a) What was its total initial kinetic energy? b) How many revolutions does it make before coming to rest? c) What torque did the ground apply to the ball?

Question

A ball begins rolling with a speed, respective to the ground, of 10 m/s. The ball is uniformly solid, has a mass of 1.5 kg and a radius of 0.05 m.  It rolls without slipping over the horizontal ground for a linear distance for 25 m before coming to rest. Answer all questions and show all work.  

Answer 1

This is a great physics review problem of applying ordinary straight-line or 1-dimensional translational displacement problem to rotational problem.  Here, we shall make good use of equivalences between straight-line distances and rotational angles, straight-line acceleration and rotational angular acceleration, and rotational kinetic energy.

Remember that for given initial and final measured velocities (v_0 and v_f) and the linear distance x traversed on frictionless surface, we can find its acceleration a from solving v_f² - v_0² = 2 a x.

Similarly, given initial and final measured angular velocities (ω_0 and ω_f) and the total angle θ rotated on frictionless surface, we can use similar formula ω_f² - ω_0² = 2 α θ to find its angular acceleration α, which is important in finding torque of rotating sphere.

The torque is needed here because torque dissipates rotational kinetic energy. Torque of any rotating object about its center of mass is its moment of inertia 2MR²/5 times angular acceleration α. Rotational kinetic energy of rotating object is M v²/2 + I ω²/2.

Now we have all the formulas needed to solve this problem.

1) solid sphere's moment of inertia is I = 2MR²/5 = 2 * 1.5 * 0.05² / 5 (kg m²).

2) The total angle rotated before final stop after traveling 25 (m) is θ = 25/(2π*0.05) = 250/π (rad).

3) The initial rotational velocity ω_0 = v_0/R = 10/0.05 = 200 (rad/s).

4) The angular deceleration to stop rotation is α = - ω_0²/(2θ) = - 200²/500/π = - 80π (rad/s²).

5) The initial rotational kinetic energy is M v_0²/2 + I ω_0²/2 = 1.5 * 10²/2 + 0.6 * (0.05*200)² = 75 + 0.6 * 10² = 135 (J).

6) The number of rotations before the sphere coming to a stop is θ/(2π) = 125/π². A related question is the time it took for the rolling sphere to come to a stop: t = - ω_0/α = 200/(80π) = 2.5/π (s).

7) The magnitude of the torque of rotating sphere is I α = 0.6 * 0.05² * 80π = 0.12π (N m).

In summary, understand well translational and rotational correspondences allows you to solve problems like this almost effortlessly. Also, be sure to append units of each calculation for completeness.