Question

A body revolves with a constant speed v in a circular path of radius r . The magnitude of its average acceleration during motion between 2 points in diametrically opposite direction is --- ??

Answer 1

see diagram..

When the body is at point P at an angle Ф with OA, the centripetal acceleration a is :
 
$Let\ a_o=\frac{mv^2}{r}\\\\At\ P:\ \vec{a}=-a_o(cos\theta\ \vec{i}+sin\theta\ \vec{j})\\\\At\ P':\ \vec{a}=a_o(cos\theta\ \vec{i}-sin\theta\ \vec{j})$

We observe that when the body is at P', the vertical y-component of acceleration is the same as that at P.  But the horizontal x -component cancels with that of the body at P.  So we take average of y-components.

Avg.\ Acceleration
$\vec{a_{avg}}=\frac{1}{\pi}\int \limits_{0}^{\pi}\ \vec{a}\ d\theta\\\\=\frac{a_o}{\pi}\int \limits_{0}^{\frac{\pi}{2}}\ (-sin\theta- Sin\theta)\ \vec{j}\ d\theta\\\\=\frac{a_o}{\pi} [2\ Cos\theta]_0^{\frac{\pi}{2}}\\\\=\frac{2a_o}{\pi}\\\\=\frac{2\ mv^2}{\pi\ r}$

Answer 2

Answer:

Average Acceleration, A= 2v²/πr

Explanation:

This is a case of Uniform Circular Motion. Now;-

[Refer to the attached image to visualize the case]

Given;-

          Speed (Constant) = v

          Radius(Circular Path)= r

          ∅(After observation)= π or 180°

Let, A avg be the magnitude of average acceleration. Then;-

We know that;-

           A avg = v²/r ×sin∅/2 / ∅/2  [By formula]

           A avg = v²/r × sinπ/2 / π/2  [Since, ∅= π]

           A avg = v²/r × 2/π

           A avg = 2v²/πr

Therefore, the magnitude of average acceleration is 2v²/πr.

Hope it helps ;-))