Question

A spherical shell of mass M and radius R is completely filled with a frictionless fluid, also of mass M. It is released from rest, and then it rolls without slipping down an incline that makes an angle θ with the horizontal. What will be the acceleration of the shell down the incline just after it is released? Assume the acceleration of free fall is g. The moment of inertia of a thin shell of radius r and mass m about the center of mass is I =2/ 3mr2; the moment of inertia of a solid sphere of radius r and mass m about the center of mass is I =2/5mr2.

(A) a = g sin θ
(B) a =3/4 g sin θ
(C) a =1/2 g sin θ
(D) a =3/8 g sin θ
(E) a =3/5 g sin θ

Answer 1

b)
`a = (3)/(4)gsin\theta`

Step-by-step explanation:

Since the liquid filled inside the shell has no friction so the liquid will not rotate along with the shell

So here we can say

`I = (2)/(3)MR^2`

now we have force equation for the sphere

`2Mg sin\theta - f = (2M)a`

now we have equation of torque given as

`fR = (2)/(3)MR^2 ((a)/(R))`

`f = (2)/(3)Ma`

now we have

`2Mgsin\theta - (2)/(3)Ma = 2Ma`

`2Mgsin\theta = (8)/(3)Ma`

`a = (3)/(4)gsin\theta`