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 inertia of the stick and wax about the pivot is 0.02kgm^2. What is the initial angular velocity of the stick.

Answer 1

Here if we consider wax + stick as a system

we can say  that there is no external torque on it so angular momentum is conserved

Initial angular momentum of stick + wax = final angular momentum of stick + wax

initial angular momentum is given by

$L_i = mvr$

here

m = 20 g

v = 5 m/s

r = 50 cm

$L_i = 0.020* 5 * 0.50 = 0.05$

final angular momentum is given by

$L_f = I\omega$

here I = 0.02 kg m^2

now by angular momentum conservation

$L_i = L_f$

$0.05 = 0.02*\omega$

$\omega = 2.5 rad/s$

so the final angular speed of wax + stick is 2.5 rad/s

Answer 2

Angular momentum = I × angular velocity  20×10^-3×5×0.5=2×10^-2× angular velocity  Angular velocity =2.5 rad/s