IPhO 1975 · Theory problem 1

Rotating rod

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A rod revolves with a constant angular velocity \(\omega\) around a vertical axis A. The rod includes a fixed angle of \(\pi/2-\alpha\) with the axis. A body of mass \(m\) can glide along the rod. The coefficient of friction is \(\mu=\tan\beta\). The angle \(\beta\) is called “friction angle”.

a) Determine the angles \(\alpha\) under which the body remains at rest and under which the body is in motion if the rod is not rotating (i.e. \(\omega=0\)).

b) The rod rotates with constant angular velocity \(\omega>0\). The angle \(\alpha\) does not change during rotation. Find the condition for the body to remain at rest relative to the rod.

Figure 1

You can use the following relations:

\[ \sin(\alpha\pm\beta)=\sin\alpha\cdot\cos\beta\pm\cos\alpha\cdot\sin\beta \]

\[ \cos(\alpha\pm\beta)=\cos\alpha\cdot\cos\beta\mp\sin\alpha\cdot\sin\beta \]

Official English text of the problem.

Answers

Take \(0\le\alpha\le\pi/2\) and \(0\le\beta<\pi/2\). Let \(L\ge0\) denote the distance along the rod from its intersection with the rotation axis to the body.

  • a) A body initially at rest remains at rest when \[ \boxed{0\le\alpha\le\beta.} \] It slides down the rod when \[ \boxed{\beta<\alpha\le\frac{\pi}{2}.} \] The case \(\alpha=\beta\) is limiting equilibrium.

  • b) The necessary and sufficient condition for rest relative to the rod is \[ \boxed{ \left|g\sin\alpha-\omega^2L\cos^2\alpha\right| \le \tan\beta\,\cos\alpha\left(g+\omega^2L\sin\alpha\right). } \] For \(0\le\alpha<\pi/2\), this is equivalently \(L_{\min}\le L\le L_{\max}\), where \[ \boxed{ L_{\min}= \frac{g}{\omega^2\cos\alpha} \max\!\left\{0,\tan(\alpha-\beta)\right\}, } \] \[ \boxed{ L_{\max}= \begin{cases} \displaystyle\frac{g\tan(\alpha+\beta)}{\omega^2\cos\alpha}, &\alpha+\beta<\dfrac{\pi}{2},\\[6pt] \infty, &\alpha+\beta\ge\dfrac{\pi}{2}. \end{cases} } \] For a vertical rod, \(\alpha=\pi/2\), rest is impossible.

Solution

Model the body as a particle constrained to a rigid rod in uniform gravity. Use Coulomb static friction: the friction force can take either direction along the rod, with magnitude at most \(\mu N\), where \(N\) is the normal reaction. We determine whether a body initially at rest relative to the rod can remain so.

a) Nonrotating rod

The rod makes an angle \(\alpha\) with the horizontal. Gravity therefore has components

\[ mg\sin\alpha \quad\text{down the rod},\qquad mg\cos\alpha \quad\text{perpendicular to the rod}. \]

There is no acceleration perpendicular to the rod, so

\[ N=mg\cos\alpha. \]

For equilibrium, friction must balance the component of gravity along the rod:

\[ mg\sin\alpha\le\mu mg\cos\alpha. \]

For \(0\le\alpha<\pi/2\), this becomes

\[ \tan\alpha\le\mu=\tan\beta, \]

and hence \(\alpha\le\beta\). At equality, static friction reaches its maximum value.

If \(\alpha>\beta\), friction cannot balance gravity, and the body released from rest slides down the rod. At \(\alpha=\pi/2\), the normal reaction and thus the available friction vanish, so the body also falls.

b) Rotating rod

Use a frame rotating with the rod. A body at rest in this frame experiences no Coriolis force; there is also no Euler force because \(\omega\) is constant. In addition to gravity and the contact forces, include the centrifugal force.

The body's distance from the vertical axis is

\[ r=L\cos\alpha, \]

so the centrifugal force is horizontally outward, with magnitude

\[ F_{\mathrm c}=m\omega^2r=m\omega^2L\cos\alpha. \]

Take the positive direction along the rod to be outward and upward. The combined component of gravity and centrifugal force along the rod is

\[ F_{\parallel} =F_{\mathrm c}\cos\alpha-mg\sin\alpha =m\left(\omega^2L\cos^2\alpha-g\sin\alpha\right). \]

Their components perpendicular to the rod act in the same direction. Consequently, the normal reaction required for equilibrium is

\[ N=mg\cos\alpha+F_{\mathrm c}\sin\alpha =m\cos\alpha\left(g+\omega^2L\sin\alpha\right). \]

Friction must balance \(F_{\parallel}\), so equilibrium is possible precisely when

\[ |F_{\parallel}|\le\mu N. \]

Substituting and cancelling \(m\) gives

\[ \left|g\sin\alpha-\omega^2L\cos^2\alpha\right| \le \tan\beta\,\cos\alpha\left(g+\omega^2L\sin\alpha\right). \]

Thus rotation changes both the force that friction must balance and the normal reaction that determines the available friction.

To express the condition as bounds on \(L\), split the absolute-value inequality into its two inequalities. Multiply by \(\cos\beta>0\), and use the supplied addition formulas. This gives

\[ \omega^2L\cos\alpha\cos(\alpha-\beta) \ge g\sin(\alpha-\beta), \]

\[ \omega^2L\cos\alpha\cos(\alpha+\beta) \le g\sin(\alpha+\beta). \]

For \(0\le\alpha<\pi/2\), both \(\cos\alpha\) and \(\cos(\alpha-\beta)\) are positive. The first inequality therefore requires

\[ L\ge\frac{g\tan(\alpha-\beta)}{\omega^2\cos\alpha}. \]

When \(\alpha\le\beta\), the right-hand side is nonpositive, so this imposes no restriction beyond \(L\ge0\). Combining these cases yields the stated \(L_{\min}\).

For the second inequality:

  • If \(\alpha+\beta<\pi/2\), its coefficient of \(L\) is positive, giving \[ L\le\frac{g\tan(\alpha+\beta)}{\omega^2\cos\alpha}. \]
  • If \(\alpha+\beta\ge\pi/2\), its left-hand side is nonpositive while its right-hand side is positive. It is therefore satisfied for every \(L\ge0\), and there is no upper bound.

Below a positive \(L_{\min}\), gravity cannot be balanced and the body tends to slide inward and downward. Above a finite \(L_{\max}\), the outward force cannot be balanced and the body tends to slide outward and upward. Equality at either bound is included as limiting equilibrium.

Finally, for \(\alpha=\pi/2\), the body lies on the rotation axis. The centrifugal force and normal reaction vanish, leaving gravity unbalanced.

The solution was checked against the official solution.