Question

A particle of mass m starts to slide down from the top of the fixed smooth sphere. What is the tangential acceleration when it breaks off the sphere?

• A. $\frac{2g}{3}$
• B. $\frac{\sqrt{5}g}{3}$
• C. g
• D. $\frac{g}{3}$

Answer 1

Answer: (B)

$\frac{\sqrt{5}g}{3}$

Let the particle breaks of the sphere (radius R) at point P i.e $N=0$ at P

From geometry: $AB=OB-OA=R-Rcos\theta$

Work-energy theorem: $mg(R-Rcos\theta )=\frac{1}{2}m{v}^2$ ........(1)

Circular motion : $\frac{m{v}^2}{R}=mgcos\theta$

From (1), $mgR\times (1-cos\theta )=\frac{1}{2}mgR$ $cos\theta$

$⟹cos\theta =\frac{2}{3}$

Thus $sin\theta =\frac{\sqrt{5}}{3}$

Now $m{a}_t=mg$ $sin\theta$ $⟹a_t=g\times \frac{\sqrt{5}}{3}$