First slide

Vertical circular motion

Question

A small body of mass m slides down from the top of a hemisphere or radius r (Fig. 2). The surface of block and hemisphere are frictionless. The height at which the body lose contact with the surface of the sphere is

Moderate

A

(3 / 2) r
B

(2 / 3) r
C

(1 / 2)g t²
D

v² / 2g

Solution

See fig. (8). The body will lose contact when centripetal acceleration becomes equal to the component of acceleration due to gravity along the radius.
Velocity at P, $v = \sqrt{2 g (r - h)}$
$(∵ v^{2} - u^{2} = 2 gx)$
Centripetal acceleration will be v²/r. It should be equal to the component of g along PO. Hence
$\frac{v^{2}}{r} = gcos θ$
or $\frac{2 g (r - h)}{r} = g \times \frac{h}{r}$
Solving, we get $h = (\frac{2 r}{3})$

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