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Question

A solid cylinder of mass (M=1 kg) and radius R=0.5 m is pivoted at its centre The axis of rotation of the cylinder is horizontal. Three small particles of mass (m=0.1 kg) are mounted along its surface as shown in figure. The system is initially at rest.

The angular speed of cylinder, when it has rotated through 90 in anticlockwise direction.
(Take g=10 m/s2)

A
5 rad/s
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B
23 rad/s
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C
10 rad/s
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D
42 rad/s
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Solution

The correct option is A 5 rad/s
M.O.I. of system (cylinder + masses) about axis passing through centre of solid cylinder is,
I=I0+I1+I2+I3
I=MR22+m1R2+m2R2+m3R2
I=R2(M2+m1+m2+m3)
or, I=(0.5)2(0.5+0.3)=0.20 kg m2


From change in configuration of system, loss in P.E has occured.
Applying mechanical energy conservation between initial and final position.
Loss in P.E. = Gain in rotational K.E.
mgR=12Iw2
(0.1)×10×0.5=12×0.20×w2
w2=10×0.1×0.512×0.20=5
w=5 rad/s

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