Two Rollers on the Same Incline: Shape Factor Beats Mass
In a practical revision session for Maharashtra HSC Physics, a teacher gives two rollers that look identical from outside. Roller A is a solid cylinder and roller B is a thin-walled hollow cylinder; both have mass 2.0 kg and outer radius 5.0 cm. They are released from rest at the same instant from the top of a 1.20 m plank inclined at 30° to the bench, and both roll without slipping. A student predicts a dead heat 'because the mass and radius are the same'. In fact the hollow roller arrives about 0.14 s after the solid one. In a second trial a waxed block of the same mass slides from rest down the same plank, which has been wiped with oil, and reaches the bottom in 0.70 s. The student is surprised that the sliding block beats both rollers and asks whether the result would change if the rollers were made of a denser metal.
1. Calculate the acceleration of each roller down the incline and the time each takes to travel 1.20 m. Hence explain, in terms of energy, why the hollow roller loses to the solid one even though their masses and radii are equal.
Read the worked answer
For rolling without slipping, Newton's laws along the incline and about the centre give
$a = \dfrac{g\sin\theta}{1+I/(MR^2)} = \dfrac{g\sin\theta}{1+k}$, with $k=I/MR^2$.
Solid cylinder: $k=\tfrac12$, so
$a_A = \dfrac{9.8\times0.5}{1.5}=3.27\ \text{m s}^{-2}$,
$t_A=\sqrt{2s/a}=\sqrt{2.40/3.27}=0.86\ \text{s}$.
Hollow cylinder: $k=1$, so
$a_B=\dfrac{9.8\times0.5}{2}=2.45\ \text{m s}^{-2}$,
$t_B=\sqrt{2.40/2.45}=0.99\ \text{s}$.
The 0.13 s gap matches the observed 0.14 s.
Energy view: from the top, $Mgh=\tfrac12Mv^2+\tfrac12I\omega^2=\tfrac12Mv^2(1+k)$. Thus $v^2=2gh/(1+k)$. The hollow roller has the larger $k$, so a larger fraction of the released gravitational energy is locked up as rotational kinetic energy and less remains as translational kinetic energy. Mass and radius cancel from the ratio, so the denser metal would give the same times as long as the shape (and hence $k$) is unchanged.
2. Compare the rollers with the block's 0.70 s. Why is the sliding block fastest, even though friction is essential for the rollers?
Read the worked answer
For the block sliding on the oiled plank, friction is negligible, so
$a=g\sin\theta=4.9\ \text{m s}^{-2}$ and $t=\sqrt{2.40/4.9}=0.70\ \text{s}$, exactly as measured.
A rolling body's acceleration is $g\sin\theta/(1+k)$, which is always less than $g\sin\theta$ because part of the energy must go into rotation. Static friction at the contact point is necessary to provide the torque that makes the body rotate, but in pure rolling the contact point is instantaneously at rest, so this friction does no work and dissipates no energy. Its role is to redistribute energy between translation and rotation, not to add energy. Consequently the rollers are slower than the frictionless slider even though friction is present. If the plank were perfectly smooth, both rollers would slide without rotating and would also take 0.70 s, tying with the block.
3. The plank is polished until the coefficient of static friction is only 0.15. Using the minimum-friction condition, decide which roller or rollers can still roll without slipping, and reconcile this with the fact that oiling the plank made the block slide easily.
Read the worked answer
For rolling without slipping, the required friction is
$f=\dfrac{k}{1+k}Mg\sin\theta$, with normal force $N=Mg\cos\theta$. Hence
$\mu_{\min}=\dfrac{f}{N}=\dfrac{k\tan\theta}{1+k}$.
Solid: $\mu_{\min}=\dfrac{0.5\times0.577}{1.5}=0.19$.
Hollow: $\mu_{\min}=\dfrac{1.0\times0.577}{2}=0.29$.
Both values exceed the available 0.15, so both rollers slip; the solid one is closer to its limit and slips less severely. This is consistent with the block: a sliding block never needs friction to maintain a rolling constraint, so oiling only reduces the retarding kinetic friction and lets it slide freely. Once a roller slips, kinetic friction acts up the incline, giving $a=g(\sin\theta-\mu_k\cos\theta)$ plus an angular acceleration $\alpha=fR/I$, so the clean 0.86 s and 0.99 s predictions no longer apply and the timing order may even change.