Oscillating disc on flat table Suppose we have a uniform disc of mass $2m$ and radius $R$, and we place a small piece of putty of mass $m$ onto the disc at a position very close to the rim.
Now let's say the disc with putty is now set vertically on edge on a horizontal table where it can roll on its rim without slipping. How do we find the period of oscillation of the disc-putty system? Specifically, I'm having trouble finding an expression for the frictional force.

Attempt: If I equate the torques, I get $I\ddot{\theta} = -mg\theta +F_{n}R$ where $I$ is the moment of inertia (MOI) of the disc about the center of mass (COM) and $F_n$ is the frictional force (and I've used the small angle approximation).
Question: I need to find an expression for $F_{n}$. The only way I can think of doing this is using the constraint $V_{cm} = \omega R$, and using $F_{net}=2m a_{cm}$. However, since the only force acting horizontally is the frictional force, this would mean that the disc isn't oscillating $-$ when in fact it should oscillate due to the opposing torque from friction. What is this analysis missing?
If anyone could clear this up, it would be much appreciated

 A: Thanks to the OP for this great question.
$$\underline{\textit{Notation and preliminaries}}$$
We refer to the disc-putty system of particles, which constitute a rigid body as the body. Let us denote

*

*the radius of the disc by $R$, the mass of the disc and putty by $M=2m$ and $m$,

*the moment of inertia (MOI) of the disc and the MOI of the body referred to the axis passing through the the center of mass (COM) of the disc by $I:=\frac{1}{2}MR^2$ and $\bar{I}:=I+mR^2=2mR^2=MR^2$,

*the signed magnitudes or component values of the linear velocity (measured positive in the direction towards the right of the screen) of the COM of the disc by $v$, the linear acceleration  of COM of the disc by $a$,

*the signed magnitudes or component values of the horizontal components of the linear tangential velocity and acceleration of the putty w.r.t. the ground by $v_p$ and $a_p$,

*the signed magnitudes or component values of the angular displacement (measured positive in the direction out of the screen), the angular velocity and the angular acceleration of the body w.r.t. the ground or inertial reference frame by $\theta$, $\dot{\theta}$ and $\alpha:=\ddot{\theta}$, wherein the displacement $\theta = 0$, corresponds to the configuration in which the putty is located vertically at the bottom of the disc,

*and finally, the force of friction (measured positive in the direction towards the right of the screen) by $F_f$.

We assume that the disc is rolling without slipping on the surface on which it is placed.
$$\underline{\textit{Analysis}}$$

*

*We assume that the putty is located at or close to the bottom of the disc initially.

*The kinematic analysis yields $v=-\dot{\theta} R$, $a = -\alpha R$ (be careful about the signs of the variables expressing these vector component values, especially in rolling without slipping phenomena, when using graphical methods to obtain equations!) where the signs correspond to the choice of the direction of the axes of the coordinate system used, $v_p=v+\dot{\theta} R \cos\theta \approx v+\dot{\theta} R \approx 0$ and $a_p=a + \alpha R \cos\theta - \dot{\theta}^2 R \sin\theta \approx a + \alpha R - \dot{\theta}^2 \theta R\approx 0$.

*The dynamical analysis uses the non-inertial reference frame attached to the body at the center of the disc (a related learning value of this post is that we need to consider pseudo or inertial forces not exclusively because the reference frame attached to the center of mass of the disc is non-inertial but inclusively along with the reason that in the analysis we are calculating torques referred to a point which is not the center of mass of the body being analyzed; also note in this context that attitudinal acceleration of the reference frame does not result in pseudo torques although there is a 'Coriolis torque' or Coriolis-effect-torque due to the rotation of the frame, even if it is not accelerating attitudinally; see the 'Learning value' section for more details) so that the net torque involves the torque due to the relevant pseudo force component $-(M+m)a\cos\theta=3m\ddot{\theta}R\cos\theta$ (reference frame is attached to point accelerating at $a$), which is calculated as $\frac{R}{3}\cdot3\ddot{\theta}R=mR^2\ddot{\theta}\cos\theta$ (the rigorous way to obtain this is to represent all vectors in the unit vector  $\hat{i},\hat{j},\hat{k}$ system of the ground or inertial reference frame, in this case as $\frac{R}{3}(\sin\theta\hat{i}-\cos\theta\hat{j})\times-(M+m)a\cos\theta\hat{i}\approx mR^2\ddot{\theta}\hat{k}$ on applying $a=-\alpha R=-\ddot{\theta}R$ and $\cos\theta\approx1$) where we note that the center of mass is located below the center of mass of the disc at a distance $\frac{R}{3}$.

*This results in the equations of motion $\bar{I}\alpha=-mgR\sin\theta+F_f R+mR^2\ddot{\theta}\approx-mgR\theta+F_f R+mR^2\ddot{\theta}$ and $F_f=Ma+m \dot{\theta}^2 R \sin{\theta} \approx Ma$, yielding the attitude dynamics $$\ddot{\theta}=-{\Huge(}\dfrac{mgR}{\bar{I}+MR^2-mR^2}{\Huge)}\theta=-{\Huge(}\dfrac{mgR}{3mR^2}{\Huge)}\theta,$$ on recalling and applying the (a learning value of this post) important kinematic relationship $a=-\ddot{\theta}R$ (note that this reads $a=\ddot{\theta}R$ if the angular displacement is measured positive into the screen with corresponding changes in signs in the preceding scalar equations, of course leading to the identical final result above) to obtain the above expression.

*Indeed, the analysis can be made much simpler by incorporating the assumption of small angular motion by explicitly assuming that the putty has negligible acceleration $a_p$ to readily arrive at the equation above. However, the complete analysis presented here gives a comprehensive picture of the actual complexity and precise knowledge of the application of approximating assumptions.

*Further, the equation above describes a simple harmonic motion (SHM) with the angular frequency (${}^c/s$) given by $\omega_\text{oscillation}=\sqrt{\dfrac{mgR}{3mR^2}}$ and the time period ($s$) given by $\dfrac{2\pi}{\omega_\text{oscillation}}$.


Note that the direction of the friction force is shown correctly in the figure in the context of the depicted state of the angular and linear acceleration of the rigid body. However, relying on variable names such as $F_f$ can be misleading and a rigorous utilization of notation is achieved by stating the complete description as follows. The force of friction is denoted by $F_f \hat{i}$, where the unit vector $\hat{i}$ is directed towards the right of the screen and is parallel to the ground shown in the figure.

$$\underline{\textit{Learning value}}$$

*

*It is important to carefully consider the signs of variables when using graphical methods to obtain dynamical equations of motion, especially in phenomena involving rolling without slipping and the rigorous way to do this is to write all vectors as components of well defined unit vectors $\hat{i},\hat{j},\hat{k}$. Further, the relationship between variables and their signs depend on the choice of the coordinate system used, that is, the system comprised by $\hat{i},\hat{j},\hat{k}$. Overall, relying on writing scalar equations to execute dynamical analysis is a bad idea!

*If the attitude dynamics of a rigid body are being calculated using a non-inertial reference frame which does not use the center of mass of the rigid body as the reference point, then the Euler equation describing the dynamics must include the moment of forces due to the pseudo force acting at the center of mass (see the book Introduction to Spaceflight Dynamics by Thomson). We note that the torque due to the pseudo force vanishes if the torques are calculated using the center of mass of the body as the reference point regardless of the reference frame being inertial or non-inertial.
A: Ignoring friction, you can see the off-center mass will create a torque that causes rotational acceleration and the disk will spin.
As the disk spins, the mass will pass the center and this time the torque will accelerate in the other direction, slowing the disk and causing oscillation.
Adding in friction here couples the rotation of the disk to the translation of the disk.  We don't need an exact value for friction as long as it is sufficient to avoid slipping.  We can equate it in the rotation and translation equations so that it disappears.
$$I\alpha = -mg\theta R + F_{\text f}R$$
$$F_{\text f} = 2ma$$
$$I\alpha + mg\theta R = 2ma$$
Then by the geometry of disc, you can relate $\alpha$ and $a$ and solve for them directly.  As the rotation takes the weight past the center, the acceleration will change sign and the disk speed will decrease, creating an oscillation.
A: If you use the fact that the energy will remain constant over time, that is, $dE/dT = 0$. Then we can find the equations of motion which directly give $I\ddot{\theta}$ (rotational KE of disc) + $mR^2\ddot{\theta}$ (translational KE of disc) + $1/2mR^2\ddot{\theta}$ (KE of putty) + $mgRsin(\theta)=0$  where $I = mR^2$. Therefore - we find that the period of the oscillating disc on the table is simply $\omega = \sqrt{g/3R}$.
