Question

A particle of mass $m$ begins to slide down a fixed smooth sphere from the top. 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

Answer: (B)

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

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 .

The particle slides off when the normal force is zero.

$\frac{m{v}^2}{r}=mgCos\theta =-2mg(1-Cos\theta )$

or $Cos\theta =\frac{2}{3}$
so $Sin\theta =\frac{\sqrt{5}}{3}$
as $si{n}^2\theta +co{s}^2\theta =1$
Now force in tangential direction,
$F=mgSin\theta =mR$
Where R = a = tangential acceleration
So, $a=gSin\theta =$$g\frac{\sqrt{5}}{3}$