Q2.8 Average acceleration (H)

A particle moves at constant speed v in a circle of radius r. What is the average acceleration over one half of one period?
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Hint

How does the the direction of the motion at some moment relate to the direction one half period later? So what has happened to the velocity in this time?
 

Solution

Reveal
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Solution

In the course of one half of a period the velocity reverses direction. The change in velocity thus has magnitude 2v.

Lets spell that argument out more fully. Let the initial direction of motion be in the direction of some unit vector, $\hat{k}$ say. Then the initial velocity is

\[ \vec{v}_{i} = v \hat{k} \]

The direction of motion after one half period is necessarily in the direction of $-\hat{k}$. The velocity is then

\[ \vec{v}_{f} =- v \hat{k} \]

The change in velocity is thus

\[ \vec{\Delta v} = \vec{v}_{f} - \vec{v}_{i} = -2v \hat{k} \]
which is, as claimed, a vector of magnitude 2v.

This change occurs in time T/2. The magnitude of the average acceleration follows from Equation 1.4 as

\[ a_{av} = \frac{2v}{T/2} = \frac{2v}{\pi r /v} = \frac{2v^2}{\pi r} \]
where T=2πr/v is the time to complete one circuit (the period).

The direction of the acceleration is the direction of Δv: it is opposite to that of the initial velocity.