Kvant Physics Problem 20

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Verdicts: SKIP + SKIP
Solve time: 1m12s
Source on kvant.digital

Problem

Fig. 2

A hole is drilled parallel to the axis of a cylinder of radius $R$ at a distance $\dfrac R2$ from its center. The radius of the hole is $\dfrac R2$. The cylinder rests on a small plank that is slowly raised at one end (Fig. 2). Find the limiting angle of inclination of the plank for which the cylinder will still remain in equilibrium. The coefficient of friction between the cylinder and the plank is $0{,}2$.

All-Union Correspondence Olympiad (1967)

Setup and Assumptions

A long cylinder of radius $R$ and uniform material density contains a cylindrical hole parallel to its axis. In the cross section, the full disk of radius $R$ has a circular hole of radius $R/2$ whose center is displaced from the cylinder center by a distance $R/2$. The cylinder rests on a rigid plank that is slowly tilted by an angle $\theta$ relative to the horizontal.

The cylinder is assumed to remain in planar cross section equilibrium, so the problem reduces to a rigid body in a vertical plane perpendicular to the cylinder axis. The contact between the cylinder and plank is a single point contact. Static friction with coefficient $\mu = 0.2$ acts at the contact point and satisfies the inequality $|f| \le \mu N$. The cylinder is assumed not to deform, and rolling without slipping is not imposed a priori; instead, equilibrium requires both force and torque balance.

The unknown is the limiting inclination angle $\theta_{\max}$ at which static equilibrium is still possible.

Physical Principles

Static equilibrium of a rigid body requires the resultant force to vanish,

$\sum \mathbf{F} = \mathbf{0},$

and the resultant torque about any point to vanish,

$\sum \mathbf{M} = 0.$

The weight acts through the center of mass of the body and has magnitude $mg$, directed vertically downward.

The normal reaction $N$ acts perpendicular to the plank at the contact point, and the friction force $f$ acts along the plank, constrained by the Coulomb condition

$|f| \le \mu N.$

The torque of a force about a point is computed as

$\mathbf{M} = \mathbf{r} \times \mathbf{F},$

where $\mathbf{r}$ is the position vector from the reference point to the point of application of the force.

The position of the center of mass of a composite lamina is determined from area subtraction,

$\mathbf{r}_{\text{cm}} = \frac{\sum A_i \mathbf{r}_i}{\sum A_i}.$

Derivation

The cross section consists of a full disk of area $A_1 = \pi R^2$ and a removed circular region of area $A_2 = \pi (R/2)^2 = \pi R^2/4$. The hole center is displaced from the cylinder center by $R/2$ along some fixed direction in the cross section.

Taking the origin at the geometric center of the full disk, the center of mass shift $\mathbf{d}$ lies along the line joining the centers. Its magnitude follows from area subtraction:

$d = \frac{A_2 \cdot (R/2)}{A_1 - A_2}.$

Substituting areas,

$d = \frac{\left(\pi R^2/4\right)\left(R/2\right)}{\pi R^2 - \pi R^2/4}.$

Simplifying numerator and denominator,

$d = \frac{\pi R^3/8}{(3\pi R^2)/4} = \frac{R}{8} \cdot \frac{4}{3} = \frac{R}{6}.$

Thus the center of mass is displaced by $R/6$ from the geometric center.

When the cylinder rests on an incline of angle $\theta$, choose axes with $x$ along the plane (up the slope) and $y$ normal to the plane (outward). The contact point is taken as the origin. The geometric center of the cylinder lies at

$\mathbf{r}_O = (0, R).$

The center of mass is displaced from $O$ by a vector of magnitude $d = R/6$ along a fixed direction in the cross section. The configuration that yields the limiting equilibrium corresponds to the displacement having a component along the slope direction, so the center of mass position is taken as

$\mathbf{r}_{\text{cm}} = (d, R).$

The weight vector is decomposed in the inclined coordinate system. Since gravity is vertical, its components are

$\mathbf{F}_g = (-mg \sin\theta, -mg \cos\theta).$

The torque of gravity about the contact point is

$M = x_{\text{cm}} F_y - y_{\text{cm}} F_x.$

Substituting $\mathbf{r}_{\text{cm}}$ and $\mathbf{F}_g$,

$M = d(-mg \cos\theta) - R(-mg \sin\theta).$

This gives

$M = -mg d \cos\theta + mg R \sin\theta.$

Equilibrium requires $M = 0$, hence

$R \sin\theta = d \cos\theta.$

Dividing by $\cos\theta$,

$\tan\theta = \frac{d}{R} = \frac{R/6}{R} = \frac{1}{6}.$

Thus the geometric torque condition yields

$\theta_1 = \arctan\left(\frac{1}{6}\right).$

Force balance along the plane gives the friction requirement

$f = mg \sin\theta,$

and the normal force

$N = mg \cos\theta.$

The no-slip condition is

$mg \sin\theta \le \mu mg \cos\theta,$

which reduces to

$\tan\theta \le \mu = 0.2.$

This produces a second limiting angle

$\theta_2 = \arctan(0.2).$

Comparing thresholds,

$\frac{1}{6} \approx 0.167 < 0.2,$

so the torque-induced loss of equilibrium occurs first.

Result

The limiting angle is determined by the torque balance condition,

$\theta_{\max} = \arctan\left(\frac{1}{6}\right).$

Substitution gives

$\theta_{\max} = \arctan(0.1667) \approx 9.46^\circ.$

Final result:

$\boxed{\theta_{\max} \approx 9.5^\circ}.$

Sanity Checks

The expression $\tan\theta_{\max} = d/R$ is dimensionless, since both $d$ and $R$ have units of length, confirming consistency.

The result depends only on the geometric asymmetry of the mass distribution and not on $m$, $g$, or friction, consistent with a torque-dominated instability mechanism.

The friction limit yields $\theta \le \arctan(0.2) \approx 11.3^\circ$, which is larger than $9.5^\circ$, so slipping would occur only after loss of rotational equilibrium, confirming that the torque condition is the controlling constraint.

In the limit of a centered hole ($d \to 0$), the condition gives $\theta_{\max} \to 0$, consistent with immediate loss of equilibrium on any incline due to symmetry requiring friction alone to support the weight component without restoring torque.