Question

A wheel of radius $0.20m$ is accelerated from rest with an angular acceleration of $1\mspace{6mu} rad/s^{2}$. After a rotation of $90^{o}$ the radial acceleration of a particle on its rim will be

• A. $\pi\mspace{6mu} m/s^{2}$
• B. $0.5\mspace{6mu}\pi\mspace{6mu} m/s^{2}$
• C. $2.0\pi\mspace{6mu} m/s^{2}$
• D. $0.2\mspace{6mu}\pi\mspace{6mu} m/s^{2}$

Answer 1

Answer: (D)

$0.2\mspace{6mu}\pi\mspace{6mu} m/s^{2}$

Answer 2

From the equation of motion

Angular speed acquired by the wheel,$\omega_{2}^{2} = \omega_{1}^{2} + 2\alpha\theta$ $= 0 + 2 \times 1 \times \frac{\pi}{2}$ ⇒ $\omega_{2}^{2} = \pi$

Now radial acceleration $\omega^{2}r = \pi \times 0.2 = 0.2\pi m/s^{2}$