Kvant Physics Problem 20

Verified: yes
Verdicts: PASS + PASS
Solve time: 12m55s
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 board, one end of which is slowly raised (Fig. 2). Find the limiting angle of inclination of the board at which the cylinder will still remain in equilibrium. The coefficient of friction between the cylinder and the board is $0{,}2$.

All-Union Correspondence Olympiad (1967)

Setup and Assumptions

The body is a long homogeneous cylinder of outer radius $R$. A cylindrical hole of radius $R/2$ is drilled parallel to the axis of the cylinder, with its axis displaced by $R/2$ from the axis of the original cylinder.

The cylinder rests on an inclined board. The angle of inclination of the board relative to the horizontal is denoted by $\alpha$. The coefficient of static friction between the cylinder and the board is

$$\mu = 0.2 .$$

The limiting angle $\alpha$ at which equilibrium is still possible is required.

The cylinder is treated as infinitely long, so the problem reduces to the geometry of its cross section. The material density is uniform. The board is raised slowly, hence all accelerations are zero and the equilibrium is static. Air resistance and deformations are neglected.

Let $O$ be the center of the original cylinder and $O_1$ the center of the drilled hole. The distance between them is

$$OO_1=\frac R2 .$$

The contact point with the board is denoted by $P$.

Physical Principles

Static equilibrium requires that the resultant force and the resultant torque acting on the body vanish.

The forces acting on the cylinder are the weight $Mg$, the normal reaction $N$, and the static friction force $F$.

Force equilibrium along and perpendicular to the plane gives

$$F=Mg\sin\alpha ,$$

$$N=Mg\cos\alpha .$$

The condition for static friction is

$$|F|\le \mu N .$$

At the limiting angle for sliding,

$$|F|=\mu N ,$$

hence

$$\tan\alpha=\mu .$$

A second possible loss of equilibrium occurs when the line of action of the weight passes through the contact point $P$. Beyond that position the gravitational torque overturns the cylinder. The corresponding angle must also be found.

The center of mass of a composite body is determined by

$$\mathbf r_C= \frac{\sum m_i\mathbf r_i}{\sum m_i},$$

with the removed material treated as negative mass.

Derivation

First determine the position of the center of mass.

Let the mass of the complete cylinder be proportional to its cross sectional area

$$A=\pi R^2 .$$

The removed hole has area

$$A_h=\pi\left(\frac R2\right)^2 =\frac{\pi R^2}{4}.$$

Take the origin at $O$ and the positive $x$-axis directed from $O$ toward $O_1$.

The complete cylinder has centroid at $x=0$. The removed cylinder has centroid at

$$x=\frac R2 .$$

Using the method of negative masses,

$$x_C= \frac{A\cdot 0-A_h\cdot (R/2)} {A-A_h}.$$

Substituting the areas,

$$x_C= \frac{-\frac{\pi R^2}{4}\cdot \frac R2} {\pi R^2-\frac{\pi R^2}{4}} = \frac{-\frac{R}{8}} {\frac34} = -\frac R6 .$$

Thus the center of mass lies on the side opposite the hole, at a distance

$$OC=\frac R6 .$$

Now consider tipping.

For equilibrium on a plane inclined by angle $\alpha$, the radius through the contact point $P$ is perpendicular to the plane. The vertical direction differs from this radius by the angle $\alpha$.

The vector from the contact point to the center of mass is

$$\mathbf{PC}=\mathbf{PO}+\mathbf{OC}.$$

The magnitude of $\mathbf{PO}$ equals $R$ and is directed along the normal to the plane. The vector $\mathbf{OC}$ has magnitude $R/6$ and is directed away from the hole.

The critical position for overturning is reached when the vertical through the center of mass passes through the contact point. Then the vector $\mathbf{PC}$ itself is vertical.

The angle between $\mathbf{PC}$ and the normal to the plane is therefore exactly $\alpha$.

From the right triangle formed by the components of $\mathbf{PC}$,

$$\tan\alpha_{\rm tip} = \frac{OC}{OP} = \frac{R/6}{R} = \frac16 .$$

Hence

$$\alpha_{\rm tip} = \arctan!\left(\frac16\right).$$

Now examine sliding.

At the threshold of sliding,

$$\tan\alpha_{\rm slide} = \mu = 0.2 = \frac15 ,$$

so

$$\alpha_{\rm slide} = \arctan!\left(\frac15\right).$$

The body can remain in equilibrium only while neither tipping nor sliding occurs. The limiting angle is therefore the smaller of these two critical angles.

Since

$$\frac16 < \frac15 ,$$

tipping occurs before sliding.

Result

The tipping condition gives

$$\alpha_{\rm tip} = \arctan!\left(\frac16\right) = 9.46^\circ .$$

The sliding condition gives

$$\alpha_{\rm slide} = \arctan(0.2) = 11.31^\circ .$$

Because

$$\alpha_{\rm tip}<\alpha_{\rm slide},$$

the cylinder loses equilibrium by overturning before the friction limit is reached.

The limiting inclination of the board is

$$\boxed{\alpha=\arctan!\left(\frac16\right)}$$

and numerically

$$\boxed{\alpha \approx 9.5^\circ }.$$

Sanity Checks

The ratio $OC/R=1/6$ is dimensionless, so the argument of the arctangent in the tipping condition is dimensionless as required.

If the hole were absent, then $OC=0$. The tipping condition would give $\alpha_{\rm tip}=0$, which means the center of mass would remain directly above the geometric center and no finite tipping angle would arise, consistent with the fact that a uniform cylinder on a rough plane is limited only by sliding.

The friction coefficient corresponds to a sliding angle of about $11^\circ$, while the displaced center of mass produces a tipping angle of about $9.5^\circ$. These values are of the same order and physically reasonable.

The most delicate step is determining the center of mass. Reversing the sign of the removed mass would place the center of mass toward the hole instead of away from it, leading to an incorrect critical angle.