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Question

A wheel is rotating with an angular speed 20 rad/sec. It is brought to rest by applying constant torque in 5 sec. If the moment of inertia of the wheel about the axis is 0.1 kgm2, then the work done by the torque in 2 sec is

A
30 J
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B
12.8 J
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C
15 J
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D
25.6 J
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Solution

The correct option is B 12.8 J
Given initial angular speed ω1=20 rad/sec.
Let final angular speed after 5 sec, ω2=0
t=5 sec
Using equation of motion for rotational motion,
ω2=ω1αt (here retardation is taking place)
So, angular retardation
α=ω1ω2t=205=4 rad/sec2

Now , angular speed after 2 sec is given as
ω3=ω1αt=204×2
ω3=12 rad/sec
By work energy theorem for rotation,
work done by the torque equals to the loss in kinetic energy.
Work done by torque in 2 sec= Loss
in kinetic energy=12I(ω23ω21)
=12×0.1[(12)2(20)2]
=12.8 J
So, work done by the torque in 2 sec is
12.8 J.

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