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Question

A horizontal square platform of mass m and side a is free to rotate about a vertical axis passing through its centre O. The platform is stationary and a person of the same mass (m) as the platform is standing on it at point A. The person now starts walking along the edge from A to B ( see figure). The speed v of the person with respect to the platform is constant. Taking v=5 m/s and a = 1 m,


A
Angular velocity of platform when the person is at A is 6 rad/s.
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B
Angular velocity of platform increases first and then decreases again as person reaches B.
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C
Angular velocity of platform decreases first and then increases again as person reaches B.
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D
Angular velocity of platform remains constant as the person goes from A to B.
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Solution

The correct options are
A Angular velocity of platform when the person is at A is 6 rad/s.
D Angular velocity of platform remains constant as the person goes from A to B.


vm=vmp+vp
=v2+ω2x2+2vωx cos(90+θ)
=v2+ω2x22vωx sinθ

By conservation of angular momentum,
Li=0=Lf
Iω+m(vvPcos(90θ)a2=0
m12(a2+a2)ω+m(vxωsinθ)a2=0
m6a2ωma24ω+mva2=0
(xsinθ=a2)

a26ω+a24ω=va2
ω=6v5a

Therefore, angular velocity of platform remains constant (D)

and ω=6×55×1=6 rad/s (A)

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