Question

A bowling ball of mass 1.5kg and radius 0.3m rolls with linear speed 2.5m/s along the ground toward a small incline. Assume rolling without slipping!

1.1) What is the moment of inertia of the bowling ball?

1.2) What is the total kinetic energy of the bowling ball?

2.3) If the bowling ball encounters a hill, to what height can it roll up the hill before stopping? (use conservation of energy)

2.4) Will a solid cylinder with the same mass and radius roll to the same height, or smaller height than the sphere and why?

2.5) Does the bowling ball have top spin or back spin? If friction exists between the ball and the ground, will the torque cause the rotation to slow down or speed up?

Thank you!

Answer 1

m = 1.5 kg

r = 0.3 m

v = 2.5 m/s

1.1) Moment of interia = $\frac{2mr^2}{5} = \frac{2*1.5*0.3^2}{5} = 0.054 kg.m^2$

1.2) $\omega =\frac{v}{r} = \frac{2.5}{0.3} rad/s$

translational KE = 0.5mv² = 0.5*1.5*2.5² = 4.6875 J

rotational KE = 0.5Iw² = 0.5*0.054*(2.5/0.3)² = 1.875 J

So total KE = 4.6875+1.875 = 6.5625 J

2.3 by conservation of energy

mgh = total KE

1.5*9.81*h = 6.5625

=> h = 0.44597 m

Height it will roll to = 0.44597 m

2.4) moment of interia of a solid cylinder = $\frac{mr^2}{2}$

So the moment of inertia for cylinder is more than that for sphere. Therefore it will have more rotational Kinetic energy, hence more total energy. As a result, solid cylinder will roll to greater height than that of sphere.

2.5) The bowling ball will have top spin while going up the incline.

The torque will lead to slow down of the rotation.