Rolling Bowling Ball (Rolling Energy)

[Gotejjeken]

Homework Statement

A 20-cm-diameter, 6.0kg bowling ball rolls along a horizontal floor at 4.0 m/s. The ball then rolls up a 2.0-m-long ramp sloped at 30 degrees up from horizontal. Rolling friction is negligible. What is the ball's speed at the top of the ramp?

Homework Equations

K(final) + U(final) = K(initial) + U(initial)
I = 2/5 * M * R^2
omega = V/R

The Attempt at a Solution

The ball starts out with all kinetic energy so the above equation becomes k(final) + U(final) = k(initial). I set up the following equation:

1/2*I*omega(final)^2 + 1/2*M*v(final)^2 + M * g * h = 1/2*I*omega(initial)^2 + 1/2*M*v(initial)^2

After substituting in the values for I and omega, the radius and mass cancel out and I am left with:

7/10*v(final)^2 = 7/10*v(initial)^2 - g * h

After doing some simple trig on the slope I found that h = 1m, and solving for v(final) I end up with:

v(final) = 1.41 m/s

Since I don't know the answer to the problem and I get credit on an all or nothing basis, I wanted to come here and have someone verify the validity of my work before I hand it in.

 

[_coyote_]

Thanks for any help!Your equations are correct, but your algebra is incorrect. You should have:1/2*I*omega(final)^2 + 1/2*M*v(final)^2 + M * g * h = 1/2*I*omega(initial)^2 + 1/2*M*v(initial)^2After substituting in the values for I and omega, the radius and mass cancel out and you are left with:1/2*M*v(final)^2 + M * g * h = 1/2*M*v(initial)^2After doing some simple trig on the slope you found that h = 1m, and solving for v(final) you end up with:v(final) = sqrt(2 * g * h + v(initial)^2)So v(final) = sqrt(2 * 9.8 * 1 + 4^2) = 5.7 m/s

 

[kerimek]

Your approach and calculations appear to be correct. It is always a good idea to have someone verify your work before submitting it for credit. Additionally, it would be helpful to include units in your calculations for clarity.