Question

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 $0.02\,kg\,{m^2}$, the initial angular velocity of the stick is
A. 1.58 rad/s
B. 2.24 rad/s
C. 2.50 rad/s
D. 5.00 rad/s

Answer 1

Calculate the initial and final angular momentum of the system. According to conservation of angular momentum, the initial angular momentum equals the final momentum of the system.

Formula used:

The linear momentum of the body of mass m and velocity v is,
$P = mv$

The angular momentum of the body rotating with angular velocity $\omega$ is,
$L = I\omega$

Here, I is the moment of inertia.

Complete step by step answer:
We assume the moment of inertia of the rod of metre scale is ${I_r}$ and moment of inertia of the mass m is ${I_m}$. Therefore, the moment of inertia of the system will be,
$I = {I_m} + {I_r}$

The linear momentum of the wax travelling horizontally is
$P = mv$

Here, m is the mass of wax and v is the linear velocity of wax.

Substitute 20 g for m and 5 m/s for v in the above equation.
$P = \left( {0.020\,kg} \right)\left( {5\,m/s} \right)$
$\Rightarrow P = 0.1\,kg\,m/s$

When the wax rest on the end of the metre stick just before it starts rotating, the initial angular momentum of the system is,
${L_i} = {m_r}{v_r}r + mvr$

Here, ${m_r}$ is the mass of the metre stick, ${v_r}$ is the linear velocity of the metre stick and r is the radius of the circular motion. The radius of the circular motion is half of the metre stick as it is pivoted at the centre. Therefore, the radius of the circular motion will be 0.5 m as the metre stick has length 1 m.

The linear velocity if the stick is zero, therefore, the above equation becomes,
${L_i} = mvr$

Substitute $0.1\,kg\,m/s$ for $mv$ and 0.5 m for r in the above equation.
${L_i} = \left( {0.1\,kg\,m/s} \right)\left( {0.5\,m} \right)$
$\Rightarrow {L_i} = 0.05\,kg\,{m^2}/s$

As the metre stick starts rotating, the final angular momentum of the system will be,
${L_f} = \left( {{I_r} + {I_m}} \right)\omega$
${L_f} = I\omega$

Here, $\omega$ is the angular velocity of the system.

Substitute $0.02\,kg\,{m^2}$ for I in the above equation.
${L_f} = 0.02\omega$

According to the law of conservation of angular momentum, the initial and final momentum of the system is the same. Therefore, we can write,
${L_i} = {L_f}$
$0.05 = 0.02\omega$
$\Rightarrow \omega = 2.5\,rad/s$

So, the correct answer is option (C).

Note: In this question, to express the moment of inertia of the system, we have used the principle of parallel axes. To solve this question, we could have used conservation of angular momentum at the first step. Initially, the system does not rotate, therefore, the angular momentum is only due to linear momentum of wax. The final angular momentum involves the term moment of inertia, which relates angular velocity of the body.