Comparing Rotational and Linear Motion - block vs. disk vs. ring

[AstraeaSophia]

Comparing Rotational and Linear Motion -- block vs. disk vs. ring

Homework Statement

A disk and a ring, both of mass M and radius R, are placed atop an incline and allowed to roll down. A block, also of mass M, is placed atop the same frictionless incline and allowed to slide down. How do the velocities of the disk, ring, and block compare at the bottom of the incline? Which one reaches the bottom first?

Homework Equations

KE_rotational = .5I$$\omega$$²
I = kMR²
$$\omega$$ = v/r

The Attempt at a Solution

We were given certain steps to accomplish before getting to the last question, which is the one I've listed here and need help solving. So far, we've come up with expressions for the kinetic energy of the disk and the ring in terms of mass, radius and translational velocity. This is what I have:

KE_disk = .5I$$\omega$$²
= .5(.5MR²)$$\omega$$²
= (.25MR²)(v²/R²)
= .25mv_disk²

KE_ring = .5I$$\omega$$²
= .5(MR²)$$\omega$$²
= (.5MR²)(v²/R²)
= .5mv_ring²

KE_block = .5Mv_block²

This is where I get stuck. I know that the disk will reach the bottom of the incline before the ring by using U_i = -K_f and solving for the velocity value on the KE side of the equation. But I don't know how rotational relates to linear in this case. It seems too easy to say that the final velocities of the ring and the block will be the same just because their KE formulas are identical. I feel like I'm missing something. Then again, I could be overthinking it. Or maybe the expressions I derived for the KE of the disk and the ring are wrong.

Can anyone help?

P.S. -- don't know why the $$\omega$$s float up near the exponent like that, but they're not powers. Just regular ol' omegas. :-)

 

[kuruman]

This problem is artificial in the sense that, if the incline is frictionless, then no rolling and only slipping will occur for all objects. We will let that pass and pretend that the disk and the ring roll while the block slips without friction.

The important thing to realize is that when the rolling object reaches the bottom, it has two kind of kinetic energy:

1. Kinetic energy of the center of mass = 0.5*M*V_CM²

2. Kinetic energy about the center of mass = 0.5*I*ω²

The two speeds are related by V_CM=ωR because the objects roll instantaneously about their point of contact.

Therefore, the initial potential energy at the top is equal to the sum of these two terms at the bottom.

 

[AstraeaSophia]

You're right about the problem being artificial. I guess the idea is just that friction doesn't affect the movement of any objects sliding or rolling down the hill. Problems out of introductory physics textbooks are streamlined for easy solving. :-)

I figured out the problem's finer details. I will say first what should have been painfully obvious -- the kinetic energies of both the disk and the ring at the bottom of the hill are the same. However, they will be traveling at different velocities when they reach the bottom. I wanted to post the result here for anyone else who might need a step in the right direction. Very quickly, though, and without much derivation so I don't give away anything huge.

The disk and the ring are both rolling down the hill, so they are subject to both K_rotational and K_linear. If you have been given the formulas for both of these values, the kinetic energy of your disk and your ring will look like this:

K_{total, ring} = .5mv² + .5I$$\omega$$²
--- knowing the equations for $$\omega$$ and I_ring allow you to substitute those expressions in and obtain the total kinetic value of the ring in terms of its mass and velocity

K_{total, disk} = the same as above, except that I_ring has now become I_disk, and its formula consequently changes.

K_box = the normal formula, .5mv².

Once you insert the necessary substitutions and factor certain things out, you can find expressions for what the kinetic energy will be for every object at the bottom of the incline and compare these expressions. There is an obvious 1st, 2nd, and 3rd order as to which object will cross the finish line first once you solve for the velocities in each resultant equation.

Thanks for the help!

-- A.S.

P.S. -- Still don't know why my omegas are trying to fly away.

 

[kuruman]

Your omegas fly away because you are mixing regular type with LateX. You can
(a) Go to https://www.physicsforums.com/blog.php?b=347 and cut and paste characters from there in regular type, or
(b) Do the entire equation in LateX.