A solid sphere of radius R and mass M is initially at rest in the position on an inclined plane, such that the lowest point of the sphere is a vertical height

*h*above the base of the plane. The sphere is released and rolls down the plane without slipping. The moment of inertia of the sphere about an axis through its center is 2MR^{2}/5.*Express your answers in terms of M, R.**h,*g,*and the angle.*a. Determine the following for the sphere when it is at the bottom of the plane:

i. Its translational kinetic energy

ii. Its rotational kinetic energy

Let's see, let's see...

OK, there's one connection: we can express one kinetic energy (rotational or translational) in terms of the other, through that wonderful pivot:

This only works for us when we're in that enviable "rolling without slipping" condition. This is a cool connection, and super-useful, but it's old hat for us at this point.

What about the other kinetic energy? Well, we could work the pivot the other way, or... think about it a little! It started with some energy:

*mgh*. We just solved to see that some fraction of that (about 28%) is invested in making the sphere roll. Where's the other five-sevenths?It didn't turn into pumpkin pie, I'll tell you that. This is where the kids were champs today. Below, you see them solving for the rotational KE - the one that they solved for first. After that?

**Boom**. Answer.

No prevarication, no lengthy algebra. "Duh, it's the rest." That's good physics thinking.

If you forget that, then you're just pushing letters.

**Never forget that the math is just another way to write down reasoning.**If you forget that, then you're just pushing letters.

Next part:

b. Determine the following for the sphere when it is on the plane.

i. Its linear acceleration

ii. The magnitude of the frictional force acting on it

There was a bit more algebra here, but... did there have to be?

...

Think about it for a second. Cheat a bit: look at the answers, and think again.

We already know that the kinetic energies are proportional to the height change from top to bottom, and we also know that gravity causes the ball to move (put the rotational axis on the point of contact between the ramp and the ball if you don't believe me) and that static friction causes it to rotate.

Here are the hardest links in the chain:

- Each kinetic energy increases at a constant rate (with respect to distance), because the forces doing the work are constant and the angle between the force and displacement is constant.

- Because the kinetic energies increase at a constant rate, the energy that ends up being 2.5 times the other (

*5/7 mgh*versus

*2/7 mgh*) must increase at 2.5 times the rate. That is, the net force down the plane is 2.5 times the friction force.

- A tiny bit of algebra or (better) some thinking about the fractions that the KE split into will reveal the forces (net force and the friction force). Each is a fraction of the weight's component down the hill:

*mg sin(theta)*.

No, it's still not done. Yes, this is too much for a problem that you only get 15 minutes for on a stress-filled AP exam day. Boo, College Board.

The solid sphere is replaced by a hollow sphere of identical radius R and mass M. The hollow sphere, which is released from the same location as the solid sphere, rolls down the incline without slipping.

c. What is the total kinetic energy of the hollow sphere at the bottom of the plane?

d. State whether the rotational kinetic energy of the hollow sphere is greater than, less than, or equal to that of the solid sphere at the bottom of the plane. Justify your answer. I'm guessing that c. is just there to clue us in if we haven't been thinking all along (instead, being mindless algebra drones). From a test pedagogy standpoint, I don't like this - the AP exam tends to lead students by the nose, rarely requiring them to really put together any sort of chain of reasoning. It makes the problems easier to score consistently, though, which is (I'm sure) the purpose.

d. is a good one, though. There are a couple of steps in the chain, which is a rare treat for the AP.

- The mass in a hollow sphere is, on average, further from the axis than in the solid sphere.

- The rotational inertia of the hollow sphere is greater than that of the solid sphere.

- If the two balls were rolling at the same speed, the hollow one would have more rotational KE, but the same translational KE.

- The two balls must, however have the same

*total*KE at the bottom (

*mgh*!).

- The hollow ball must therefore have a great fraction of its total KE as rotational KE (and less as translational, so it's going slower).

Try a race! Bigger and more hollow will slow down a ball; racquetballs lose to pool balls and marbles!

This is a great post, making clear lots of connections. The energy approach is always cleaner, it seems, but let's not forget who's king (I just had to throw that in, of course).

ReplyDeleteThis semester I've been working with students on analyzing a billiards masse shot. It was the first time they'd ever had to really deal with rolling WITH slipping. They've done a great job, with their model matching their data very well. What's cool is that the friction works towards rolling without slipping every time.