Question 467: Centre of Mass and Rotational Motion

Question

A thin rod of mass 0.9 kg and length 1 m is
suspended, at rest, from one end so that it can
freely oscillate in the vertical plane. A particle
of move 0.1 kg moving in a straight line with
velocity 80 m/s hits the rod at its bottom most
point and sticks to it (see figure). The angular
speed (in rad/s) of the rod immediately after the
collision will be _________.

Accepted Answer

The given situation is shown in the following figure,
Applying law of conservation of angular momentum about pivotal point,
$${L_i} = {L_f} \Rightarrow mvl = I\omega $$
$$ \Rightarrow mvl = \left( {{1 \over 3}M{L^2} + m{L^2}} ight)\omega $$ ..... (i)
Given, m = 0.1 kg, v = 80 ms
$$-$$1
, M = 0.9 kg, l = 1 m
Substituting these values in eq. (i), we get
$$0.1 	imes 80 	imes 1 = \left( {{{0.9 	imes {1^2}} \over 3} + 0.1 	imes {1^2}} ight)\omega $$
$$8 = \left( {{3 \over {10}} + {1 \over {10}}} ight)\omega  \Rightarrow 8 = {4 \over {10}}\omega  \Rightarrow \omega  = 20$$ rad/s

Centre of Mass and Rotational Motion

Choose the correct answer: