First slide

Vertical circular motion

Question

A small block is placed at the top of a sphere. It slides on the smooth surface of the sphere. What will be the angle made by the radius vector of the block with the horizontal when it leaves the surface ?

Moderate

A

$52^{\circ}$
B

$48^{\circ}$
C

$42^{\circ}$
D

$38^{\circ}$

Solution

Here $\frac{{mv}^{2}}{R} = mgsin θ . . . (1)$
The speed is acquired by the block after falling through a distance h. Hence
$\frac{1}{2} {mv}^{2} = mgh or v = \sqrt{2 gh}$ …(2)
$∴ \frac{m (2 gh)}{R} = mgsin θ = mg \frac{(R - h)}{R}$
or $2 h = R - h or h = R / 3$
From figure,
$\begin{array}{l} \sin θ = \frac{R - h}{R} = \frac{R - R / 3}{R} = \frac{2}{3} \\ θ = \sin^{- 1} \frac{2}{3} = 42^{\circ} \end{array}$

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