Question

6.
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 'o', 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
2 R
R
30
(A)
13
500
(C)
13
40
(B)
13
90
(D)
13​

Answer 1

Answer

Moment of inertia of cone about the axis shown in figure is = 3/10mR square

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 square.

So, initially the moment of inertia of system is

Ii = 3/10mR square+0

= 3/10mR square

Finally the moment of inertia of the system

I f =3/10mR square+mR square

=13/10mR square

So according to conservation of angular momentum

I iw=I fw'

⟹ω'= I'i/ If w

ω'=(3/10)mR square/

(13/10)mR square

⟹ω =3w'/13