Question

A sphere of mass M and radius R is attached by a light rod of length l to a point P. The sphere rolls without slipping on a circular track as shown. It is released from the horizontal position. The angular momentum of the system about P when the rod becomes vertical is?

• A. $M\sqrt{\frac{10}{7}gl}[l+R]$
• B. $M\sqrt{\frac{10}{7}gl}[l+\frac{2}{5}R]$
• C. $M\sqrt{\frac{10}{7}gl}[l+\frac{7}{5}R]$
• D. $M\sqrt{\frac{10}{7}gl}[l-\frac{2}{5}R]$

Answer 1

The correct option is B $M\sqrt{\frac{10}{7}gl}[l+\frac{2}{5}R]$

Let at bottom, $v$=speed and $\omega =$ angular speed

From conservation of energy-

$Mgl=\frac{1}{2}M{v}^2+\frac{1}{2}I{\omega }^2$

$⟹Mgl=\frac{1}{2}M{v}^2+\frac{1}{2}\frac{2}{5}M{R}^2{\omega }^2$

And $v=R\omega$

$⟹\frac{7}{10}M{v}^2=Mgl$

$⟹v=\sqrt{\frac{10gl}{7}}$

Now, the momentum about P is given by-

$L=I\omega +Mvl=\frac{2}{5}M{R}^2\omega +Mvl$

$L=I\omega +Mvl=\frac{2}{5}MRv+Mvl$

$⟹L=M\sqrt{\frac{10gl}{7}}[\frac{2R}{5}+l]$

Answer-(B)