Rolling or skidding? A sphere that never starts to roll
A solid steel sphere of radius 2.0 cm and mass 50 g is released from rest at the top of a board 2.00 m long inclined at 30 degrees to the horizontal. The board carries a thin uniform coating whose coefficients of static and kinetic friction with steel are 0.12 and 0.10 respectively. From a = 5g sin(30)/7 = 3.5 m/s^2 a student predicts a descent time of 1.07 s and adds that 'the sphere rolls, because it is round and the surface is not slippery'. A phone video gives 0.99 s, and frame-by-frame inspection shows the sphere sliding throughout - a painted stripe on its surface turns far too slowly for rolling. The student is puzzled: friction is dissipative, so slipping ought to make the descent slower, not faster. He also wonders whether the sphere will eventually 'catch up', start rolling, and finish the run without slipping.
1. Test the pure-rolling assumption quantitatively. What minimum coefficient of static friction does rolling require at this angle, and what does that imply for this run?
Read the worked answer
For a body rolling without slipping down an incline, the friction needed is
f = (I/R^2) a, a = g sin(theta)/(1 + I/mR^2).
For a solid sphere I/mR^2 = 2/5, so a = (5/7) g sin(theta) and
f = (2/5) m a = (2/7) m g sin(theta).
Normal reaction N = m g cos(theta), so
mu_min = f/N = (2/7) tan(theta) = 0.2857 x 0.5774 = 0.165.
The available static friction coefficient is 0.12, which is less than 0.165, so static friction cannot supply the required torque. The sphere must slip from the instant of release; the student's 1.07 s prediction assumes a constraint the surface cannot enforce. Note that mu_min is independent of m, R and g - it depends only on the angle and the mass distribution - so a heavier or smaller sphere would behave identically.
2. Compute the actual acceleration, the descent time and the fraction of the released gravitational energy that ends up as heat. Explain the student's paradox.
Read the worked answer
With kinetic friction f = mu_k m g cos(theta) = 0.10 x 0.05 x 9.8 x 0.866 = 0.0424 N:
a = g(sin(theta) - mu_k cos(theta)) = 9.8(0.5 - 0.0866) = 4.05 m/s^2. t = sqrt(2L/a) = sqrt(4.00/4.05) = 0.994 s - matches the video's 0.99 s.
Rotation: alpha = fR/I with I = (2/5)mR^2 = 8.0e-6 kg m^2, so alpha = 0.0424 x 0.02/8.0e-6 = 106 rad/s^2, and alpha*R = 2.12 m/s^2.
Energy at t = 0.994 s: v = 4.03 m/s, omega = 105 rad/s. Translational KE = 0.406 J; rotational KE = 0.5 x 8.0e-6 x (105)^2 = 0.044 J; total 0.450 J. Released: mgh = 0.05 x 9.8 x 1.00 = 0.490 J. Heat = 0.040 J (8 percent).
Check: slip distance = 0.5(4.05 - 2.12)(0.994)^2 = 0.95 m, and f x slip = 0.040 J.
Paradox resolved: pure rolling forces 2/7 of the drop into rotation and only 5/7 into translation. Sliding uses a smaller friction force, so less energy is spun into the sphere and more is left as translational kinetic energy - the centre of mass accelerates faster. The 8 percent dissipation is much smaller than the redistribution, so slipping wins on time.
3. Does the sphere ever reach rolling? How would the answer change on the Moon, where g is about one sixth?
Read the worked answer
Define the slip velocity s = v - omega*R. While slipping,
ds/dt = a - alpha*R = g sin(theta) - mu_k g cos(theta)(1 + 5/2) = g[sin(theta) - 3.5 mu_k cos(theta)].
For theta = 30 degrees, mu_k = 0.10: 0.5 - 3.5 x 0.10 x 0.866 = 0.5 - 0.303 = 0.197, so ds/dt = 1.93 m/s^2 > 0. The slip velocity grows from zero, so v is always greater than omega*R: the sphere never rolls during the run.
General condition for the slip to close: sin(theta) < 3.5 mu_k cos(theta), i.e. mu_k > (2/7) tan(theta) = 0.165 - which is exactly the rolling condition. Since the surface cannot provide it, rolling is impossible on this incline.
On the Moon every acceleration is scaled by g, so the slip rate becomes 1.93/6 = 0.32 m/s^2 and the descent time grows as 1/sqrt(g) to about 2.4 s. The inequalities contain no g, so the verdict (slips all the way) is unchanged; only the timescale changes. This is the point of the question: whether rolling occurs is a geometric/frictional question, not a gravitational one.