Exam question, rotational motion.

[Kelschul]

Homework Statement

Two identical soccer balls are released from rest from the top of ramp consisting of a straight section connected to a circular section having the shape shown at right (height of ramp from where ball is released to bottom of curve is 4.0m. The radius of the curve is 1.5m). The end of the circular section of track is vertical. One ball slides down the ramp, while the other ball rolls without slipping. A soccer ball can be considered a thin-walled spherical shell. What is the speed of each ball at the bottom of the curve?

Homework Equations

I=(2/3)MR(^2)

The Attempt at a Solution

I have to solve one ball using rotational motion concepts, and one just regularly, right?
I think this is what I have to do, but without having mass, or any other variable I just can't figure out what equation to start with.
The only equation I can find that has height in it, which seems the only logical place to start for me is the equation for work done by non-conservative forces. I started to use this equation, but then I realized it still wouldn't help me find the speed at the bottom of the hill because I don't know mass, velocity, or the Wnc.
I'm scanning my notes like crazy trying to find an equation!

Maybe I just found something...
PE=mgh
Initial Pe is m(9.8)(4.0), Final is 0 because there is no height...
39.2m=.5mvf(^2)-0 (no initial velocity)
vf=8.9m/s for the sliding ball? right?

 

[Kurdt]

Its definitely to do with energy considerations. One ball will have all its energy transferred to kinetic energy while the other will have a sum of kinetic energy and rotational energy. I think the last part of your attempt has touched on what you need to do and also shows that the masses cancel. Do you know the equation for rotational energy? It is very similar to translational kinetic energy.

 

[m2003]

Yes, you are correct. The equation for potential energy, PE = mgh, can be used to find the final velocity of the sliding ball. Since the ball starts from rest, its initial velocity is 0 and its final velocity at the bottom of the ramp can be found using the equation you have mentioned, vf = √(2gh).

For the rolling ball, we can use the concept of conservation of energy. At the top of the ramp, the ball has potential energy, which gets converted into kinetic energy as it rolls down the ramp. At the bottom of the ramp, the kinetic energy is equal to the rotational kinetic energy of the ball. The equation for rotational kinetic energy is given by KE = (1/2)Iω^2, where I is the moment of inertia of the ball and ω is the angular velocity. Since the ball is rolling without slipping, we can use the relationship between linear and angular velocity, v = ωr, where r is the radius of the ball.

Combining these equations, we get KE = (1/2)(2/3)MR^2 (v/r)^2, where M is the mass of the ball and R is the radius of the ball. Equating this to the potential energy at the top of the ramp, we get mgh = (1/2)(2/3)MR^2 (v/r)^2. Solving for v, we get v = √(3gh/2).

Plugging in the values for g, h, and R, we get v = √(3*9.8*4/2*1.5) = 6.3 m/s.

So, the final velocities of the sliding and rolling ball at the bottom of the ramp are 8.9 m/s and 6.3 m/s, respectively.