Question

A body revolves with constant speed v in a circular path of radius r, the magnitude of its average acceleration during motion between two points in diametrically opposite directions is:

Answer 1

Step 1: Given data

1. Velocity of the body is v.
2. The radius of the circular path is r.

Step 2: Average acceleration

1. Acceleration is defined as the time rate of change in velocity.
2. The average velocity, $A_{avg}=\frac{changeinvelocity}{totaltimetken}.$

Step 3: Finding the average acceleration

The two given points are placed at the two ends of diameter on the circular path.

Let's assume the velocity at point c is $-v$ and at point E, the velocity is $V$ as shown in the figure.

Now, the change in velocity of these two particles is,

$$\begin{gathered} A=V-\left(-V\right)=2V \\ orA=2V................\left(1\right) \end{gathered}$$

where A is the acceleration.

Now, for circular motion, the period of revolution is,

$T=\frac{\pi r}{v}.................\left(2\right)$

where r is the radius of the circular path and v is the velocity of the particle.

So, from equations (1) and (2)

$$\begin{gathered} A_{avg}=\frac{2V}{{\displaystyle \frac{\pi r}{V}}} \\ or{A}_{avg}=\frac{2V^2}{\pi r} \end{gathered}$$

DIagram

Therefore, the average acceleration during motion between two points in diametrically opposite directions is $\frac{2V^2}{\pi r}$.