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How does the mass of a ball effect its kinetic energy when it is rolled down a ramp?
I am doing a simple science experiment by changing the weight of a plastic ball by filling it with heavier and lighter substances and then rolling it down a ramp to see how far it can push a shoe box at the end. How is the gravitational potential energy effecting the amount of kinetic energy the ball has and how does this give heavier object more energy to push the box further
A rolling ball has two sources of KE: its linear velocity v and its angular velocity w. When there is no slippage we can show that v = wr; where v is both the linear velocity (e.g., the velocity at the rotation axis or hub) and the tangential velocity (e.g., the velocity of the "rim" where the mass touches the surface). In math talk we can write the total KE as...
KE = ke(v) + ke(w) = 1/2 mv^2 + 1/2 Iw^2, which is the linear plus angular kinetic energies any rolling mass (like a ball or wheel) will have. Now here's the cool thing.
Inertia I = kmr^2; where k is a number depending on the shape and density distribution of the object with mass m. So ke(w) = 1/2 Iw^2 = 1/2 kmr^2w^2 = 1/2 kmv^2 as v = wr when no slippage is assumed. Which means, ta da...
KE = 1/2 mv^2 + 1/2 kmv^2 = 1/2 mv^2(1 + k) is the total kinetic energy of any and all non slipping rolling objects. If you know the mass and can look up k in the tables found on the Internet [See source.], you can find the total kinetic energy of that ball and just about any other regular shape you can come up with.
Now to answer your question. Assuming the same release height h for each roll.
PE = mgh = 1/2 mv^2(1 + k) = KE, where the potential energy at height h is converted at the bottom of the ramp into kinetic energy. This is a reflection of the conservation of energy law.
As you can see, this is important, the m's cancel out between the PE side and the KE side; so we have 2gh = v^2(1 + k) and v^2 = 2gh/(1 + k). And it makes no difference how big or small the mass is because mass is not a factor in determining v = sqrt(v^2) = sqrt(2gh/(1 + k)).
But, h, the release height is a big time player. For example, take the ratio V/v = sqrt(2gH/(1 + k))//sqrt(2gh/(1 + k)) = sqrt(H/h) so that V = v sqrt(H/h). As you can see if H = 2h, the height is doubled, V = v sqrt(2) and (V/v)^2 = 2; so the kinetic energy when released from H would be KE(H = 2h) = 2 ke(h) or twice that from h.
CAUTION. Having said mass makes no difference, I must amend that assertion a bit. While m mass makes no difference, the distribution of m in a rolling body does.
The distribution of m determines in part the k value for the rolling body. As you are filling the hollow ball with mass, you are changing not only m, but also how that m is distributed inside the ball. And as v = sqrt(2gh/(1 + k)), and KE depends on v^2, any change in the k value between experiments will change the v and consequent KE value.
My point is this... do not be surprised if the kinetic energies do not come out the same for your changing masses. Changing k values will do it. That should be mentioned in your lab report.
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