Question

A particle of mass m begins to slide down a fixed smooth sphere from the top with negligible initial velocity. What is its 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

The correct option is B $\frac{\sqrt{5}g}{3}$

If a particle starts from the top,the speed of the particle will be given by energy conservation law.It is given as

change in the potential energy = gain in kinetic energy

$mgR(1-cos\theta )=\frac{1}{2}m{v}^2$

from force equation now

since it break off at this position so normal force will be zero

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

now we will use the value from above two equations

$mgcos\theta =2mg\times (1-cos\theta )$

by solving above equation

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

now for tangential acceleration as we know that

$a_t=\frac{F_m}{m}$

$a_t=\frac{mgsin\theta }{m}$

$a_t=gsin\theta$

from above equation

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

so tangential acceleration will be

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

$a_t=7.45\frac{m}{s^2}$