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A uniform horizontal circular platform of mass `200 kg` is rotating at 10 rpm about a vertical axis passing through its center. A boy of mass `50 kg` is standing at its edge. If the boy moves to the center of the platform, the frequency of rotation would becomeA. `7.5` rpmB. `12.5` rpmC. 15 rpmD. 20 rpm |
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Answer» `[(1)/(2)(200)r^(2)+50r^(2)](10)=(1)/(2)(200)r^(2)omega` `omega=15rp m` |
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