Question

A particle is in circular motion in a horizontal plane. It has angular velocity of $10\pi rad/s$ at the end of 2 s and angular velocity $15\pi rad/s$ at the end of 4 s. The angular acceleration of particle is

• A. $5\pi rad/s^2$
• B. $2.5\pi rad/s^2$
• C. $7.5\pi rad/s^2$
• D. $2\pi rad/s^2$

Answer 1

The correct option is B $2.5\pi rad/s^2$

Initial angular velocity, ${\omega }_o=10\pi rad{s}^{-1}$

Final angular velocity, $\omega =15\pi rad{s}^{-1}$

Time difference, $\Delta t=4-2=2\sec$

$\omega ={\omega }_o+\alpha t$

$\alpha =\frac{\omega -{\omega }_o}{t}=\frac{15\pi -10\pi }{2}=\frac{5\pi }{2}=2.5\pi rad{s}^{-2}$

Hence, angular acceleration is $2.5\pi rad{s}^{-2}$