Question

A small ball of mass $m$ and radius $r=\frac{R}{10}$ rolls without slipping along the track shown in the figure. The radius of circular part of the track is $R$ . If the ball starts from rest at a height of $8R$ above the bottom, the normal force on the ball at the point $P$ is

• A. $\frac{100}{3}mg$
• B. $\frac{100}{9}mg$
• C. $\frac{200}{9}mg$
• D. $\frac{50}{7}mg$

Answer 1

Answer: (B)

Using Work-Energy Theorem
$7mgR=\frac{1}{2}{\left(mv\right)}^2+\frac{1}{2}\left(\frac{2}{5}{\left(mr\right)}^2\right)\times {\left(\omega \right)}^2$
since the sphere is rolling without slipping $v=r\omega$
$\therefore 7mgR=\frac{1}{2}\times \left(\frac{7}{5}{\left(mr\right)}^2\right)\times \frac{v^2}{r^2}$
$\Rightarrow v^2=10gR$
at point $P$ , we can consider sphere as a point object doing circular motion in a circle of the radius $(R-r)$ with velocity $v$ ,
Hence at point $P$
$N=\frac{{\left(mv\right)}^2}{\left(R-r\right)}=\frac{10mgR}{\left(R-\frac{R}{10}\right)}=\frac{10mgR}{\frac{9R}{10}}$ ( given $r=\frac{R}{10}$ )
or $N=\frac{100}{9}mg$