Question

A small body is placed on the top of a smooth sphere of radius $R$. Then the sphere is imported a constant acceleration $w_0$ in the horizontal direction and the body begins sliding down.

Find the angle ${\theta }_0$ between the vertical and the radius vector drawn from the centre of the sphere to the break-off point; calculate ${\theta }_0$ for $w_0=g$.

• A. ${\theta }_0={17.6}^{\circ }$
• B. ${\theta }_0={14.6}^{\circ }$
• C. ${\theta }_0\approx {14}^{\circ }$
• D. ${\theta }_0\approx {17}^{\circ }$

Answer 1

The correct option is D ${\theta }_0\approx {17}^{\circ }$

From the free body diagram of the sphere, balancing forces along the normal,

$m{a}_n=mg\cos \theta -N-mg\sin \theta$

Also the loss in potential energy is equal to gain in kinetic energy,

loss in potential energy along y-axis $=mg(R-R\cos \theta )$

loss in potential energy along x-axis $=m{\omega }_{0}R\sin \theta =mgR\sin \theta$

Hence, $\frac{1}{2}m{v}^2=mgR[1-\cos \theta +\sin \theta ]$

But, $a_n=\frac{m{v}^2}{R}$

Substituting these in first equationa and putting $N=0$(for body to lose contact) gives

$\cos \theta -\sin \theta =\frac{2}{3}$

Also we know: $co{s}^2\theta +si{n}^2\theta =1$

$⟹\theta ={17}^{\circ }$