Question

A particle is moving in a circular path with an angular velocity of $2\text{rad/s}$ and angular acceleration of $\alpha =\frac{5}{2}\omega$, where $\omega$ is the angular velocity at any particular instant. Find the angular velocity after the particle has moved through an angle of $3\pi$.

• A. $20.64\text{rad/s}$
• B. $15.52\text{rad/s}$
• C. $25.56\text{rad/s}$
• D. $28.25\text{rad/s}$

Answer 1

The correct option is C $25.56\text{rad/s}$

Given angular acceleration is
$\alpha =\frac{5}{2}\omega$
$\Rightarrow \alpha =\frac{d\omega }{dt}=\frac{5}{2}\omega$

Using chain rule, we get,

$\frac{d\omega }{d\theta }\cdot \frac{d\theta }{dt}=\frac{5}{2}\omega$

$\Rightarrow \omega \cdot \frac{d\omega }{d\theta }=\frac{5}{2}\omega$

$\Rightarrow 2\cdot d\omega =5d\theta$

Integrating with limits on both sides, we get,

$2{\int }_2^{\omega }d\omega =5{\int }_0^{3\pi }d\theta$

$\Rightarrow 2(\omega -2)=5(3\pi -0)$

$\Rightarrow \omega =\frac{15\pi +4}{2}\approx 25.56\text{rad/s}$