Question

A rod of length $L_A=10\text{m}$ and mass $M=1\text{kg}$ is pivoted about its centre. A piece of wax of mass $50\text{g}$ travels horizontally and perpendicular to the rod at $5\text{m/s}$. It strikes and adheres to one end of the stick so that the stick starts rotating in a horizontal circle. The angular velocity of the rod after the strike is ${\omega }_A$. If we reduced the length of the rod to $L_B=5\text{m}$, the angular velocity of the rod after the strike will be ${\omega }_B$. [Assume rod to be of uniform density]
The corrrect statement(s) is(are) :

• A. ${\omega }_A<{\omega }_B$
• B. ${\omega }_A>{\omega }_B$
• C. ${\omega }_A={\omega }_B$
• D. Can't say

Answer 1

The correct option is A ${\omega }_A<{\omega }_B$

Given, $L_A=10\text{m},M_A=1\text{kg},m=0.05\text{kg}$
$I=$ Moment of inertia of (rod +wax) after strike
$=\frac{M_A{L}_A^2}{12}+m{\left(\frac{L_A}{2}\right)}^2=\frac{1\times {10}^2}{12}+0.05\times {\left(\frac{10}{2}\right)}^2$
$=\frac{115}{12}{\text{kg-m}}^2$
As we know, angular momentum
$L=I\omega =mv\left(\frac{L_A}{2}\right)=I{\omega }_A$
$\Rightarrow 0.05\times 5\times \left(\frac{10}{2}\right)=\frac{115}{12}\times {\omega }_A$
$\Rightarrow {\omega }_A=\frac{3}{23}\text{rad/s}=0.1304\text{rad/s}$
For rod $B$:
$L_B=\frac{L_A}{2}=\frac{10}{2}=5\text{m}$
Mass density is uniform, hence
$M_B=\frac{M_A}{2}=\frac{1}{2}=0.5\text{kg}$
$I$ after the strike $=I_{\text{rod}}+I_{\text{wax}}$
$I=\frac{M_B{L}_B^2}{12}+m{\left(\frac{L_B}{2}\right)}^2=\frac{0.5\times 5^2}{12}+0.05\times {\left(\frac{5}{2}\right)}^2=\frac{65}{48}{\text{kg-m}}^2$
As we know,
$L=I\omega \Rightarrow mv\left(\frac{L_B}{2}\right)=I{\omega }_B$
$\Rightarrow 0.05\times 5\times \left(\frac{5}{2}\right)=\frac{65}{48}\times {\omega }_B$
$\Rightarrow {\omega }_B=\frac{6}{13}\text{rad/s}=0.461\text{rad/s}$
Hence ${\omega }_B>{\omega }_A$