Question

A particle is undergoing circular motion of radius $R$ with angular speed $\omega$. It is brought to rest in $T$ time and then again accelerated to the same angular speed in $\frac{T}{2}$ time. The angular displacement of the particle in this time interval of $\frac{3T}{2}$is (consider uniform acceleration and retardation)

• A. $\frac{\omega T}{4}$
• B. $\frac{\omega T}{2}$
• C. $\omega T$
• D. $\frac{3\omega T}{4}$

Answer 1

The correct option is D $\frac{3\omega T}{4}$

By using equation:

$0=\omega +{\alpha }_{1}T$

$\Rightarrow {\alpha }_1=-\frac{\omega }{T}$

Now ${\theta }_1=\omega T+\frac{1}{2}\left(\frac{-\omega }{T}\right)T^2$

$\Rightarrow {\theta }_1=\frac{1}{2}\omega T$

And $\omega =0+{\alpha }_2\left(\frac{T}{2}\right)$

$\Rightarrow {\alpha }_2=\frac{2\omega }{T}$

Now, ${\theta }_2=0+\frac{1}{2}\left(\frac{2\omega }{T}\right){\left(\frac{T}{2}\right)}^2$

$=\frac{\omega T}{4}$

$\Rightarrow \theta =\omega T(\frac{1}{2}+\frac{1}{4})=\frac{3\omega T}{4}$

Hence, Option $\left(C\right)$ is the correct answer.