CHAPTER 17: MOTION IN A CIRCLE AQA

Question 1

What is uniform circular motion?

Answer 1

When an object is rotating at a steady rate/constant linear speed.

Question 2

How is the speed of a point on the perimeter calculate?

Answer 2

Speed v, = circumference / time for one rotation

using frequency = 1/T

Circumference = 2πr

v = 2πr/T

Question 3

what is the conversion from rad to degrees?

Answer 3

360 degrees = 2π radians

1 rad = 57.3

Question 4

How is angular displacement calculated?

Answer 4

θ = Angular displacement 2πft

Time = T

Therefore 2πt/T

Question 5

What is angular speed? How is it calculated?

Answer 5

Angular displacement per second

ω = 2π/T or 2πf or v/r

since frequency = 1/T

Its unit is rads^-1

Question 6

What happens to the velocity of an object moving in a circle at a constant speed?

Answer 6

The velocity is always changing direction, therefore the object is accelerating.

The velocity of an object in uniform circular motion at any point along its path is along the tangent to the circle at that point.

Question 7

What is centripetal acceleration? It’s equation?

Answer 7

The acceleration of a body in uniform circular motion towards the center of the circle.

a = v^2/r

since v =wr^2

then a also = w^2r

Question 8

What is centripetal force?

Answer 8

The resultant force on an object moving around a circle at constant speed is called centripetal force because it acts towards the centre of the the circle

Question 9

What is the centripetal force acting on an object whirling on a string? a satelittle moving around earth?

Answer 9

Tension provides centripetal force in the string

Gravtity between the satellite and the earth is the centripetal force

Question 10

What is the equation for centripetal force?

Answer 10

For an object moving at constant speed v, along a circular path of radius r, a =v^2/r or w^2r

Therefore after applying newtons second law

F = mv^2/r or mw^2r

Question 11

Describe the reasoning behind why a car on a roundabout does not slip.

Answer 11

As a car moves around a roundabout the centripetal force is provided by the sideways force of friction between the tyres and the road therefore

Force of friction F = mv^2/r

for no skidding to occur the force of friction between tyres and the road surface must be less than the limiting value Fo that is proportional to the vehicle weight mg

therefore it must be less than a particular value Vo which is given in the equation

limiting force of friction, Fo =mVo^2/r

since Fo is proportional to vehicle weight Fo =μmg

where μ is coefficient of friction

if μumg = mv0^2/r then

maximum speed for no slipping v0 = √ μgr

Question 12

Explain how there can be no sideway friction on a the tyres on the road on a banked track?

Answer 12

For there to be no sideways friction on the tyres from the road, the horizontal component of the support forces N1 and N2 must act as the centripetal force.

Resolving the forces into horizontal components = (N1 + N2) sin θ ) and vertical component = (N1 + N2) cos θ)

Since the horizontal component must act as the centripetal force
(N1 + N2) sin θ ) = mv^2/r

because the vertical component balances the weight (N1 + N2) cos θ) = mg

therefore tanθ = mv^2/mgr

tan θ = v^2/gr

there will be no sideways friction if the speed is such that
v^2 =grtanθ

Question 13

How do Big dipper amusement park rides make use of centripetal force?

Answer 13

It pushes you in your seat as you pass through the dip.
The difference between the support force acting upwards and your weight acts as the centripetal force S -mg = mv^2/r

the support force S = mg + mv^2/r

the extra force = mv^2/r

Question 14

In an amusement park swing how does energy transfer help show the extra support force due to circular motion?

Answer 14

a person of mass m on a very long swing of length L, os releashed from height H. The maximum speed occurs when the swing passes throguh the lowest point

1/2mv^2 = mgh

where v is swing speed at lowest point

v^2 =2gh

if swing is on circular parth of radius L at the lowest point the support force is in the opposite dirction to the persons weight mg act towards the center and provides the centripetal force

S - mg = mv^2/L

since v^2 =2gh then, S -mg = 2mgh/L

where 2mgh/L shows the extra support force the person experiencs due to circular motion

Question 15

Explain how the centripetal force in a big wheel rollercoster works.

Answer 15

At maximum height the reaction reaction R from the wheel on each person acts downwards, the resultant force at this position is equal to mg + R The reaction force and weight are what provides the centripetal force.

At highest position the wheel speed is v,

mg + R = mv^2/r

therefore R = mv^2/r -mg

at a particular speed v0 such that v0^2 = gr then R = ) so there would be no force on the person due to the wheel.

Question 16

Why does the speed of an object in circular motion not change?

Answer 16

The net force is acting perpendicular to the direction of motion(velocity) which means that there is no work done by the force hence no change in speed.

Question 17

How do you change degrees to radians?

Answer 17

Multiply by π/180

Question 18

How do you change radians to degrees?

Answer 18

Multiply by 180/ π

Question 19

What direction is centripetal acceleration?

Answer 19

centripetal acceleration and is perpendicular to the direction of the linear/tangential speed(velocity)

Question 20

What direction are centripetal force and acceleration acting in?

Answer 20

• Towards the center of the circle
• Perpendicular to velocity/direction of

Question 21

What is happening to the velocity in uniform circular motion?

Answer 21

The velocity is constantly changing.

Question 22

In uniform circular motion, the linear velocity is always changing but the speed is constant how is this possible? How can acceleration occur?

Answer 22

Velocity is a vector quantity, while the magnitude (speed) remains the same, the direction is always changing.

centripetal Acceleration is causing the change in direction.

Question 23

How does Newton’s second law relate to centripetal acceleration?

Answer 23

The centripetal acceleration is caused by a centripetal force of constant magnitude that also acts perpendicular to the direction of motion (towards the centre)
This is a result of Newton’s Second Law which states that the force acting on an object with constant mass is proportional to its acceleration and takes place in the same direction. f=ma

Question 24

For a car on a hill, what is the maximum velocity the car should not exceed or it will lose contact with the road?

Answer 24

Vo = √gr

Question 25

3 Explain why a circular athletics track is banked for
sprinters but not for marathon runners.

Answer 25

Normal reaction provides a component which increases centipetal force, as well as friction. - As sprinters move at a higher velocity, a higher centripetal force is required to mantain stablility/traction.

The banking of the track means that the normal reaction of the track pushing up on the sprinter helps with his/her centripetal force as it has a component towards the centre of the circle. A flat track just pushes upwards and the sprinter needs to rely more on friction to get the centripetal force.