Subject: Re: two wheels with the same mass roll down an incline plane differently

Date: Thu Dec 30 12:06:07 1999
Posted by Ricky J. Sethi
Position: PhD

Hi Joel,

You've just stumbled across a very interesting phenomenon. You're right in suspecting that there might be an equation or two that describes this very strange behaviour. In order to derive these equations, however, we're going to have to make a few assumptions and idealizations. Nature readily lends itself to interpretation but only if the analysis is idealized and straightforward; even the simplest natural phenomenon defies a complete analysis by our meagre abilities.

The first idealization we'll introduce is to assume that the moments of inertia, I, for our two wheels are actually a little simpler than reality. Let's assume that the moment of inertia for the wheel with the mass concentrated near the axle is the same as that for a solid cylinder and that the moment of inertia for the wheel with the mass around the edge is that of a hollow cylinder.

Quick justification: since the moment of inertia of a complex object is simply the addition of the moments of inertia of its simpler, component objects, we can say that, in the first case, the wheel is broken down into an axle component and an outer wheel component where we'll neglect the mass of the outer wheel; similarly, in the latter case, we'll again break the wheel down into an axle and an outer wheel component, only this time we'll neglect the mass of the inner axle. If all of this seemed like some strange dialect of ancient Greek, you might want to review the basics of rotation, torque, and angular velocity at either this or this great site.

Now we're ready to tackle the question of the total energy, both at the top and at the bottom. As you pointed out, both wheels start out with the same Potential Energy. And Conservation of Energy tells us that they must both end up with the same total energy they started out with. If we call the bottom of the plane the 0 point of height, then all the potential energy from the top must have been converted to Kinetic Energy at the bottom. The only tricky part in this analysis is that the wheels have both rotational AND translational kinetic energy at the bottom; let's work out the rotational kinetic energy for each wheel first:

(1) KErot = ½ Iw2

where w is the angular velocity of the wheel and I is the moment of inertia of the wheel. Now we only need to account for the translational kinetic energy of the center of mass of each of the wheels. This is simply:

(2) KEtrans = ½ M v2

where M is the total mass of the wheel. Using these equations, conservation of energy yields:

(3) PEinitial = KEROTATIONAL-final + KETRANSLATIONAL-final

Now, if we use the relation between translational and rotational velocity, namely v=rw since we're assuming rolling without slipping, then this becomes:

(4) Mgh = ½ I w2 + ½ M v2 = ½ I (v/r) 2 + ½ M v2

Since we're really interested in only the final velocity for each of the wheels, this yields:

(5) v = sqrt[2Mgh/(I/r2 + M)]

So the final velocities, for each of the above wheels, assuming an initial height of 1m become:

 Type of Cylinder I of Cylinder Velocity of Cylinder Hollow MR2 Sqrt[10] m/s Solid ½ MR2 Sqrt[13.3] m/s

Interestingly, for our simplified cases, the M and r cancel out of Eqn. 5. So we see that the solid cylinder (i.e., the one with the mass concentrated closer to the center) has a lower moment of inertia and so has a higher final velocity (translational). This might seem counter-intuitive at first but one way to look at it is that less of the initial energy is needed to get the wheel with the lower moment of inertia rolling along at the necessary w = v/r so more energy is available for the translational kinetic energy contribution, hence resulting in a higher final translational velocity.

As for the final stopping distance of each wheel, one way to approach this is to assume that friction with the ground is the only retarding force. And, since the masses of both wheels are equal, the frictional force will be the same for both (as Ffr=mkN). And since force is also the time rate of change of momentum, the same retarding force will take longer to stop the wheel with the higher momentum, i.e., the higher velocity. I hope this explanation didn't end up making things seem even more complicated. This is actually a very good question that's addressed in almost all undergraduate physics texts. A couple of good ones to refer to (and that should be readily available at your local library) are any of the texts by Beiser or Giancoli (personal favourites of mine). And, of course, if you'd like to really get a handle on just about any physics topic, a great source is The Feynman Lectures, notably the first volume for our present problem. Also, if you find any errors or particularly murky portions in this answer, please feel free to drop me a line at rickys@sethi.org and I'd be happy to discuss this further.

Best regards,

Rick.