SHM of a rigid body

This is one of those cases where you have to pay attention to the kinematics of the problem. The pendulum has 1 DOF, let's call it the swing angle $theta$.

The motion of the rigid body depends on this variable and its derivatives.

$$begin{aligned} omega & = dot{theta} & dot{omega} & = ddot{theta} v_t & = (R+ell) dot theta & dot{v}_t & = (R+ell) ddot{theta} v_r & = 0 & dot{v}_r &= -(R+ell) dot{theta}^2 end{aligned}$$

where $v_t$ is the tangential velocity and $omega$ the rotational velocity. Also, consider the mass moment of inertia about the sphere center $I_C=tfrac{2}{5} m R^2$, as well a the tangential and radial acceleration of the center.

Now you have a choice about which point to form the equations of motion. You can choose between the center of mass and the pivot (since it is not moving).

• Pivot - Find the MMOI about the pivot $I_P = I_C + m (R+ell)^2$ and form the equations of motion. Since the pivot does not move, only the rotational equation is needed $$ -(R+ell) m g sin theta = I_P ddot{theta} tag{1} $$ which is solved for $$ddot{theta} =- frac{m g(R+ell)}{I_C + m(R+ell)^2} sin theta $$

• Center - Now we have to consider the pivot forces and the translating motion of the center of mass $$ begin{aligned} F_r - m g cos theta & = m dot{v}_r F_t - m g sin theta & = m dot{v}_t -(R+ell) F_t & = I_C ddot{theta} end{aligned} tag{2}$$ which is solved for

  $$ddot{theta} = - frac{m g (R+ell)}{I_C +m (R+ell)^2} sin theta $$

As you can see, both methods yield the same result, as long as you account for everything that needs to be accounted for. By default resolve the equations about the center of mass, unless like in this case it simplifies the problem considerably. You can only pick points that are inertial or the centers of mass to write the equations of motion.

For both cases though you have to resolve the kinematics first to establish the motion of each center of mass as a function of the degrees of freedom.