# What work, and why, is being done on this system of two rotating circular disks

A uniform circular disc with mass $$m_1$$ and radius $$R_1$$ is fixed at its midpoint and rotates at angular velocity $$w_1$$. Another uniform circular disc with mass $$m_2$$ and radius $$r_2$$ is carefully lowered without rotating down to disc 1. Disc 2 is prevented from moving in horizontal direction via a friction free groove.

Initially, the two discs will slide against each other, but after a while the friction between the discs will ensure that they roll without sliding. Calculate the angular velocity of disc 1 as the discs roll without sliding.

Let $$w_1(t)$$ and $$w_2(t)$$ be the angular velocity of body 1 and body 2 respectively at time $$t$$. Let $$t=0$$ be the time when body 2 starts making contact with body 1 and let $$t=a$$ be the time when the discs stop sliding.

Let $$I_{G1}$$ and $$I_{G2}$$ be the moment of inertia for body 1 and body 2 respectively around their center of mass, $$G_1$$ and $$G_2$$.

We are trying to find $$w_1(a)$$ as a function of the known parameters and constants.

At $$t=0$$, it's easy to realize that the total energy for (body 1 + body 2) is pure rotational energy, because their center of mass is not moving and there are no springs attached to this system. But this is true for any time $$t$$, so we let $$T(t)$$ be the total rotational energy for (body 1 + body 2) at a given time $$t$$: $$T(t) = \frac{1}{2}I_{G1}w_1^2(t) + \frac{1}{2}I_{G2}w_2^2(t)$$

At $$t=0$$, body 2 is not rotating so we have $$T(0) = \frac{1}{2}I_{G1}w_1^2(0)$$

At $$t=a$$, body 1 and body 2 are both spinning, so we have $$T(a) = \frac{1}{2}I_{G1}w_1^2(a) + \frac{1}{2}I_{G2}w_2^2(a)$$

Now, using the work-energy-theorem we get $$(1):= U= T(a)-T(0) = \frac{1}{2}I_{G1}w_1^2(a) + \frac{1}{2}I_{G2}w_2^2(a) - \frac{1}{2}I_{G1}w_1^2(0)$$

where $$U$$ is the work done by the forces acting on the system (body 1 + body 2) between the time $$t=0$$ and $$t=a$$. The only forces acting on the system is the reaction forces at $$G_1$$ and $$G_2$$, but these points are not moving so the work done is zero: $$U = 0$$

$$U=0$$ in eq. $$(1)$$ gives us the equation $$(1)':= 0 = \frac{1}{2}I_{G1}w_1^2(a) + \frac{1}{2}I_{G2}w_2^2(a) - \frac{1}{2}I_{G1}w_1^2(0)$$

which is one equation with two unknowns, $$w_1(a)$$ and $$w_2(a)$$. But we know the relationship between these two, since at time $$t=a$$ when the discs stop sliding their tangential velocity must be the same: $$w_1(a)R_1 = w_2(a)R_2$$

solving for $$w_2(a)$$ we get $$w_2(a)= \frac{w_1(a)R_1}{R_2}$$

and putting this into eq. $$(1)'$$: $$0 = \frac{1}{2}I_{G1}w_1^2(a) + \frac{1}{2}\frac{I_{G2}w_1^2(a)R_1^2}{R_2^2} - \frac{1}{2}I_{G1}w_1^2(0)$$

and we solve for $$w_1^2(a)$$: $$w_1^2(a)= \frac{I_{G1}}{I_{G1}+\frac{I_{G2}R_1^2}{R_2^2}}w_1^2(0)$$

The moment of inertia for the bodies are known as $$I_{G1}= \frac{m_1R_1^2}{2}$$ and $$I_{G2}= \frac{m_2R_2^2}{2}$$ respectively, so the final answer can be written as $$(2):= w_1^2(a)= \frac{m_1}{m_1+m_2}w_1^2(0)$$

The correct solution to this problem, however, is $$(3):= w_1^2(a)= \frac{m_1^2}{(m_1+m_2)^2}w_1^2(0)$$

Comparing $$(2)$$ and $$(3)$$, we see that $$(3) < (2)$$ due to the logic that $$x^2 < x$$ when $$x < 1$$. Here is $$x= \frac{m_1}{m_1+m_2} = 1 - \frac{m_2}{m_1+m_2} < 1$$ so $$(3) < (2)$$ must be true. That is, the system should have less rotational energy than my solution implies.

This implies that either

1. There is work being done from forces acting on the system (body 1 + body 2) which I have missed.
2. The total energy is not solely made of rotational energy.
3. Calculation error.

Of these three, 2 and 3 is highly unlikely. This means that 1 is true, work is being done on the system. Is this true, and what forces are doing work on the system?

We only care about the work done from forces acting on the system, not forces inside the system like frictional forces and normal forces between the disks, since these forces inside the system doesn't change the total energy of the system.

• Between $t=0$ and $t=a$, the discs are still sliding, so there is still work done by friction, hence $U\neq 0$. Jul 20, 2018 at 19:55

Option 1 is false. No work is being done on either disk by an external force. Option 2 is correct. Total energy is not only rotational.

Your statement that internal forces do not change the total energy of the system is correct - provided that you include all forms of energy. Friction between the two disks reduces the kinetic energy of the system, but dissipates an equal amount of heat and sound.

Suppose the friction force is $F$. Then the torque on each disk is $FR$ where $R$ is the radius of each disk. These torques act for the same time $t$ and therefore cause a change in angular momentum $FRt=\Delta L=I\Delta \omega$. If the friction force varies then the integral of torque wrt time for each disk is still equal to the total change in angular momentum for that disk.

$F, t$ are the same for both disks, so $\Delta L$ for each disk is in proportion to the radius of the disk.

You need one other relation to find $\Delta \omega$ for each disk. That relation is provided by the no-slip condition between the disks at the point of contact.

• So $$U_{body1}= -\int_{t=0}^{t=a} Fw_1(t)R_1dt$$ $$U_{body2}= -\int_{t=0}^{t=a} (-F)w_2(t)R_2dt$$ Is the work done by the frictional force for body 1 and body 2 respectively? Then adding the two we get the total work done by the friction between the disks: $$U= U_{body1} + U_{body2}= \int_{t=0}^{t=a} (F)w_2(t)R_2dt -\int_{t=0}^{t=a} Fw_1(t)R_1dt$$ where I have changed the sign of the force in the integral for body 2 since two negatives gives a positive. The integral can also be calculated with respect to distance, if we wish. Jul 20, 2018 at 20:54
• Clarification to my first comment: To get the work done I integrated the power, which is the dot product of force and velocity, over the time interval $t=0$ and $t=a$. I used the fact that frictional force is ortogonal to the velocity, and velocity can be written as the product of angular velocity and radius, to write the integrand as scalar: $$FwRdt$$ The direction of the velocity for the bodies contact point is the same, but the direction of the frictional force is opposite, so we get opposite signs for each term in the sum $U=U_{body1} + U_{body2}$ Jul 20, 2018 at 21:04
• That looks right, but it is not practically useful for solving the problem. Jul 20, 2018 at 21:24
• Can the problem be solved using work-energy-theorem? Is the energy lost to heat and sound due to friction accounted for in my expression for $U$? I understand how to solve it using your prescribed method, but I am mainly interested in solving the problem using work-energy theorems. Jul 20, 2018 at 21:27
• Yes it is also possible to solve the problem using the work-energy theorem. See Rolling sphere placed on horizontal surface or Worked Example 8.7 : Rotating Cylinder. The same method can be used in your problem. Jul 20, 2018 at 21:51