Question

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

• A. $1.58\text{rad/s}$
• B. $2.24\text{rad/s}$
• C. $2.50\text{rad/s}$
• D. $5.00\text{rad/s}$

Answer 1

The correct option is C $2.50\text{rad/s}$

Given,
Mass of wax = $m=20\text{g}$
velocity of wax =$v=5\text{m/s}$
Moment of inertia of rod+wax $I=0.02{\text{kg/m}}^2$

As we can see, there is no external torque on the system. So, the angular momentum of the system about axis of rotation will remain same. Hence, $L_i=L_f$
Initially:


Initial angular momentum
$L_i=mvr=mv\left(\frac{h}{2}\right)$

Finally:


After collision, wax sticks to the stick. Stick is pivoted about its centre. Hence, it will start rotating about its centre.
So, Angular momentum of the system after the collision,
$L_f=I\omega$

From, $L_i=L_f$
$mv\left(\frac{h}{2}\right)=I\omega$
$0.02\times 5\times \frac{1}{2}=0.02\times \omega$
{meter stick, hence length $\left(h\right)=1\text{m}$}
$\Rightarrow \omega =2.5\text{rad/s}$