Find final rotational kinetic energy without knowing radius?

[matxjos]

Homework Statement

A 0.125kg basketball is rolling w/out slipping on a horizontal table at 4.50 m/s when it rolls off the edge and it falls to the floor, 1.10 m below. What is the rotational kinetic energy of the ball right before hitting floor?

Homework Equations

KE rot: .5 I w^2
KE translational: .5 m v^2
PE: mgh

The Attempt at a Solution

Initial: I figured out KE translational and PE for the initial conditions on the horizontal surface, these are easy.

I do not know how to get K rotational energy!

Final: I used vertical constant acceleration equations to find the final velocity. Then I got the kinetic energy right before hitting the floor. Potential energy is 0.

No idea how to get K rot energy!

I've been working on this problem forever (all morning). The answer to the problem is apparently 0.506J.

HOW?!

 

[gneill]

Once the ball leaves the table there's nothing to provide torque to change its rotation, so ignore everything that happens after it leaves the table.

Write an expression for the rotation rate of the ball as it rolls. That's the $\omega$ for your rotational kinetic energy expression. What expression will you use for the moment of inertia?

(P.S. I feel that the problem author used an incorrect moment of inertia; a basketball is a thin spherical shell, not a solid sphere).

 

[matxjos]

Once the ball leaves the table there's nothing to provide torque to change its rotation, so ignore everything that happens after it leaves the table.

Write an expression for the rotation rate of the ball as it rolls. That's the $\omega$ for your rotational kinetic energy expression. What expression will you use for the moment of inertia?

(P.S. I feel that the problem author used an incorrect moment of inertia; a basketball is a thin spherical shell, not a solid sphere).

Well, a solid sphere's moment of inertia is 2/5MR^2. Is this the expression I use? I still don't know the radius...

I'm not even sure what you mean to write an expression for the rotation rate. Like in rad/s? I know how to get that from rev/min, or even m/s, but I don't know the radius. :(

Thanks for looking at this problem!

 

[matxjos]

Once the ball leaves the table there's nothing to provide torque to change its rotation, so ignore everything that happens after it leaves the table.

Wait, if there is no rotation after the ball leaves the table, does that mean the rotational energy is zero? Or just the same as the initial KE rot?

 

[haruspex]

Well, a solid sphere's moment of inertia is 2/5MR^2.

Solid basketballs?! That's serious training.

I still don't know the radius...

You'll find you don't need to. Just put in an unknown for it and it should cancel out later.

I'm not even sure what you mean to write an expression for the rotation rate. Like in rad/s?

Yes. Create another unknown for this, and an equation relating it to linear speed and radius.

if there is no rotation after the ball leaves the table

gneill said there'd be no change in rotation rate.

 

[matxjos]

Solid basketballs?! That's serious training.

You'll find you don't need to. Just put in an unknown for it and it should cancel out later.

Yes. Create another unknown for this, and an equation relating it to linear speed and radius.

gneill said there'd be no change in rotation rate.

Wow, thanks, I got the answer! I didn't realize things cancel out...physics is most confusing when that has to happen. But very interesting.

Thanks haruspex and gneill!