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Question

A disc having radius r=1 m and m=2 kg is rotating about an axis passing through its COM and perpendicular to the plane of disc. Initial velocity of a point A at the edge of disc is 6 m/s. A retarding torque of magnitude 2 Nm starts acting on disc about its axis of rotation. Find the velocity of point A after t=2 s.


A
1 m/s
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B
3 m/s
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C
4 m/s
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D
2 m/s
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Solution

The correct option is D 2 m/s

Moment of inertia of disc, I=mr22
=2×122 kg-m2
=1 kg-m2

We know that, V=ω×r
Vo=ωo×r
ωo=Vor=61=6 rad/s

Also we know,
|τ|=I|α|
|α|=|τ|I=21=2 rad/s2

As mentioned in the question, torque is retarding in nature.
α will be in the opposite direction to initial angular velocity.
α=2 rad/s2
Let (ω) be the angular velocity of disc after t=2 s.
Using equation of motion;
α=ωωot
2=ω62
ω=2 rad/s

Again using,
V=ω×r for point A
V=2×1=2 m/s

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