A hemispherical bowl of radius r is rotated with an angular velocity omega about a vertical axis passing through its centre. An object with the same angular velocity remaining attached with the inner surface of the bowl. The straight line joining the body and the centre of the bowl makes an angle 45^(@) with the vertical. The coefficient of friction between the body and the bowl is 0.2, Assuming the body is just about to move down along curved surface of the bowl, prove that omega^(2) r = 0.943 g.

A hemispherical bowl of radius r is rotated with an angular velocity o

1.	A hemispherical bowl of radius r is rotated with an angular velocity omega about a vertical axis passing through its centre. An object with the same angular velocity remaining attached with the inner surface of the bowl. The straight line joining the body and the centre of the bowl makes an angle 45^(@) with the vertical. The coefficient of friction between the body and the bowl is 0.2, Assuming the body is just about to move down along curved surface of the bowl, prove that omega^(2) r = 0.943 g.
Answer» Solution :The hemisphere is rotated about the vertical AXIS AO with an angular velocity `omega`. The forces acting on the BODY at point P are shown in the . When the body is just about to move down. mg = n cos`THETA + mu "n cos " theta` or, `""mg = n(cos theta + mu sin theta)""`...(1) and `"" m omega^(2) r sin theta = n sin theta - mu" n sin" theta ` or,`""m omega^(2) r = n(1 - cot theta)""`...(2) From (1) and (2), we get ` (omega^(2) r)/(g) = (1 - mu cot theta)/( cos theta + mu sin theta)` = `(1 - 0.2 cot 45^(@))/(cos45^(@) + 0.2 sin 45^(@) ) [ because mu = 0.2 and theta = 45^(@) ]` `therefore "" omega^(2) r = 0.943 g`