A2 Physics Unit 4: Motion in a Circle

Uniform Circular Motion

Centripetal Acceleration

On the Road

At the Fairground

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  • Created by: lazywolf
  • Created on: 30-11-13 21:44

Uniform Circular Motion

Circumference = 2(pi)r

Frequency of rotation, f = 1 / T, where T = time for 1 revolution

Speed in circle, v = 2(pi)r / T = 2(pi)rf

Angular Displacement: The angle an object turns through in radians in time t = 2(pi)ft = 2(pi)t / T

Angular Speed (w, omega): The rate of change of angular displacement in radians per second = 2(pi) / T = 2(pi)f

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Centripetal Acceleration

An object at constant speed in circular motion is accelerating because the velocity is continuously changing direction and acceleration is change of velocity per second

Centripetal Acceleration is always towards the center of the circle

a = velocity squared / radius of circle = w (omega) squared x radius of circle

Centripetal Force is the resultant force that causes the velocity to change direction and is towards the center of the circle therefore causing centripetal acceleration

F = m(v squared) / r = m(w squared)r

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On the Road

On the top of a hill or humped bridge a car's weight and normal reaction force of the road equal the centripetal force. So mg - R = m(v squared) / r

On a roundabout or bend the friction force from the tyres is the centripetal force. So Fr. = m(v squared) / r

On a banked track, for there to be no friction, the horizontal component of the normal reaction force must be the centripetal force.
Considering components, RsinA = m(v squared) / r and RcosA = mg so dividing the horizontal component by the vertical component tanA = (v squared) / gr. Therefore there is no sideways friction if (v squared) = gr tanA

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At the Fairground

On a ride that takes you through a big dip, at the bottom the difference between the normal reaction force and your weight acts as the centripetal force. So R - mg = m(v squared) / r

On a very long swing of length L is released from height h above the equilibrium point. Maximum speed is at the lowest point and can be worked out by: 0.5m(v squared) = mgh, therefore (v squared) = 2gh
The difference of forces at this point equals the centripetal force so R - mg = m(v squared) / L. But  (v squared) = 2gh, therefore R - mg = 2mgh / L and when h = L the centripetal force equals twice the person's weight.

On a big wheel at the highest point the reaction of the seat and the weight of the person both act towards the center of the circle, so R + mg = m(v squared) / r, and therefore R = [m(v squared) / r] - mg. If v = gr, the R = 0 and there will be no force from the seat.

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