Question

The centre of mass of a uniform rod of length $10\text{m}$ is moving with translational velocity of $50\text{m/s}$ on a frictionless horizontal surface as shown in the figure and the rod rotates about its centre of mass with an angular velocity of $5\text{rad/sec}$. Find out the velocities of points $A$ and $B$.

• A. $\overrightarrow{V_A}=75\hat{i}$
• B. $\overrightarrow{V_B}=25\hat{i}$
• C. $\overrightarrow{V_A}=25\hat{i}$
• D. $\overrightarrow{V_B}=75\hat{i}$

Answer 1

Answer: (C), (D)

$\overrightarrow{V_B}=75\hat{i}$

Given,
$\overrightarrow{V_{com}}=50\text{m/s}\left(\hat{i}\right)$
Angular velocity about COM $\omega =5\text{rad/s}\left(\hat{k}\right)$ (by right hand thumb rule)
Length of the rod $L=10\text{m}$
As we know
$\overrightarrow{V_B}=\overrightarrow{V_{com}}+\overrightarrow{\omega }\times \overrightarrow{r_{BO}}$
$\overrightarrow{V_B}=50\hat{i}+5\hat{k}\times 5(-\hat{j})$
$\overrightarrow{V_B}=50\hat{i}+25\hat{i}\{\because \hat{k}\times \hat{j}=-\hat{i}\}$
$\overrightarrow{V_B}=75\hat{i}$
Similarly,
$\overrightarrow{V_A}=\overrightarrow{V_{com}}+\overrightarrow{\omega }\times \overrightarrow{r_{AO}}$
$\overrightarrow{V_A}=50\hat{i}+5\left(\hat{k}\right)\times 5\left(\hat{j}\right)$
$\overrightarrow{V_A}=50\hat{i}+25(-\hat{i})\{\because \hat{k}\times \hat{j}=-\hat{i}\}$
$\overrightarrow{V_A}=25\hat{i}$