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As in the title, our Physics teacher gave us this brain teaser after learning introductory motion. "If I have a dumbbell [with bigger weights than handle] that is rolling down an incline with only its handle rolling (the weights are hanging off the sides of the incline), what would happen to it's speed when it tocuhes the ground and keeps rolling? Does it increase, decrease or stay constant?"

At first I thought the answer was simple, it would decrease in speed as it would expend energy during its impact with the ground (I don't think this is a valid reason though). However, the teacher later gave the explanation that the dumbbell would in fact increase in speed as when the handle was rotating, the weights would be rotating at the same speed and thus when it impacts, the weights which have larger circumference would have the same rotations/min as the handle (which has a smaller circumference) and thus as it is larger, with the same rotation rate, would increase speed. This satisfied me until I heard rumors that the teacher (who enjoys being messing with students) said that the speed remains constant and was lying in class.

I am confused on the real answer, so I would really appreciate some concrete answers.

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Let $\omega_-$ be the angular velocity before the dumbbell hits the ground, and let $\omega_+$ be the angular velocity after the dumbbell hits the ground. Let $v_-$ and $v_+$ be the corresponding linear velocities. Let $r$ be the radius of the handle, and $R$ be the radius of the weights. Let $m$ be the mass of the dumbbell, and $I$ its moment of inertia.

When the dumbbell hits the ground, there's going to be a frictional force on the dumbbell pointing to the right. The net effect of this force is to create an impulse, $\Delta p=F\Delta t$. After this impulse, the dumbbell is again rolling without slipping.

A diagram of the problem

We know from Newton's second law that $$ \Delta p = mv_+-mv_- $$ This force also creates a torque. Since the force acts on the edge of the dumbbell, we know that $\Delta L=R\Delta p$. Thus,

$$ R\Delta p = -I\omega_++I\omega_- $$ (note that the initial angular momentum is $-I\omega_-$, not $I\omega_-$, if we use the convention that positive angular momentum points out of the page. Similarly, the final angular momentum is $-I\omega_+$.)

Now, we know before and after the force acts, the dumbbell rolls without slipping. Thus, $\omega_-=v_-/r$, and $\omega_+=v_+/R$. Plugging these in, we get two equations involving $\Delta p$, $v_+$, and $v_-$.

$$ \Delta p = mv_+-mv_- $$ $$ R\Delta p = -Iv_+/R+Iv_-/r $$ Combining them gives $$ Rm(v_+-v_-) =-Iv_+/R+Iv_-/r $$ Rearranging, we get $$ (Rm+I/R)v_+ =(Rm+I/r)v_- $$ or $$ \frac{v_+}{v_-} = \frac{Rm+I/r}{Rm+I/R} $$ Now, since $r<R$, we have that $Rm+I/r>Rm+I/R$, and so $v_+>v_-$. The dumbbell speeds up.

Of course, you could also predict the dumbbell sped up just from the diagram I drew. The direction of the frictional force clearly points to the right, which means that the center of mass of the dumbbell will accelerate to the right, speeding up. But if you want to know how much it will speed up by, you have to do the math!

Note that if you do the math, you'll find that while $v_+>v_-$, $\omega_+<\omega_-$. So the rotation slowed down--it just didn't slow down enough to make $v_+$ smaller than $v_-$. And note that there WAS energy lost in this collision; it just wasn't enough to make $v_+$ smaller than $v_-$. It decreased the rotational kinetic energy, but still increased the translational kinetic energy. So all of the explanations you got were wrong. It doesn't slow down (although it does lose energy!) and it doesn't rotate at the same speed as it was rotating before (although it have a larger velocity!).

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  • $\begingroup$ Your solution can't be right: if the collision is inelastic, there is clearly a dependance on the angle of the slope. Indeed, for a horizontal slope, $v_-=v_+$, while for a vertical slope, only the rotationnal part of the initial total energy can be converted into translationnal energy, so you can easily find that $v_+ < v_-\frac{R}{r}$, which is contradicting your last equation (except for elastic collision). $\endgroup$
    – Spirine
    Jul 15 '17 at 17:51
  • $\begingroup$ @Sprine I guess I'm being a little sloppy with what I'm calling the collision. I am assuming the moment the dumbbell hits the ground is elastic (equivalently, I'm assuming the sharp discontinuity where the ramp meets the ground is actually smooth, so that the normal force is always perpendicular to the motion). This is a standard assumption in intro physics courses. $\endgroup$ Jul 15 '17 at 18:38
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    $\begingroup$ @Sprine However, I'm not assuming that energy is conserved during the whole process, because it isn't. The dumbbell speeds up because of friction; friction is not a conservative force, so you can't apply conservation of energy to the problem. $\endgroup$ Jul 15 '17 at 18:39
  • $\begingroup$ @Sprine You assumed that the energy of the dumbbell when it was rolling without slipping down the ramp was equivalent to the energy of the dumbbell when it was rolling without slipping on the ground. What you neglected was that between these two processes, the dumbbell was rolling and slipping, thus losing kinetic energy. This is avoided if you consider the entire impulse due to the kinetic friction, since then you can solve the problem without conservation of energy. $\endgroup$ Jul 15 '17 at 18:41
  • $\begingroup$ If the floor is perfectly rough then friction is a conservative force. Think of cogwheels. $\endgroup$ Jun 6 '19 at 4:36
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Let's call $\omega_-$ (resp. $\omega_+$) the angular speed before (resp. after) the dumbbell touches the ground, and $v_-$ (resp. $v_+$) the translation speed before (resp. after) it touches the ground. Let's assume that the collision is elastic, or in other words, the energy is conserved. We then get this equation:

$$ Mv_-^2 + Jw_-^2 = Mv_+^2+J\omega_+^2$$

where $M$ is the mass of the dumbbell and $J$ its moment of inertia.

However, considering that the dumbbell rolls without sleeping before and after the collision, we get the following kinematic constraints:

$$\begin{cases} \omega_- = \frac{v_-}{r}\\ \omega_+ = \frac{v_+}{R} \end{cases}$$

where $r$ (resp. $R$) is the radius of the handle (resp. the weights). Finally, the first equation becomes

$$\frac{v_+^2}{v_-^2} = \frac{M + \frac{J}{r^2}}{M + \frac{J}{R^2}}$$

Now, since $r<R$ for a usual dumbbell, $v_+ > v_-$: it speeds up.

You could wonder why I stated that the collision is elastic. First of all, using a not so inclined plane, you can reach any speed while losing a very low amount of energy during the collision, so it seems reasonable to consider that the energy loss is negligible, compared with the energy transfer between kinetic and rotationnal energy. Secondly, by using different materials for the slope and the dumbbell, different speed or by changing the inclination of the slope, one can achieve to lose almost all of the energy of the dumbbell during the collision: for the problem to have a solution, the collision must be elastic (or at least not too much inelastic).

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  • $\begingroup$ This is not correct. You can't use conservation of energy, because the collision between the dumbbell and the ground is necessarily not elastic. You need to explicitly consider the forces and torques involved. $\endgroup$ Jul 15 '17 at 16:40
  • $\begingroup$ I should be more careful: The instantaneous collision between the dumbbell and the ground can be elastic, but the process by which the dumbbell speeds up involves sliding friction, so kinetic energy is not conserved. $\endgroup$ Jul 15 '17 at 18:45
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The kinetic energy in the dumbell's handle is transferred throughout the whole dumbell, and the dumbell slows down because of it's sides being stopped by the ground, then the energy that is used to stop the dumbell is distributed throughout the whole of it.

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