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Motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower point is |
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Answer» simple harmonic motion Here, in motion of ball, small angular displacement OCCUR and hence it has angular simple harmonic motion. According to figure (b), when the angle between TWO maximum displacement is `theta` then restoring force acts on it is, `F= -mg sin theta` `THEREFORE ma= -mg sin theta` `therefore a= -G sin theta` `therefore (d^(2)x)/(dt^(2)) = -g sin theta` `= -gxx (x)/(R )""[therefore sin theta approx theta = (x)/(R )]` `therefore a = -(g)/(R )*x` Motion of ballis simple harmonic motion so its acceleration `a= -omega^(2) x` comparing with it, `omega^(2) = (g)/(R )` `therefore omega = sqrt((g)/(R ))` and `(2pi)/(T)= sqrt((g)/(R ))` `therefore T= 2pi sqrt((g)/(R ))` So, motion of the ball is SHM and periodic.
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