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Question

A turntable rotates about a fixed vertical axis, making one revolution in 10 s. The moment of inertia of the turntable about the axis of rotation is 1200 kg.m2. A man of mass 80 kg, initially standing at the centre of the turntable, runs outwards along the radius. What is the approximate angular velocity of the turntable when the man is at a distance of 2 m from the centre?

A
1.25 rad/s
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B
2.75 rad/s
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C
0.50 rad/s
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D
0.1 rad/s
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Solution

The correct option is C 0.50 rad/s
Let I0 be the initial moment of inertia of the system (man + table)
I0=Iman+Itable
=0+1200=1200 kg.m2
(Iman=0 as the man is at the axis of rotation)
I=final moment of inertia of the system
I=Iman+Itable
I=mr2+1200
I=80(2)2+1200=1520 kg.m2

weight of disc is passing through it's centre and weight of man is acting to the axis of rotation hence we can conserve angular momentum as τext=0 about axis of rotation,

By applying P.C.A.M:
I0ω0=Iω ....(i)
ω0=2πT0=π5 rad/s
From Eq. (i)
ω=I0ω0I=1200×π1520×50.50 rad/s

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