# SHM of a rigid body

In the analysis of SHM of a point sized bob oscillating with small angular displacement we can analyse the SHM in both linear and angular terms and arrive at the same answer and this should be true for other rigid bodies executing angular SHM since in all cases $$\theta \approx x_{cm}/d$$, where d is the distance of COM from point of suspension so if $$\theta$$ varies as a SHM so is $$x_{cm}$$ going to.

I tried the same for a solid sphere of radius $$R$$ and mass $$m$$ connected with a massless string of length $$l$$. On displacing the bob by an angle $$\theta$$ the restoring force acting on the COM is going to be:

$$F_r = mgsin(\theta)$$

Also, $$sin(\theta)\approx \theta = x/(l+R)$$ where $$x$$ is the linear displacement of the COM . So: $$F_r = mgx/(l+R)$$

So, the Time period is going to be: $$T = 2\pi\sqrt{(l+R)/g}$$

So I end up getting the same equation as that of the point sized bob...which I guess is wrong because when we analyse this scenario using angular SHM, as I was taught, we get a different answer:

So the torque about the point of suspension P is going to be: $$\tau_p = mgsin(\theta)\times (l+R) \approx mg(l+R)\theta= I_p \alpha$$ So using $$T = 2\pi\sqrt{I_p/C}$$, where $$C = mg(l+R)$$

So, the time period here is going to be: $$T= 2\pi\sqrt{\frac{2mR^2/5 + m(R+l)^2}{mg(l+R)}}$$ which of course is different from my previous analysis. So I am clearly missing something ,help will be really appreciated.

Ps: I have tried this for other bodies (like rod hinged at its end) and it works... I am having issues with rigid bodies connected by a string to their point of suspension but in cases where the bodies have the hinge point lying on them there this linear SHM analysis works when we take hinge forces into account.

• Isn't $I_p= 2mR^2/5 + m l^2$, since $l$ is taken to go all the way to the center of the ball? This still doesn't answer the difference... – ohneVal Jan 5 at 9:40
• I took the l as the length of the string from the point of the suspension to the point where it touches the sphere. – Dhrxv Jan 5 at 9:43
• Then you have to change the COM equation in the first approach and include $R$. Also the torque equation will be changed, since the torque is made at the COM. – ohneVal Jan 5 at 9:43
• I am thinking perhaps the assumption is what is wrong, namely why do we expect both treatments to agree? The COM treatment assumes point-like objects, so unless one introduces some corrections, the COM treatment knows nothing about the size/shape of the object. I expect then the real observation to match better with the torque treatment. – ohneVal Jan 5 at 10:03
• I dont think it should matter how we approach a problem..the final answer should be the same, but here it is not which means I am missing something in my force analysis. This was my original question, what am I missing? – Dhrxv Jan 5 at 10:20

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.

• As I pointed out I was able to do the force analysis just like you did for bodies which had the point of suspension on them as I could take the hinge forces Fr,Ft into account and arrived on the correct answer but the problem in this case was that acording to me there is no Ft in this case acting on the bob since the bob is connected by a massless string so there is only Fr=Tension , which the string can exert on the ball...I think my assumption that the string can only exert tension/ Fr on the ball is wrong – Dhrxv Jan 6 at 3:48
• The lateral acceleration means there is also tangent force. It is not like the simple pendulum where tension is only along the string. The pivot forces are in two directions depending on centripetal motion and rotational acceleration. – John Alexiou Jan 6 at 13:22

In your question, you deal more carefully with the angular approach leading to $$mg(l+R)sin \theta=I\ddot\theta$$ than you do with the linear approach. Are you assuming that the linear approach leads to the following (for small displacements)? $$mgx=m\ddot x$$ The $$m$$ that appears on the right hand side is mass in its inertial role. But not all the bob has the same acceleration. The top of the bob moves with a smaller acceleration than the bottom, simply because it is moving in a smaller arc. This is the subtlety that is taken into account by the 'angular treatment' using a moment of inertia that includes $$\tfrac 25 m R^2$$, but not by the simple linear treatment, which is slightly flawed for a bob of finite size.

• I know that all of the bob does not have the same acceleration which is why I tried to analyse only the motion of COM point because it's motion is determined only by the net force on the body...So I think we can fully analyse the motion of the bob's COM with the knowledge of only the net force on the body, as I pointed out in the end of my answer I tried the COM approach for other rigid bodies using only force analysis and was able to reach on the same answer. – Dhrxv Jan 5 at 17:12
• (a) "I tried to analyse only the motion of COM point because it's motion is determined only by the net force on the body." I think not. The C of M is acted upon by forces from the other parts of the bob. Without forces between neighbouring parts of the bob, then because different parts of the bob have different accelerations, the bob would come apart into many pieces. (b) "I tried the COM approach for other rigid bodies using only force analysis and was able to reach on the same answer." What, exactly do you mean? – Philip Wood Jan 5 at 17:36
• (a) I think yes, for a rigid body $\vec F_{net} = m\vec a_{com}$..I am providing a link for you to check:en.wikipedia.org/wiki/Rigid_body_dynamics. (b)I mean I tried this approach for other rigid bodies which had the point of suspension on them and arrived at the same answer by both this linear approach and angular approach – Dhrxv Jan 5 at 17:54
• (a) I had mistaken your meaning. You are quite right, of course, that $\vec F_{net} = m \vec a_{com}$. So, as you've said, we seem to have a discrepancy. I think we resolve it like this. The string must also provide a torque on the bob, as the bob turns (and has an angular acceleration) about its centre as the pendulum swings. This implies that the string, if extended into the bob, would not pass through the centre of the bob. This in turn complicates the linear motion equations... – Philip Wood Jan 5 at 18:54