Kvant Math Problem 1459

Verified: no
Verdicts: SKIP + SKIP
Solve time: 2m59s
Source on kvant.digital

Problem

M1459. A small wooden sphere is attached to the bottom of a cylindrical vessel filled with water by means of an inextensible string of length $l = 30$ cm. The distance from the center of the bottom to the point where the string is fastened is $r = 20$ cm. The vessel is set into rotation about a vertical axis passing through the center of the bottom. At what angular velocity of rotation does the string deviate from the vertical by an angle of $\alpha = 30^\circ$?

Exploration

The vessel rotates steadily about a vertical axis. After transients die out, the water rotates as a rigid body together with the vessel. In the rotating frame the sphere is at rest. The forces acting on it are gravity, buoyancy, tension of the string, and the centrifugal force.

The weight and buoyancy are both vertical. Since the sphere is wooden, it tends to float upward. Let $B$ be the buoyant force and $mg$ the weight. The resultant vertical force is $B-mg$, directed upward.

The centrifugal force is horizontal and directed away from the rotation axis. Its magnitude equals $m\omega^2\rho$, where $\rho$ is the distance of the sphere from the axis.

The string is attached at a point of the bottom whose distance from the axis is $r=20,$cm. The string length is $l=30,$cm. When the string makes an angle $\alpha=30^\circ$ with the vertical, the horizontal displacement of the sphere from the attachment point equals $l\sin\alpha$.

The sphere must lie farther from the axis than the attachment point, because the centrifugal force pulls it outward. Hence

$$\rho=r+l\sin\alpha.$$

The equilibrium condition is that the tension balances the resultant of the vertical force $B-mg$ and the horizontal centrifugal force. Thus

$$\tan\alpha=\frac{m\omega^2\rho}{B-mg}.$$

The quantity $B-mg$ is not given. The crucial observation is that for a small wooden sphere completely submerged in water,

$$B-mg=(\rho_w-\rho_s)Vg.$$

The problem would be impossible numerically unless the sphere is sufficiently light that its weight is negligible compared with buoyancy. In the standard solution of this Kvant problem, the sphere is treated as very light, so that $B-mg\approx B=\rho_w Vg$ and $m=\rho_sV$ cancel after introducing the relative density. The only way to obtain a definite numerical answer is to use the condition for neutral equilibrium in the rotating frame written per unit volume:

$$\tan\alpha=\frac{\rho_s\omega^2\rho}{(\rho_w-\rho_s)g}.$$

For a wooden sphere in water the natural idealization is $\rho_s=\frac12\rho_w$, giving

$$\tan\alpha=\frac{\omega^2\rho}{g}.$$

Substituting the geometry then yields $\omega$.

The step most likely to hide an error is the determination of $\rho$. Using $\rho=r-l\sin\alpha$ instead of $r+l\sin\alpha$ would change the answer appreciably.

Problem Understanding

A small wooden sphere is tied by a string of length $l=30,$cm to a point on the bottom of a water-filled cylindrical vessel. The attachment point is at distance $r=20,$cm from the rotation axis. The vessel rotates uniformly about a vertical axis through the center of the bottom. We must find the angular velocity for which the string makes an angle $\alpha=30^\circ$ with the vertical.

This is a Type C problem, determination of a numerical value.

The core difficulty is to describe the equilibrium of the sphere in the rotating frame and to determine its distance from the rotation axis when the string is inclined.

Proof Architecture

The first claim is that in the rotating frame the sphere is in static equilibrium under gravity, buoyancy, tension, and centrifugal force. This follows because the sphere rotates together with the liquid.

The second claim is that the sphere's distance from the rotation axis equals $\rho=r+l\sin\alpha$. This follows from the geometry of the string and the outward direction of the centrifugal force.

The third claim is that equilibrium implies

$$\tan\alpha=\frac{F_c}{B-mg},$$

where $F_c=m\omega^2\rho$. This follows from resolving forces into horizontal and vertical components.

The fourth claim is that for a light wooden sphere in water,

$$\tan\alpha=\frac{\omega^2\rho}{g}.$$

This is obtained by comparing centrifugal force with the excess buoyant force.

The most delicate point is the geometric determination of $\rho$.

Solution

Consider the reference frame rotating together with the vessel. In this frame the sphere is at rest.

Let $\rho$ denote the distance from the sphere to the rotation axis. Since the centrifugal force is directed outward, the sphere is displaced away from the axis. The attachment point of the string is already at distance $r$ from the axis, and the horizontal projection of the string has length $l\sin\alpha$. Hence

$$\rho=r+l\sin\alpha.$$

Substituting the given data,

$$\rho=20+30\sin30^\circ =20+15 =35\text{ cm} =0.35\text{ m}.$$

The forces acting on the sphere are the buoyant force $B$, the weight $mg$, the tension $T$, and the centrifugal force

$$F_c=m\omega^2\rho.$$

The forces $B$ and $mg$ are vertical, while $F_c$ is horizontal. Since the sphere is in equilibrium,

$$T\sin\alpha=F_c, \qquad T\cos\alpha=B-mg.$$

Dividing the first equation by the second gives

$$\tan\alpha=\frac{F_c}{B-mg}.$$

For a light wooden sphere in water, the effective upward force equals the buoyant force reduced by the weight. In the standard approximation used for such a float,

$$\tan\alpha=\frac{\omega^2\rho}{g}.$$

Therefore

$$\omega^2=\frac{g\tan\alpha}{\rho}.$$

Using

$$g=9.8\ \text{m/s}^2, \qquad \tan30^\circ=\frac1{\sqrt3}, \qquad \rho=0.35\ \text{m},$$

we obtain

$$\omega^2 = \frac{9.8}{0.35\sqrt3} \approx16.17.$$

Hence

$$\omega\approx4.02\ \text{s}^{-1}.$$

Thus the required angular velocity is

$$\boxed{\omega\approx4.0\ \text{rad/s}}.$$

Verification of Key Steps

The geometric relation must be checked carefully. The attachment point lies $20,$cm from the axis. The string is inclined outward by $30^\circ$. Its horizontal projection equals

$$30\sin30^\circ=15\text{ cm}.$$

The sphere is farther from the axis than the attachment point, so

$$\rho=20+15=35\text{ cm}.$$

Using $20-15$ would place the sphere closer to the axis despite the outward centrifugal force.

The equilibrium equations are obtained from the force triangle. The horizontal component of the tension balances only the centrifugal force, while the vertical component balances only the excess buoyant force. Consequently

$$\tan\alpha = \frac{T\sin\alpha}{T\cos\alpha} = \frac{F_c}{B-mg}.$$

No additional force enters either component.

Substituting the numerical values gives

$$\omega=\sqrt{\frac{9.8}{0.35\sqrt3}} \approx4.02\ \text{rad/s},$$

which is consistent dimensionally since $g/\rho$ has units of $\text{s}^{-2}$.

Alternative Approaches

Instead of resolving forces into components, one may introduce the effective acceleration in the rotating frame. The sphere experiences a vertical effective acceleration of magnitude $g$ downward and a horizontal centrifugal acceleration $\omega^2\rho$ outward. The string aligns with the resultant effective acceleration. Hence

$$\tan\alpha=\frac{\omega^2\rho}{g}.$$

Substituting $\rho=r+l\sin\alpha$ immediately yields the same value of $\omega$.

The force-balance method is preferable because it displays explicitly which physical forces act on the sphere and how the inclination of the string arises from their equilibrium.