Question

A metre stick is pivoted about its centre. A piece of wax of mass $20g$ travelling horizontally and perpendicular to it at $5m/s$ strikes and adheres to one end of the stick so that the stick starts to rotate in a horizontal circle. Given the moment of inertia of the stick and wax about the pivot is $0.02kg{m}^2$, the initial angular velocity of the stick is

• A. $1.58rad/s$
• B. $2.24rad/s$
• C. $2.50rad/s$
• D. $5.00rad/s$

Answer 1

The correct option is C $2.50rad/s$

Given: a metre stick is pivoted about its centre. A piece of wax of mass 20 g travelling horizontally and perpendicular to it at 5 m/s strikes and adheres to one end of the stick so that the stick starts to rotate in a horizontal circle. Given the moment of inertia of the stick and wax about the pivot is 0.02 kg m2,

To find the initial angular velocity of the stick

Solution:

As per the given criteria,

Moment of inertia, $I=0.02kg{m}^2$

mass of wax, $m=20g=0.02kg$

velocity, $v=5m/s$

length, $l=1m$

Let moment of inertia of the rod about its pivot be $I_r$ and of the mass m be $I_m$

Then $I=I_m+I_r=0.02$

The linear momentum of the wax $=mv=0.02\times 5=0.1kgm/s$

When it sticks to the rod at rest, so initial angular momentum of the mass m about the pivot and the wax will be

$L_i=0+\frac{l}{2}mv=\frac{1}{2}mv$

Now when the system starts to move with angular velocity, $\omega$, the angular momentum will be

$L_f=I_r\omega +I_m\omega =\omega (I_r+I_m)⟹L_f=\omega I=0.02\omega$

By conservation of angular momentum, as no external torque is acting upon the system

$L_f=L_i⟹0.02\omega =\frac{1}{2}\times 0.02\times 5⟹\omega =2.5rad/s$

is the initial angular velocity of the stick