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Suppose we have a solid disk of mass $M$ and radius $R$ that is spinning at an angular velocity of $\omega_0$ about an axis going out its cm. It is brought to touch a stationary disk of mass $m$ and radius $r$. How would you find the final angular velocities in this scenario?

This is what I was thinking might work:

$$ \frac{1}{2} \omega_0 M R^2 = \frac{1}{2} M R^2\frac{v_\text{final}}{R} - \frac{1}{2}mr^2 \frac{v_{final}}{r}$$ $$v_\text{final} = \frac{\omega_0 MR^2}{M R - m r}\,.$$ I solved it for the tangential velocity there and would just use $v=\omega r$ to find the actual angular velocity of each disk.

Would this approach work? What is the best way to solve this problem? Conservation of angular momentum or dynamics?

If the two disks slip at first when next to each other, do we need to adjust for that or can we ignore that stage and look at them when they are not slipping and at final speeds?

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closed as off-topic by John Rennie, Kyle Kanos, user36790, ACuriousMind, Daniel Griscom Jan 27 '16 at 1:26

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  • $\begingroup$ can you explain your equations step by step. They are not apparent to me.You should solve the problem by using conservation of angular momentum. $\endgroup$ – Bruce Lee Jan 26 '16 at 5:04
  • $\begingroup$ The first equation is how I tried to set up conservation of angular momentum. The moment of inertia for a spinning disk is (1/2)MR^2 so Iw = (1/2)MR^2 w_initial on the left hand side. The right hand side is the new angular momentum. I expressed it in terms of the tangential velocity because that should be the same and used the fact that v=wr when there is no slippage. $\endgroup$ – Paula Jan 26 '16 at 5:09
  • $\begingroup$ then the equation you wrote is wrong as there is a plus sign instead of the minus you wrote. $\endgroup$ – Bruce Lee Jan 26 '16 at 5:16
  • $\begingroup$ Would the tangential velocities be opposite so the angular velocities would subtract? $\endgroup$ – Paula Jan 26 '16 at 5:18
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    $\begingroup$ Since they are right next to each other like this: OO. One will spin clockwise and the other counterclockwise? $\endgroup$ – Paula Jan 26 '16 at 5:25
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When you drop a stationary disc onto a rotating one there must be a time when there is relative motion between the discs as you cannot have an infinite acceleration.

If there is no friction then nothing much happens and the spinning disc carries on spinning and the other disc just sits still on top of it.

To get an interaction between the discs you need frictional forces. As soon as you have frictional force between two surfaces and relative movement between them you get heat generation which in this case means that the kinetic energy of the system (both discs) decreases. So you cannot use conservation of kinetic energy to solve this problem.

Eventually there is no relative movement between the discs and the rotate at the same rate.

If there are no external torques acting then you can use the conservation of angular momentum as mentioned above.

$I_1 \omega_i = (I_1 + I_2) \omega_f$

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yes i think your assumption is correct based on the fact that in equilibrium the discs shouldn't slip on each other. but then you need to change your equation for angular momentum. use parallel axis theorem to write down the moment of inertia for the stationary disc to get the answer in the equation. since you have chosen your axis to be passing through the rotating disc, so in order to account for the motion of the stationary disc, you need to use the parallel axis theorem by shifting the axis through $R + r$.

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  • $\begingroup$ This is wrong. The discs will always slip at first because nothing can accelerate at infinite rate. Slippage means friction and friction means CoE does not apply. $\endgroup$ – Gert Jan 26 '16 at 16:04
  • $\begingroup$ @Gert i didn't dispute that anywhere.. i didn't also say anywhere that CoE is needed. please check the comments section too where at first I misinterpreted the question, but all the way along I was using conservation of angular momentum only. $\endgroup$ – Bruce Lee Jan 26 '16 at 20:36
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Rotating disc.

In order for the spinning disc to set the stationary one in motion, friction forces need to act between the contacting surfaces:

Friction forces.

That requires a Normal force $F_N$ to act between them. Using the kinetic friction coefficient $\mu_k$ we can then state:

$$F_F=\mu_k F_N$$

The friction force on the $m,r$ disc causes angular acceleration $\alpha=\frac{d \omega}{dt}$:

$$F_F r=I_1 \alpha$$

So:

$$\omega=\frac{F_F}{I_1}t$$

But the friction also causes a decelerating torque on the $M,R$ disc:

$$F_F R=-I_2 \alpha$$

So that:

$$\omega=\omega_0-\frac{F_F}{I_2}t$$

Friction stops when both spin at the same rate:

$$\frac{F_Fr}{I_1}t=\omega_0-\frac{F_FR}{I_2}t$$

From this the time $t$ and with substitution also the final $\omega$ can be calculated.

In the special case $I_1=I_2$, $R=r$ then the final angular speed is:

$$\omega=\omega_0 \frac{I_1}{2F_FR}$$

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Yes you have to take the frictional force F into account so this exerts a negative impulse moment RFdt on the right hand disk with radius R as well as rFdt on the left hand disk with radius r. So the delta of the rotational moments have the ratio r/R. Assuming the left hand disk spins with rim velocity v0 before impact and after impact they have equal rim velocities corresponding to angular velocities v/R and -v/r respectively, we thus have MRv/2 - MRv0/2 = -mrv/2 giving the correct answer v = v0/(1+mr/(MR)). Thus for equal disks each will obtain half the angular velocity of the original.

I might add that to avoid considering frictional heat loss one could imagine each disk as a gear with perfectly fitting teeths that match the final speed. The first teeth that interact can then be viewed as colliding elastically immediately transferring momentum to the stationary disk without heat loss.

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