

10.84 m/s if the car is moving faster than this speed if it's making a turn of

radius 20 M it can it will skid or it could slide off the road a car makes a circular turn of

radius 15 M at 40 m per second at what angle should the road be banked so that no friction is required

so let's draw a picture so let's say this is the car it has a normal

force and it has a weight Force mg and let's say this is the angle of

the road now the normal force has a horizontal component that provides the centripetal

force and it also has a vertical component that supports the weight Force so the angle Theta is equal to

this angle so this side is the normal force times cosine

Theta that supports the weight force and on this side we have the normal force time sin

Theta now if you want to see why that's the case let's prove why these angles are

equal to each other first because some of you out there are probably wondering why those two angles

are equal to each other so let's say this angle is 30 that means this angle is

60 which means that this angle here is 30 which means that this is

60 because this is a right angle the normal force is always perpendicular to the

surface and if that's 60 this is 30 so therefore we could say this is Theta so let's prove why this is FN sin

Theta and this is FN cosine Theta so here's a normal force it has a vertical component and a

horizontal component and here's Theta so let's say this

is let's call the side x and let's call the Side Y cosine Theta according to soaa is

equal to the adjacent side ID the hypotenuse the adjacent side is a side next to the angle which is y

the hypotenuse is a side across the 90° angle or across the box and that's the normal

force so if you solve for y you need to multiply both sides by the normal force so as you can

see Y is equal to FN cosine Theta so this is equal to FN cosine Theta now what about the other side

let's do the same thing so we're going to have to use S instead of cosine so according to sooa which is

s c ah t o a s is opposite over the hypotenuse so sin

Theta is equal to the opposite side which is is X divid by the hypotenuse which is the

normal force so if we multiply both sides by the normal force we can see that the

normal force time sin Theta is equal to the horizontal side which we have here so now you know where those equations

come from just in case you had any doubt so what you need to know is that this component of the normal force

supports the entire weight of the object so mg is equal to FN cine Theta therefore the normal force for this type

of problem is mg / cosine now the horizontal component provides the centripetal force because

the car is making a turn on a bank road so on a bank surface dependent on the

angle a car can make the turn without the help of

friction because the normal force can provide the cental force or at least the component of a normal force can provide

the centripetal force required to make the turn so we're going to set FC equal to

FN sin Theta so MV ^2 R is equal to the normal force times sin Theta so now somehow we got to find the angle

at which the road should be banked so that no friction is required to make this

turn so we can't use this equation because we don't know the mass or at least we can't use it in form

right now nor can we use this one either because we don't have M so what we need to do is replace the

normal force with mg over cosine Theta so mv^2 R is equal to

mg over cosine Theta times the sin Theta so if we divide both sides by

m we can get rid of it so it's v^2 r and S / cosine is tangent so now let's multiply both sides

by R so v^2 is equal to R actually I'm forgetting something and that's G so this is

tangent times G so what I want to do actually at this point is divide both sides by

G divide diving both sides by G is the same as multiplying by one over G so these two

cancel so V ^2 over RG is equal to tangent Theta so therefore if we take the

inverse tangent of both sides Theta is equal to the inverse tangent V ^2

RG this is the equation that you want to find the angle at which the road should be bank so that no friction is required

to make the turn so now all we need to do is plug in the information to get the

angle so Theta is equal to the inverse tangent V ^2 which is 402 / the radius of 50 * 9.8

so the angle is 72.9 7° let's try this problem a car accelerates from rest at a uniform rate

to 40 m/s in 8 seconds on a circular track of radius 600 M what is the magnitude of the tangental

ACC acceleration the radio acceleration and the net acceleration of the car so let's say if we have an object

moving in a circle now if it's turning it has a radial or a cbal

acceleration and if it's speeding up as it moves around a circle it has a tangental acceleration

now if you combine these two vectors let's say this is the tangental acceleration and this is the cental

acceleration then the hypotenuse represents the net acceleration whenever you wish to add

two vectors that are perpendicular to each other you can use the Pythagorean theorem so the net acceleration

is equal to the square root of the cental acceleration squared which is the radial acceleration plus the square of

the tangental acceleration so first let's calculate the tangental

acceleration the equation that we can use to get it is this equation V final equals V initial plus

a so the initial speed is zero because the car accelerates from rest we're looking for the acceleration the time is

8 seconds and the final speed is 40 so 40 ID 8 is 5 so this is the tangental acceleration now let's calculate the

radial acceleration or the centripetal acceleration when the speed is

40 so let's use this equation so when the speed is 40 the centripetal

acceleration is 2.67 m/s squ so now we could find the net acceleration so it's going to be the

square root of 2.67 S Plus 5^ 2 so the net

acceleration is about 5 6 7 m/s

squared so that's what you can do to find the net acceleration if the object is speeding up while traveling in a

circle a th000 kgam car travels on the road at a speed of 20 m/ second calculate the normal force exerted by

the road at points A and B if the radius of curvature is 100 m okay so let's

begin so while the car is moving at Point a the normal force has to apply a force

to change the direction of the car notice that the normal force points towards the center of this

circle so in that case the normal force provides the Cent tribal Force but not only does it provide the centri force it

must also support the weight of the car so it has to work harder in that case so at the

bottom the car is going to feel heavier because the normal force has to support the weight of the car and it has

to change the direction of the car so it has to provide the C tribal Force so at Point a the normal force is the sum of

the weight force and the cental force at point B the normal force doesn't have to apply

a force to cause the card to go this way as it moves it's simply going to fall in that

direction if it's not moving too fast so in this case the normal force is the difference between the weight force

and it's in tribal Force if it's not moving at all then the normal force supports the

weight Force it's going to be equal to the weight Force but if it is moving gravity is going to assist it in

falling down so at the top the normal force is the difference between the weave force

and the cental force those are the two equations that you need to know for this situation so let's calculate the normal

force at Point a and at point B for those of you who want to see the free body diagram the weight force is

directed downward and the centripetal force is directed

upward towards the the center of the circle for this one the normal force is directed

upward the weight force is directed downward and the centripetal force is also directed downward towards the

center of the circle just in case if you want to see the free body diagram in

action so for point a the normal force is going to be equal to the mass which is a th000 Time gravity which is 9.8

plus MV ^2 R which is a th000 Time the speed which is 20 2 divid by the radius of

100 so the weight Force is 9800 and the inent tribal force is going

to be 20 s which is 400 time 1,00 / 100 which is 4,000 so the normal force at Point

a is the sum of 9800 and 4,000 so that's going to be equal to 13,000 800

Newtons now at point B is the difference between those two numbers it's the weight Force which is

9800 minus decent tributal Force which is 4,000 so that would give you a normal force of

5,800 now how can we answer the last part what is the maximum speed that the car can travel at point B without losing

contact with the road so once it loses contact with the road there's no more normal

force normal force is a contact force if the road is not in contact with the car the road cannot exert a normal force

on the car so to find the maximum speed we need to set the normal force equal to zero and that's at the

top so mg minus MV ^2 R must equal to zero so if we add mv^2 R to both sides we can see that mg is = to mv^2 R and if

we divide by m we get this equation so G is equal to v^2

R so now let's multiply both sides by R so these cancel so RG is equal to v^2 now let's take the square root of both

sides so the velocity is going to be the square root of RG so this equation gives us the maximum

speed at which the car can travel at point B without losing contact with the road if the speed exceeds this value

it's going to fly off the road so let's see what that speed is the radius is 100 gravitational acceleration

is 9.8 so 100 * 9.8 is 980 and the squ < TK of

980 is 31.3 m/s now let's try another problem at what minimum speed should a roller

coaster be traveling when upside down at the top of a circle a radius 15 M if the passengers are not to fall

out so let's say if we have a circle now the normal force at the bottom of the circle as you mentioned

before is going to be the sum of the weight force and the cental force but the normal force at the

top is the difference between the centripetal force and the weave

Force this time it's going to be in a reverse order in the other example the car was above

the road it was basically above the circle but in this example the roller coaster is below the

circle and so if it's moving fast enough the curvature of the tracks will provide the

centripetal force causing the roller coaster to turn if it's moving too slow then

gravity wins and the passengers will just fall down so imagine if you have a ball and

if you roll it if the ball moves too slow it's just going to fall down but if it moves faster enough it's going to

round a circle and exit the circle so there's a minimum speed in which any object moving in this vertical

Circle will maintain contact with the upper part of the circle and it will not fall straight

down how can we calculate that minimum speed in which the roller coaster won't lose contact with the tracks and the

passengers will not fall out the threshold is when the normal force is

zero if the roller coaster is moving fast enough the centripetal force will exceed the wave Force so let's say if

the centripetal force is a th000 and the weight force is 700 that means there's a normal force of 300 that means that the

tracks is exerting a force on the roller coaster so it's that means the roller coaster is still in contact with the

tracks but if the weight Force exceeds the Cent tributal force that would indicate a

negative value which means that the roller coaster is no longer in contact with the normal force because the normal

force can't be negative so to find the minimum speed let's set the normal force equal to zero

and let's solve for r i mean not for R but for V so let's add mg to both sides so mg is equal to MV ^2

R so we can cancel M and if we multiply both sides by R we'll get the equation that we had in the last

example so the minimum speed is simply the square root of the radius times the gravitational

acceleration in this example the radius is 15 G is still 9.8 and the minimum speed

is 12.1 m/s now let's move on to another topic

let's talk about gravity gravity is a force that brings matter together let's

say this is the Sun and this is the earth gravity is a force of attraction and the gravitational force

is significant between very large planetary objects so the sun exerts a gravitational force

on the Earth and the Earth exerts a gravitational pole on the sun now how can we quantify this force

of gravity the equation that we need is this one f is equal to G * M1 * M2 / R 2 so G is the gravitational constant

it's equal to 6.67 time 10us 11 M1 and M2 represents the mass of the two objects in this case the

mass of the Sun and the mass of the Earth and R is the distance between the center of the sun and the center of the

earth the mass of the Sun is about 2 * 10 30 kilog the mass of the Earth

is about 5.98 * 10 24

kilog and the distance between them is about 1.5 * 10 11

M using this information go ahead and calculate the gravitational force between the Earth and the Sun so all we

got to do is plug in these numbers so we have G we have the mass of the Sun that's 2 *

10 30 and the mass of the earth divided by the distance between

them and don't forget to square it now let's type this in so you should get a force of about

3.5 5 * 10^ the 22 Newtons so that's the gravitational force between the Earth and the

Sun now let's say if the mass of the Earth was doubled what effect will that have on the

gravitational force between the Earth and the Sun if you double the mass of the

Earth the gravitational force will increase if you triple the mass of the Sun the gravitational force will triple

as well now what happens if you double the distance between the Earth and the Sun

Well the gravitational force increase or decrease if you increase the distance it will decrease if you double the distance

it will decrease by a factor of four or it's going to be 1/4 of its original value now what if you triple the

distance remember to answer these questions plug into one for everything that stays the same if we triple the

distance it's 3^2 which is 9 or one of 9 so the gravitational force will be 1 nth of its original value or it's going to

decrease by a factor of n now let's say if you triple the mass of of the earth quadruple the mass of

the Sun and decrease the Distance by a half what effect will it have on the

gravitational force between the Sun and the Earth so 3 * 4 is 12 1^ 2 is 1 2^ 2 is 4 12 / 1/4 is

48 so in this case the gravitational force will be 48 * as great so if you decrease the distance

between the two uh large objects the force will increase so if you decrease the Distance

by a third 1 over 1 over 9 that's n the force will increase by a factor of nine so if

you wish to increase the gravitational force you can either increase the masses of the objects or decrease the distance

between them the mass of the Earth and Moon are the numbers given we also have the

radius of the Earth and Moon using this information go ahead and calculate the acceleration due to gravity at the

surface of the Earth and the moon so first we need to come up with an equation now the force of

gravity is equal to the weight Force because the weight force is based on gravitational acceleration

so let's say f is the force of gravity between let's say a person standing on Earth's surface and the Earth itself so

that basically is the weight force and gravitational force is G M1 M2 / R

2 whereas the weight force of a person is simply mg so if you cancel one of the masses in this case the mass of the

person you can get the gravitational acceleration of the

planet so it's G * the mass of the Planet / R 2 this is the equation you want to

use so the gravitational acceleration of the earth which we know to be 9.8 we can calculate it using this equation so it's

G * the mass of the Earth / the radius of the Earth squared so G is

6.67 * 10-1 the mass of the earth is 5.98 * 10 24 kg and the radius of the

Earth is 6. 38 * 10 6 M and let's Square it so using this information you should

should get 9799 which is approximately 9.8 m/s squared so that's the gravitational

acceleration of the earth now go ahead and use the same equation to calculate the gravitational

acceleration of the moon so it's going to be GM over R 2

again so G is going to be the same the mass of the Moon is much less than the mass of the Earth so we should

expect that the gravitational acceleration should be weaker now the radius is also smaller

though as the mass of the planet increases the gravitational acceleration increases as the radius of a planet

increases the gravitational acceleration decreases so the gravitational acceler is directly proportional to the mass of

the planet but it's inversely related to the square of the radius so let's see what the answer is

going to be so you should get a gravitational acceleration of about

1.62 m/s squared so that's the gravitational acceleration on the moon it's much weaker than that of the earth

so if you're on the moon you'll feel a lot lighter so you could jump higher now what about the gravitational

acceleration 4,000 km above Earth's surface feel free to pause the video and work on that

example so let's say this is the earth we know what the radius of the Earth the radius of the Earth is 6

.38 okay let's write that better time 10 6 M now we want to find the gravitational

acceleration at a point 4,000 km above the surface of the Earth so the R value in the equation is

the sum of the radius and the altitude of 4,000 km now we can't simply add the radius of

the Earth and the 4,000 km because the units doesn't match up the radius is in meters and the altitude is in kilm so

let's convert kilm to meters one kilometer is about 1,000 M so 4,000 * 1,000 is 4 million which is

about 4 * 10 6 M so if we take the altitude and add it to 6.38 * 10 6 this will give us

10.38 * 10 6 M which is the same as 1.38 * 10 7 m so this is the r value that we want to

use so as you move away from Earth's surface the gravitational acceleration will decrease but let's see how much

let's go ahead and get the answer so G is going to be the same the mass of the earth is still 5.98

* 10 24 but now R is a different value it's 1.38 * 10 7 and let's not forget to

square it so we should expect an answer that's much less than

9.8 so the answer that I got is a gravitational acceleration of 3.7 m/s squ so that's the gravitational

acceleration 4,000 kilometers above Earth's surface calculate the gravitational

acceleration of a planet that has a mass two times that of Earth so the gravitational acceleration

of the earth we know it's 9.8 and equation for gravitational acceleration is the gravitational

constant time the mass of the Planet / R 2 so if we double the mass what's the effect on gravitational acceleration if

we plug in two for M and one for everything else we get a value of two so doubling the mass will double the

gravitational acceleration so for part A the gravitational acceleration of let's say Planet X is twice the value of that

of Earth so it's going to be 19.6 and it's meters perss squared now what if the radius was

increased by a factor of three so if the planet had a radius that was three times greater than that of Earth

what is the gravitational acceleration so what is the effect of tripling the

radius so it's going to be 1 over 3^ s which is 1 over9 so it's going to decrease by a factor of 9 so it's going

to be 9.8 / 9 so the gravitational acceleration is now

1.09 m/s squ now what what if we reduce the distance or the radius of this planet by

half so if we plug in 1/2 into the equation it's 1 over 1/4 which is equal to

4 so it's going to be four times as great so it's going to be 9.8 * 4 and the gravitational acceleration will be

uh 39.2 m/s squ so as you can see if you increase

the mass the gravitational acceler ation will increase if you decrease the radius it

will increase as well but if you increase the radius of the planet the gravitational acceleration will

decrease calculate the speed and period of a satellite 3,000 km above Earth's surface so let's say this is the

Earth and here's the satellite the satellite is in freef Fall with the Earth

Earth exerts a gravitational force on the satellite so the satellite is Falling

Towards the Earth however the satellite does have a tangental velocity and as you've learned earlier

in this video when the velocity and force vectors are perpendicular to each other the object

turns and the satellite can maintain its orbit if it has the right speed if it's moving too

fast it can escape Earth's gravity if it's moving too slow it could eventually fall into the surface of the

Earth but if it has the right speed it can maintain its orbit so how can we calculate the speed

of the satellite if we have the height so let's say it's 3,000 kilm above the surface of

the Earth what can we do so you need to understand

that Earth exerts a force of gravity on a satellite and because that gravitational force keeps the satellite

moving in a circle gravity provides the centripetal force so that tells us that we need to set the gravitational force

equal to the cental Force now we know the gravitational force is going to be the gravitational

constant time the mass of the satellite time the mass of the Earth / the radius squar now the centripetal force is equal

to mv^2 R since the satellite is the one that's moving in a circle around the earth the mass in the cental force is

going to be the mass of the satellite time V ^2 R so we can cancel the mass of the satellite and we can cancel one of

the RS so we can see that the gravitational constant times the mass of the earth divided by the distance

between the center of the satellite and the center of the earth that's going to equal the square of the velocity of the

satellite so what we need to do is take the square root of both sides so now we have the speed of a

satellite in orbit above the Earth it's equals to the square root of the gravitational constant time the mass of

the Earth divid the distance between the two objects so we know the value of G that's

6.67 * 10- 11 and from the other problems we know that the mass of the earth is about

5.98 * 10 24 Now 1 km is 1,000 M and 1,00 is basically 10 3r

so that's the altitude 3,000 * 1,000 is 3 million which is 3 * 10 6

M now the radius of the Earth is 6.38 * 10 6 M and we have an

altitude that's the distance between the satellite and Earth's surface is 3,000 km above Earth's surface so that's 3 *

106 M to get the distance between a satellite and the surface or rather the center of the

earth we need to add these two values so 3 + 6.38 that's going to be

9.38 * 10 6 so let's see what the answer is going to be

so I got a speed of 6,521 m/ second that's how fast the satellite is

moving at this orbit now the next thing that we need to do is calculate the period of the

satellite we need to find the time it takes for it to make one revolution around the

earth so we know that the speed is equal to the circumference divided by the time which

is the period you've seen this equation earlier in a video so what we can do is we can

multiply both sides by T and that's going to give us VT is equal to 2i

R and if we divide both sides by the velocity we can see that the period is equal to 2 pi time the orbital

radius relative to the center of the earth divided by the speed so we know what R is we know it's

9.38 * 10 6 so that's the altitude plus the radius of the earth divided by the speed of the satellite which is

6521 m/s so the period of the satellite is 9,3

37.9 seconds so if we want to convert that to hours we know that there's 60 seconds in

a minute and 60 minutes in an hour so we need to divide

9,379 by 3600 and so the period of the satellite is about 2.51

hours so that's how long it's going to take to make one revolution around the Earth thees the distance between the

Earth and the Sun is 1.5 * 10 11 M and I forgot to put the unit there use this information to calculate the mass of the

Sun so if that's all we have how can we find the mass of the Sun now we know it's about 2 * 10 30

kg what equations can we use to find it so let's see what the situation is so let's say this is the

Sun and and we have the Earth revolving around it so the Earth behaves as a satellite

around the sun and the sun exerts a gravitational force on the earth and that

gravitational force on the earth provides the cental force that keeps the Earth in orbit around the

Sun so knowing that we could set the gravitational force equal to the cental force

the gravitational force between two objects it's g m so that's the mass of the Sun Times the mass of the Earth

divid by the distance between them which is 1.5 * 10 11 M the centripetal force is the mass of the object moving

in a circle in this case that's the mass of the Earth times the cental acceleration

AC now we could use m v^2 R but that's going to give us the speed of the satellite which if we want to find the

speed of the Earth around the Sun we could if we want to but we're not going to do

that so cental acceleration we can replace it with v ^2 r or we could replace it with 4 pi^ 2 R

over T ^2 which we obtained this equation earlier in a

video so let's use uh that for the cental acceleration so it's 4 pi^ 2 R /

t^2 so we can cancel the mass of the Earth let's multiply both sides by R 2 over

G if we do that on the left side G will cancel and R 2 will cancel so so on the left we have the

mass of the sun which is equal to 4 pi^ 2 *

RB divided GT ^2 so now let's plug in the information that we

have so so R is 1.5 * 10^ 11 that's the distance between the Earth and the Sun G is the gravitational

constant and what's T So T is the period which is the time it takes for

the earth to make one revolution around the Sun and we know that's equal to one year

it takes one year for the earth to travel around the Sun but we need to plug the the period using

seconds so we can't just plug in one year let's convert this to seconds so we know that there's 365 days in a single

year and there's 24 hours in a single day and there's 60 Minutes in a single hour and there's 60 seconds in the

minute so it's going to be 365 * 24 * 60 * 60 which is a large

number so it's going to be 315 36 0 0 0 so basically that's 31,5 36,000

seconds oh let's not forget to raise R to the thir power so now let's plug this information in and let's see what the

mass of the Sun is going to be Also let's not forget to square the period so you should get

2.01 time 10^ 30 kg which is approximately the mass of the sun

by the way if you're not getting the answer try type in this information on the bottom using parentheses and close

the whole thing in parenthesis because depending on your calculator sometimes based on the way you you type it in the

calculator might be dividing by this number and then multiplying by the second number to avoid that just enclose

everything in the bottom using parentheses and you shouldn't have that problem

here's another problem for you calculate the speed and height above the Earth's surface of a geosynchronous satellite a

geosynchronous satellite is one that stays above the same point on the equator of the earth so the period of

this satellite is approximately one day because it takes one day for the Earth

to rotate around its own axis so all we have is the period how can we find the height and

the speed so using the equation that we had before we said the centripetal force

equal to the gravitational force so here we have the gravitational force between the satellite and the

Earth and R represents the distance between the center of the satellite and the center of the

earth now let's say this is the Earth the satellite revolves around the

earth so therefore the cpal force acts on the satellite because the satellite is

moving in a circle so the mass is going to be the mass of the satellite and the cental

force equation now we're going to replace the cental acceleration with 4 pi^ SAR R

over t^2 so just like before we can cancel the mass of the satellite instead of the

mass of the Earth looking at the last problem so our goal is to solve for R so if we cross multiply 4i ^ 2 R * R 2 is

4i ^ 2 RB and GM * t^2 is G * the mass of the Earth time the period

squar so now let's divide by 4 pi^ 2 so the distance

between the Earth and the satellite which is R can be found using this

equation what we need to do now is take the cube root of both sides so R is going to be equal to the cube

root of G which is 6.67 time 10- 11 let's get rid of

this time the mass of the Earth which is 5.98 * 10 25 4 time the period / 4 pi^

2 so what is the period in seconds so we know the period of the satellite is one day and it's 24 hours in a day and it's

3600 seconds in an hour because it's going to be 60 minutes per hour 60 seconds per minute 60 * 60 is 3600 so 24

* 3600 gives us a period of 8 6,400 seconds and let's not forget to square

the period so let's plug everything in so you should get an R

value of 42 m and the satellite so if we want to find the

height above the surface of the Earth we need to subtract the total

distance by the radius of the Earth so the height is going to be 4225 * 10 7 minus the radius of the

Earth which is 6.38 * 10 6 m so the height is about

3.5 87 * 10 7 m if we divide it by a th000 we can get the height in kilometers which is about

35,8 70 km so that's the height of a geosynchronous satellite above Earth's

surface so now let's calculate the speed of the satellite now that we've have the height above the surface of the

Earth so the speed of the satellite is going to be the gravitational constant times the mass of the earth divided by

the distance between the satellite and the Earth which is this number so it's going to be G which is

6.67 * 10- 11 time the mass of the Earth which is 5.98 * 10

24 ID the distance between them which is 4225 * 10 7 m so you should get a speed of about

3, 72.6 m per second so that's the speed of a geosynchronous satellite in orbit

the distance between Earth and the Sun is 1.5 * 101 M so let's say this is the Sun and let's say this is planet

Earth and Earth can revolve around the Sun the distance between the Sun and Mars is 2.28 * 10 11

M so let's say Mar Mars is somewhere over here and it too can revolve around the

Sun how many Earth days does it take Mars to make one revolution around the Sun so we need to use kepa's laws to

figure this out his third law states that the ratio of the squares of the periods of any two

planets revolving about the sun is equal to the ratio of the cubes of their mean distances from the Suns

so basically this equation T1 over

T2 raised to the second power that's the ratio of the square of the periods of the two planets is equal

to R1 / R2 raised to the 3 power and it makes sense because Mars has Mars is fur further away from

the Sun so the distance the orbital radius of Mars is significantly larger than Earth so Mars should take a longer

time to revolve around the Sun so the period should be significantly greater than that of

Earth so now let's use subscripts to represent Earth and Mars so our goal is to find the period of Mars and days so

instead of writing T1 we're going to write TM for Mars te for Earth RM for Mars that's the orbital

radius of Mars around the Sun or that's basically the distance between Mars and the Sun and then re e the distance

between Earth and the Sun so we're looking for TM the period of the Earth in days is

365 it takes 365 days or one year for the earth to revolve around the Sun the distance between Mars and the

Sun is 2.28 * 10^ 11 and the distance between the Earth in the sun is 1.5 * 10^

11 so we can get rid of of the 10 11 those two things they cancel so

2.28 / 1.5 that's 1.52 and let's raise that to the third power so the right side of the

equation is 3.51 1181 t^2 is just t^2 and 365

2 that that's 133,50 so let's cross multiply so 1 * T ^2 is

t^2 and 133,50 time

35181 that's about 467,000 calculate the period of Mars around the

Sun so the answer is about 684 Earth days so that's one year on planet Mars that's how long it takes for Mars to

revolve around the sun go ahead and try this problem it takes Mercury 88 Earth days to revolve

around the Sun how far is Mercury from the Sun and we have the Earth Sun distance so this time we need to solve

for R so t for mercury divided by the period of the

earth squared this is going to be equal to the radius or the distance between Mercury and the sun divided by the

distance between between Earth and the Sun raised to the thir power so TM that's 88 that's the time it

takes Mercury to revolve around the Sun and for the Earth is 365 days r e that's going to be 1.5 * 10^

11 and we're going to raise that to the third power and then we need to solve for this variable

so let's cross multiply so what we have is 1.5 * 10^ 11 raised to the

3r time 88 squared which equals

365 squar time R Cub if we cross multiply so 1.5 * 10 11 raised to the 3

power that's 3.375 * 10 33 if you multiply that by 88 squ you get 2.61 36 * 10 37 and then divide that by 365

squ so what we now have is R the 3 is equal to 1 962 * 10

32 so at this point we could take the cube root of both sides if you can't find the cube root

function raise both sides to the 1/3 so then you get a distance of 58 * 10^ 10 m

so that's the distance between Mercury and the sun now let's go over the elevator

problem so let's say a person is in the elevator and they're standing on a

scale what would the scale read if the elevator is at rest so if the elevator is not moving at

all the scale is going to read the weight of the person which is simply mg so that's going to be 50 *

9.8 and that's going to be 490 Newtons so that's the answer for part A now what about Part

B when it's moving Upward at a constant velocity what will the scale read now imagine if you're in an

elevator when do you feel that unusual feeling in an elevator is it when the elevator begins

to accelerate upward or while it's moving or while it's about to stop you can feel the acceleration you

can feel a different weight Force when the elevator begins to accelerate upward or if it begins to accelerate

downward or if it's moving up as it begins to just when it's about to stop you feel it but while it's moving up at

a constant speed or if it's moving down at a constant speed you don't feel a thing that's because the normal

force of the floor inside the elevator it doesn't change at constant speed or if the elevator is at rest only

when the elevator accelerates upward or downward does the normal force

change and you feel that effect in the elevator so the answer for Part B and part C is simply

490 the normal force exerted on You by the ground will be equal to the wave Force regardless if the elevator is at

rest or if it's moving at constant speed now what happens when it accelerates

upward will the scale read a force that's greater than the weight force or

less the scale is going to read the normal force that it exerts on you the normal force is an upward Force the

weight force is a downward Force so if the elevator is accelerating upward the normal force has to exert a

force to support your weight plus it has to lift you upward so it has to be greater than the weight it takes more

effort to support not only the weight of the object but also to lift it to a higher

position the net force therefore is the difference between the normal force and the weight

Force the net force according to Newton's second law is mass time acceleration the weight force is mg so

if we add mg to both sides we will get this equation M A+ mg is equal to the normal

force so if we take out the mass we can see that the normal force is dependent on

acceleration so as the elevator accelerates upward the normal force is going to be

50 time the upward acceleration of pos2 plus the gravitational

acceleration so it's going to be 11.8 * 50 so the scale is going to read a normal force of 590

Newtons now let's calculate the normal force as the elevator accelerates downward so the normal force is going to

be 50 times an acceleration of-5 plus the gravitational acceleration of 9.8 so 9.8 minus 5 that's

4.8 so 50 * 4.8 is 240 Newtons so as it accelerates downward

the scale is going to read a normal force that's less than the weight Force

Heads up!

This summary and transcript were automatically generated using AI with the Free YouTube Transcript Summary Tool by LunaNotes.

Generate a summary for free

Related Summaries

Newton's Laws of Motion Explained with Real-Life Examples

Explore Newton's three laws of motion, including concepts of inertia, force, and action-reaction pairs. Learn how forces affect motion with practical examples and problem-solving techniques for physics applications.

Understanding Orbital Motion: The Role of Gravity and Speed

This video explores the principles of orbital motion, focusing on how objects can remain in orbit around the Earth. It discusses the significance of gravity and the necessary speed for maintaining an orbit, dispelling common myths about gravity in space.

Understanding Gravitational Forces and Their Applications in Physics

This video lecture delves into the principles of gravitational forces, their intensity, and their applications in various physical scenarios. It covers key concepts such as gravitational field intensity, Newton's law of gravitation, and the relationship between mass and gravitational attraction, making it essential for students preparing for competitive exams like NEET.

Complete Guide to Motion: Distance, Velocity, Acceleration & Projectile Physics

Explore fundamental physics concepts including distance vs. displacement, speed vs. velocity, acceleration, and motion graphs. Learn to solve typical exam questions on free fall and projectile motion with clear explanations and practical examples.