What is the acceleration of the center of mass of the rolling hollow sphere?

[chrismcr]

A hollow spherical shell with mass 1.50kg rolls without slipping down a slope that makes an angle of 40.0degree angle with the horizontal.
-Find the magnitude of the acceleration of the center of mass of the spherical shell

I am really confused on how to go about this problem. I know it has to do with inertia, but wow, i just don't know where to start.

 

[NTUENG]

U used the term 'rolls without slipping'. I thought of using rotational mechanics to solve it but u are not given the radius of the spherical shell. Thus, i shall treat it like a case of linear motion down the slope. Resolving forces and applying Newton's 2nd law on the shell, we have Normal reaction force = mg cos 40 and mg sin 40 = ma. Thus a = g sin 40 = 9.8 sin 40. If u want to find angular acceleration of the shell, u need to have the shell's radius and use the relation, linear acceleration = spherical radius x angular acceleration.

 

[danni7070]

No, you don't need the radius for this one. This is a hollow shell and it's moment of inertia is 2/3mr^2 = I
Fy = 0
Fx = mgsin(theta) - f = ma_cm

I get from Fx=mgsin(theta)-f=ma_cm (1)
and fR = 2/3mra_cm (2)

Solve f from (1).
Substitute that into (2) and then solve for a_cm

I get a_cm = (3gsin(theta))/5