Question

A particle is sticked to a disc at a distance $\frac{r}{2}\text{m}$ from centre. Disc has initial angular velocity of ${\omega }_o=6\hat{j}\text{rad/s}$ and constant angular acceleration of $\alpha =-\hat{j}{\text{rad/s}}^2$ as shown in figure. If radius of disc is $1\text{m}$, find the speed of particle after $t=2\text{s}$.

• A. $4\text{m/s}$
• B. $3\text{m/s}$
• C. $6\text{m/s}$
• D. $2\text{m/s}$

Answer 1

The correct option is D $2\text{m/s}$



Given:
${\omega }_o=6\hat{j}\text{rad/s}$
$\alpha =-\hat{j}{\text{rad/s}}^2$
$r=1\text{m}$
Let angular velocity of disc after time $t=2\text{s}$ is $\omega$
Using
$\omega ={\omega }_o+\alpha t$
$\omega =6\hat{j}-\hat{j}\times 2=4\hat{j}\text{rad/s}$

Using property of rigid body, we can say particle is sticked to disc therefore it will perform circular motion about centre of disc sharing same angular velocity as of disc.

$\therefore V_{particle}$ after $2\text{s}$ is
$V_{particle}=\omega \times \frac{r}{2}=4\hat{j}\times \frac{1}{2}\hat{i}\text{m/s}=-2\hat{k}\text{m/s}$
$\therefore$ Speed of particle after $2\text{s}$ is $2\text{m/s}$