A bowling ball encounters a 0.76 m vertical rise on the way back to the ball rack, as the drawing illustrates. Ignore frictional losses and assume that the mass of the ball is distributed uniformly. If the translational speed of the ball is 5.40 m/s at the bottom of the rise, find the translational speed at the top.
Do you use the equation for total mechanical energy and if so how do you get the mass and radius?
Physics - Count Iblis, Monday, November 3, 2008 at 10:20am
What happens is that in the answer the mass and the radius drops out.
The total kinetic energy is:
Ekin = 1/2 m v^2 + 1/2 I omega^2
If the ball rolls without slipping, then:
omega = v/R
The moment of intertial of the ball is:
I = 2/5 m R^2
So, the kinetic energy is:
Ekin = 1/2 m v^2 + 1/5 m v^2 =
7/10 m v^2
The potential energy is mgh, so the total energy of the ball is at height h and velocity v is
E(h,v) = m g h + 7/10 m v^2 =
m [g h + 7/10 v^2]
The total energy is conserved, so you can find the velocity at the top by solving:
E(h=0.76 meter, v ) = E(0, v = 5.40 m/)
The unknown mass m drops out.