Question

A uniform rod of mass $m=2kg$ and length $l$ is kept on a smooth horizontal plane. If the ends $A$ and $B$ of the rod move with speeds $v$ and $2v$ respectively perpendicular to the rod, then

• A. angular velocity of the rod is $\omega =\frac{3v}{l}$
• B. linear velocity of $CM$ of the rod is $V_C=\frac{v}{2}$
• C. kinetic energy of the rod (in Joule) is $K=\frac{m{v}^2}{2}$
• D. kinetic energy of the rod (in Joule) is $K=\frac{m{v}^2}{4}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A angular velocity of the rod is $\omega =\frac{3v}{l}$
B linear velocity of $CM$ of the rod is $V_C=\frac{v}{2}$
C kinetic energy of the rod (in Joule) is $K=\frac{m{v}^2}{2}$

As the velocities of the ends $A$ and $B$ of the rod are given as $2v$ and $v$ respectively, we need to find the velocity of centre of mass of rod and its angular velocity as follows.
Since ${\overrightarrow{v}}_A={\overrightarrow{v}}_{AC}+{\overrightarrow{v}}_C$
We know ${\overrightarrow{v}}_{AC}=-\frac{1}{2}\omega \hat{i}$ and ${\overrightarrow{v}}_C=v_C\hat{i}$
But we have velocity of $A;v_A(=-v)=v_C-\frac{l}{2}\omega \left(i\right)$
Similarly, we have $v_B(=2v)=v_C+\frac{l}{2}\omega \left(ii\right)$
Solving $\left(i\right)$ and $\left(ii\right)$, we have
$\omega =\frac{3v}{l}and{v}_C=\frac{v}{2}$
Then substituting $v_C$ and $\omega$ in the equation
$K=\frac{1}{2}m{v}_C^2+\frac{1}{2}{I}_C{\omega }^2$
We have $K=\frac{1}{2}m{\left(\frac{v}{2}\right)}^2+\frac{1}{2}m{\left(\frac{ml}{12}\right)}^2{\left(\frac{3v}{l}\right)}^2$
This gives $K=\frac{m{v}^2}{2}$