Question

A particle of mass $m$ begins to slide down a fixed smooth sphere from the top as shown. What is its acceleration when it breaks off the sphere?

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

Answer 1

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

Suppose the particle of mass $m$ slides down a smooth sphere of radius $R$, starting from rest at the top. Suppose the particle leaves the sphere at angle $\theta$ with vertical.
The particle slides off when the normal force is zero.
Using Newton's second law,
$\frac{m{v}^2}{r}=mg\left(\cos \theta \right)$

from WE theorem
$\frac{m{v}^2}{2}=mgr(1-\cos \theta )$

from simplification,

$mg\cos \theta =2mg(1-\cos \theta )$
$3\cos \theta =2$
$\cos \theta =\frac{2}{3}$
Now tangential acceleration is $g\sin \theta =g\sqrt{1-{\left(\frac{2}{3}\right)}^2}=\frac{\sqrt{5}g}{3}$