Question

The sphere shown in figure lies on a rough plane when a particle of mass m traveling at a speed $v_0$ collides and sticks with it. If the line of motion of the particle is at a distance 'h' above the plane, find the value of 'h for which the sphere starts pure rolling on the plane. Assume that the mass M of the sphere is large compared to the mass of the particle so that the center of mass of the combined system is not appreciably shifted from the centre of the sphere.

• A. $\left(\frac{5}{7}\right)R$
• B. R
• C. $\left(\frac{7}{5}\right)R$
• D. $\left(\frac{7}{3}\right)R$

Answer 1

The correct option is C

$\left(\frac{7}{5}\right)R$

Take the particle plus sphere as a system.

Using conservation of linear momentum:

$m{v}_0=(m+M)v$

v=$\frac{m{v}_0}{\left(mM\right)}$

We shall use conservation of angular momentum about the centre of mass which is to be taken at the centre of the sphere. Angular momentum of the particle before collision is:

$m{v}_0(h-R)$

If the system rotates with angular speed ωafter collision,

Angular momentum of the system becomes:

$(\frac{2}{5}M{R}^2+m{R}^2)\omega$

Hence,

$m{v}_0(h-R)=(\frac{2}{5}M{R}^2+m{R}^2)\omega$

$\omega =\frac{m{v}_0(h-R)}{(\frac{2}{5}M+m)R}$

Take the particle plus the sphere as the system. The sphere will start rolling just after the collision if

$v=\omega R,i.e.,\frac{m{v}_0}{(m+M)}=\frac{m{v}_0(h-R)}{(\frac{2}{5}M+m)R}$

giving, $h=(\frac{7}{5}M+2m)R\approx \frac{7}{5}R$