Question

After you pick up a spare, your bowling ball rolls without slipping back toward the ball rack with a linear speed of vi=2.62 m/s. To reach the rack, the ball rolls up a ramp that rises through a vertical distance of h=0.47m.

Part A

What is the linear speed of the ball when it reaches the top of the ramp?

Part B

If the radius of the ball were increased, would the speed found in part A increase, decrease, or stay the same?

Answer 1

Using conservation of of energy

Total energy of rotating ball before climbing ramp = total energy of rotating ball after climbing

0.5 mu^2 + 0.5 Iwo^2 = 0.5 mv^2 + mgh + 0.5 I w^2 ... (i)

Moment of inertia of sphere

I = (2/5) mr^2

wo = u/r

w = v/r

Putting in... (i)

0.5m* 2.62^2 + 0.5 (0.4 mr^2) ( 2.62/r)^2 = 0.5 mv^2 + m*9.8*0.47 + 0.5(0.4 mr^2) (v/r) ^2

3.432 + 1.373 - 4.606 = v^2( 0.5 + 0.5*0.4)

v = 0.533 m/s

======

As the final velocity is independent of radius, so it remains same.

=====

Goodluck