Question

For a body in a 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. $\frac{2}{\pi }$
• B. $\frac{\pi }{2}$
• C. $\pi$
• D. $2$

Answer 1

The correct option is A

$\frac{2}{\pi }$

Step 1: Given

The body is in a circular motion

The revolution is half

Step 2: State the values

The magnitude of the instantaneous acceleration, $a_i=\frac{v^2}{r}$

Magnitude of the average acceleration for half revolution,

$$\begin{gathered} a_{av}=\frac{v-(-v)}{{\displaystyle \frac{t}{2}}} \\ =\frac{2v}{{\displaystyle \frac{t}{2}}} \\ =\frac{4v}{t} \end{gathered}$$

The time is given by, $t=\frac{2\pi r}{v}$

Step 3: Find the average acceleration

$$\begin{gathered} a_{av}=\frac{4v}{t} \\ =4v\times \frac{v}{2\pi r} \\ =\frac{2v^2}{\pi r} \end{gathered}$$

Step 4: Compare average and instantaneous accelerations

$$\begin{gathered} \frac{a_{av}}{a_i}=\frac{{\displaystyle \frac{2v^2}{\pi r}}}{{\displaystyle \frac{v^2}{r}}} \\ =\frac{2}{\pi } \end{gathered}$$

Hence, option A is the correct option.