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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
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B
2.24 rad/s
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C
2.50 rad/s
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D
5.00 rad/s
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Solution

The correct option is C $$2.50\ rad/s$$
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, $$I=0.02kgm^2$$
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
$$L_i=0+\dfrac l2mv=\dfrac 12mv$$
Now when the system starts to move with angular velocity, $$\omega$$, the angular momentum will be
$$L_f=I_r\omega+I_m\omega=\omega(I_r+I_m)\\\implies L_f=\omega I=0.02\omega$$
By conservation of angular momentum, as no external torque is acting upon the system
$$L_f=L_i\\\implies 0.02\omega=\dfrac 12\times0.02\times5\\\implies\omega=2.5rad/s$$
is the initial angular velocity of the stick

Physics

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