Question

For a body in circular motion with a constant angular velocity, the magnitude of the average acceleration over a period of half a revolution is ............times the magnitude of its instantaneous acceleration.

a) 2/pie
b) pie/2
c) pie
d) 2

Answer 1

d) 2

Explanation:

The average acceleration is the difference between the velocity at final and initial positions divided by the time.

Consider, at t =0, the circular motion begins. As the angular velocity is constant, only the direction would vary. The direction of the velocity of the body in a uniform circular motion is given by the tangent. Suppose, the body is moving in clockwise direction. Then, at the top, the velocity is towards right and after half a revolution, its direction is towards left. consider, magnitude of velocity as v, left direction as negative and rightwards as positive.

The average acceleration would be:

$a_{avg} = \frac{v_f-v_i}{\Delta t}\\a_{avg}=\frac{-v-v}{\Delta t} = \frac{-2v}{\Delta t}\\|a_{avg}|=\frac{2v}{\Delta t}$

The instantaneous acceleration would be the velocity at that point divided by elapsed time. After, half revolution, the velocity is -v and time is Δt.

$a_{ins} =\frac{-v}{\Delta t}\\|a_{ins}| =\frac{v}{\Delta t}$

On comparing the two:

$a_{avg} =2\times a_{ins}$

Hence, correct option is (d).

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