Rolling/non-rolling motion of unsymmetrical bodies

[raghavgupta93]

I have a sphere (mass m, radius r) whose centre of mass is r/2 distance away from its geometric centre. If it is kept on a smooth horizontal surface with the centre of mass at the same horizontal level as the geometric centre, what exactly is going to happen?

Also, I need to find out the velocity of the COM as the sphere turns by an angle (say, theta).

Given the the moment of inertia of the sphere about an axis passing through its COM and perpendicular to the plane of motion is mr²/3.

It's very tricky to me when it's NOT given that the sphere rolls without sliding (most problems related to unsymmetrical bodies have this mentioned).

 

[LightHero]

In such a case, the sphere will not move at all. It will remain stationary and its centre of mass will remain at the same horizontal level as the geometric centre. The only motion the sphere will experience will be a rotational motion about the centre of mass due to the unbalanced moment of inertia. The angular velocity of the sphere will be determined by the equation of conservation of angular momentum. The angular velocity (ω) will be equal to the torque (τ) divided by the moment of inertia (I). For this case, the torque is equal to the product of the mass (m) and the distance between the COM and the geometric centre (r/2), and the moment of inertia is mr2/3. Therefore, the angular velocity of the sphere will be given by: ω = (mr/2) / (mr2/3) = 3/4 radians/second.