Simple Kinetic Energy Problem?

Question

A solid 1.0 kg ball of radius 0.50 m sits at the top of a ramp. It rolls down, ending up 10.0 m lower than where it started.

1) What is the TOTAL kinetic energy of the ball at the bottom?

2) How fast is the ball moving at the bottom of the ramp?

3) What is the ball's angular velocity and angular momentum at the bottom?

Now assume there is no friction. How would this change the above answers?

Answer 1

TE = ke(l) + ke(a) = 1/2 mv^2 + (1/5) mv^2 = (7/10) mv^2 is the total energy at the bottom of the ramp; this includes linear KE and angular (rotational) KE; assuming inertia I = 2/5 mr^2 for the solid ball of mass m and radius r.

PE = mgh; where m = 1 kg, g = 9.81 m/sec^2 and h = 10 m, the drop in elevation upon reaching the bottom.

1) From the conservation of energy, we have PE = mgh = KE at the bottom of the ramp.  KE is the total kinetic energy.  You can plug the numbers.

2) From KE = 7/10 mv^2 = mgh; solve for v^2 = (10/7) gh and v = sqrt(10gh/7) where v is both the tangential velocity along the contacted surface of the sphere and the velocity of the center of mass at the hub.

3) W = v/r is the angular velocity; where v you found and r = .5 m.  Angular momentum = I W = 2/5 mr^2 (v/r) = 2/5 mvr; you can plug the numbers.

Ooops missed your tack on assumption.  Without the friction, the sphere would slip, not roll, down the ramp.  It would be like falling through a vacuum, but at an angle rather than straight down.  The PE would still be mgh; so the KE at the bottom would still be KE = mgh.  The difference is that KE = ke(l) linear kinetic energy only.  Thus 1/2 mv^2 = mgh and v^2 = 2gh rather than 10gh/7 when there is rotational KE.