Question

An object at rest at point $A$ slides down on a smooth surface ending at point $B$ at a fixed hemisphere as shown in the figure. Determine the angle $\theta$ at which the object will leave the hemisphere.

• A. ${\text{cos}}^{-1}\left(\frac{4}{7}\right)$
• B. ${\text{cos}}^{-1}\left(\frac{5}{6}\right)$
• C. ${\text{cos}}^{-1}\left(\frac{1}{2}\right)$
• D. ${\text{cos}}^{-1}\left(\frac{1}{3}\right)$

Answer 1

The correct option is B ${\text{cos}}^{-1}\left(\frac{5}{6}\right)$


When the block leaves the surface then the normal force between the block and the surface becomes zero.
$N=0$
At point $C$,
$\Rightarrow \text{mg cos}\theta =\frac{m{v}^2}{R}+N$
$\Rightarrow \text{mg cos}\theta =\frac{m{v}^2}{R}$
[ as $N=0$ at $C$ ]
$\Rightarrow v^2=Rg\cos \theta$ ...............(1)
By using law of conservation of mechanical energy,
Loss in Potential energy $=$ Gain in Kinetic energy
$mg(R+h)-mgR\cos \theta =\frac{m{v}^2}{2}-\frac{m{v}_0^2}{2}$
$\Rightarrow g(R+h)-gR\cos \theta =\frac{Rg\cos \theta }{2}-\frac{v_0^2}{2}$
[ from (1) ]
$\Rightarrow \frac{3gR\cos \theta }{2}=g(R+h)+\frac{v_0^2}{2}$
$\Rightarrow \cos \theta =\frac{2}{3gR}\left(g\right(R+h)+\frac{v_0^2}{2})$
$\Rightarrow \cos \theta =\frac{2}{3\times 10\times 4}(10\times (4+1)+\frac{0^2}{2})$
$\Rightarrow \cos \theta =\frac{5}{6}$
$\Rightarrow \theta ={\text{cos}}^{-1}\left(\frac{5}{6}\right)$
Hence option $\left(b\right)$ is correct.