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Question

A meter 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 which 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, the 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,
Mass of wax = m=20 g
velocity of wax =v=5 m/s
Moment of inertia of rod+wax I=0.02 kg/m2

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, Li=Lf
Initially:


Initial angular momentum
Li=mvr=mv(h2)

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,
Lf=Iω

From, Li=Lf
mv(h2)=Iω
0.02×5×12=0.02×ω
{meter stick, hence length (h)=1 m}
ω=2.5 rad/s

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