A hollow ball has a mass of 0.310 kg and radius 0.0340 m. The ball is thrown vertically upwards from rest. It rises through a height of 1.40 m then drops down again. When it is released, it is moving upwards at 5.24 m s−1 and rotating at 2.70 revolutions per second.
For the following two situations, explain whether the height to which the ball rises will be less than, greater than, or the same as 1.40 m.
1) The ball is not rotating, but is given the same linear speed when it is released
2) The ball is solid instead of hollow, but has the same mass and radius. The same amount of total work is done to give the ball its linear and rotational motion, and it has the same angular speed.
2. Relevant equations
Ek(lin) = 1/2mv^2
Ek(rot) = 1/2Iw^2
Ep = mgh
I = [itex]\sum[/itex]mr^2
3. The attempt at a solution
1) Since ball is not rotating, Ek(rot) = 0. Ek(lin)=1/2mv^2, so if speed (v) same, Ek(lin) same. Total Ek has decreased.
2) Since ball is solid (mass distributed closer to center) and has same mass, rotational inertia (I) is smaller (I = [itex]\sum[/itex]mr^2). Since Ek(rot) = 1/2Iw^2, less Ek(rot) is gained. Since work done is the same, same total Ek gained. More of this Ek will be as Ek(lin) than Ek(rot) compared to before.
Where I am stuck is about energy conversion between Ep and Ek. At first I thought that since Ep gained = Ek lost, it won’t matter what form that Ek is in (so for Q1, since total Ek is less Ep gained will be less and height is less, and for Q2 since total Ek is same, Ep gained will be same and height is the same).
But when you think about it intuitively, it seems like only Ek(lin) contributes to the gravitational Ep gained, since that’s the actual movement of the ball upwards? The Ek(rot) is just the ball spinning.
The only solution I can think of to that is that the Ek(rot) is converted into some other form of Ep which isn’t gravitational?
If this is the case (only Ek(lin) contributes to Ep(grav)), then for Q1 it will reach the same height and for Q2 it will reach a higher height?