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 the figure. The cone is rotating with angular speed ${}^{\prime }{\omega }^{\prime }$, about the axis of symmetry. If a particle of mass 'm' is released from the apex of the cone, to slide along the groove, then the angular speed of cone when the particle reaches to the base of the cone is?

• A. $\frac{3\omega }{13}$
• B. $\frac{4\omega }{13}$
• C. $\frac{5\omega }{13}$
• D. $\frac{9\omega }{13}$

Answer 1

Answer: (A)

$\frac{3\omega }{13}$

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

Here, 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

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

So according to conservation of angular momentum

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

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

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

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