Question

A disc having radius $r=1\text{m}$ and $m=2\text{kg}$ is rotating about an axis passing through its COM and perpendicular to the plane of disc. Initial velocity of a point $A$ at the edge of disc is $6\text{m/s}$. A retarding torque of magnitude $2\text{Nm}$ starts acting on disc about its axis of rotation. Find the velocity of point $A$ after $t=2\text{s}$.

• A. $1\text{m/s}$
• B. $3\text{m/s}$
• C. $4\text{m/s}$
• D. $2\text{m/s}$

Answer 1

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


Moment of inertia of disc, $I=\frac{m{r}^2}{2}$
$=\frac{2\times 1^2}{2}{\text{kg-m}}^2$
$=1{\text{kg-m}}^2$

We know that, $V=\omega \times r$
$\therefore V_o={\omega }_o\times r$
$\Rightarrow {\omega }_o=\frac{V_o}{r}=\frac{6}{1}=6\text{rad/s}$

Also we know,
$|\tau |=I|\alpha |$
$|\alpha |=\frac{|\tau |}{I}=\frac{2}{1}=2{\text{rad/s}}^2$

As mentioned in the question, torque is retarding in nature.
$\therefore \alpha$ will be in the opposite direction to initial angular velocity.
$\therefore \alpha =-2{\text{rad/s}}^2$
Let $\left(\omega \right)$ be the angular velocity of disc after $t=2\text{s}$.
Using equation of motion;
$\alpha =\frac{\omega -{\omega }_o}{t}$
$-2=\frac{\omega -6}{2}$
$\Rightarrow \omega =2\text{rad/s}$

Again using,
$V=\omega \times r$ for point $A$
$V=2\times 1=2\text{m/s}$