Question

A uniform solid cone of mass m, base radius 'R' and height 2R, has a smooth groove along its slant height as shown in figure. The cone is rotating with angular speed '$\omega$', about the axis of symmetry. If a particle of mass 'm' is released from apex of cone, to slide along the groove, then angular speed of cone when particle reaches to the base of cone is _______.

Answer 1

Moment of inertia of cone about the axis shown in figure is $=\frac{3}{10}m{R}^2$

Hence, R=radius of cone

m=mass of cone

When the particle is at apex it has no moment of inertia about axis of rotation .

When the particle is at the base of cone it has a moment of inertia which is equal to $m/s^2$.

So, initially the moment of inertia of system is

$I_i=\frac{3}{10}m{R}^2+0$

$=\frac{3}{10}m{R}^2$

Finally the moment of inertia of the system

$I_f=\frac{3}{10}m{R}^2+m{R}^2$

$=\frac{13}{10}m{R}^2$

So according to conservation of angular momentum

$I_i\omega =I_f{\omega }^{\prime }$

$⟹{\omega }^{\prime }=\frac{I_i^{\prime }}{I_f}\omega$

${\omega }^{\prime }=\frac{\left(\frac{3}{10}\right)m{R}^2}{\left(\frac{13}{10}\right)m{R}^2}\omega$

$⟹{\omega }^{\prime }=\frac{3{\omega }^{\prime }}{13}$