Question

A wooden rod rotates with a constant angular speed $\omega$ about a vertical axis, such that the rod makes a constant angle $\alpha$ with the horizontal plane. Along the rod there is a ring of mass 'm' which can slide up and down along the rod. The coefficient of friction between the ring and the rod is equal to 'tan $\alpha$'. Find the maximum and the minimum angular speed $\omega$ so that the ring stays at a constant distance L measured from the lower end of the rod.

Answer 1

• Minimum angular speed: $\omega_{\min}=0$
• Maximum angular speed: $\omega_{\max}=\sqrt{\dfrac{2g\sin\alpha}{L\cos2\alpha}}$

Answer 2

We will show that in the rotating frame the ring (of mass m) experiences two “effective” forces:

1. Gravity, $\vec{W}=mg$ downward.
2. A “centrifugal force” (per unit mass) which is $\omega^2r$ pointing radially outward from the rotation axis.

Because the rod makes a fixed angle $\alpha$ with the horizontal (so that its plane contains the vertical and the radial direction) the point on the rod at a distance $L$ from the lower end has:

• A horizontal (radial) distance $r=L\cos\alpha$.
• A vertical height $z=L\sin\alpha$.

Thus, when we take “up along the rod” as the positive s‐direction (from lower end to upper end), a unit vector along the rod is

$\hat{e}=(\cos\alpha,\sin\alpha)$

(with the first component in the horizontal (radial) direction and the second vertical upward).

Step 1. Find the Components of the Two Forces Along the Rod

1. Centrifugal Force

The centrifugal force (in the rotating frame) is

$\vec{F}_c = m\omega^2r \hat{r} = m\omega^2 (L\cos\alpha)(\text{radial unit vector}).$

Since the rod is along the radial direction (horizontally) and upward by $\sin\alpha$, its component along the rod is

$F_{c,\text{rod}} = m\omega^2 (L\cos\alpha)\cos\alpha = m\omega^2L\cos^2\alpha.$

It is directed upward along the rod.

2. Weight

Weight is vertical:

$\vec{W}=(0,-mg).$

Its component along the rod is obtained by projecting onto $\hat{e}$. (Recall that the vertical component of $\hat{e}$ is $\sin\alpha$.) Thus,

$F_{w,\text{rod}} = -mg\sin\alpha.$

Here the negative sign indicates that it acts downward along the rod.

Thus the net effective force along the rod is

$F_{\text{net}} = m\omega^2L\cos^2\alpha - mg\sin\alpha.$

Step 2. Friction Force Condition

Since the ring is free to slide but is held in place by friction, the friction must balance any net tendency to slide along the rod. However, static friction has a maximum value

$f_{\max} = \mu N,$

where the coefficient of friction is given as

$\mu = \tan\alpha.$

The normal force $N$ is produced by the components of the centrifugal force and weight perpendicular to the rod. Choose a unit vector perpendicular to the rod (in the plane of the motion) as

$\hat{n}=(-\sin\alpha,\cos\alpha).$

Now, compute the perpendicular components:

• From the centrifugal force:

The centrifugal force is horizontal:

$\vec{F}_c=(m\omega^2L\cos\alpha,0).$

Its component perpendicular to the rod is

$F_{c,\perp} = \vec{F}_c\cdot \hat{n} = m\omega^2L\cos\alpha(-\sin\alpha) = -m\omega^2L\cos\alpha\sin\alpha.$

• From the weight:

$\vec{W}=(0,-mg),\quad F_{w,\perp}=\vec{W}\cdot \hat{n} = -mg\cos\alpha.$

Thus, the total normal force (taking magnitude) is

$N = m\cos\alpha\left(g+\omega^2L\sin\alpha\right).$

So the maximum friction is

$f_{\max}= \mu N = \tan\alpha\, m\cos\alpha\left(g+\omega^2L\sin\alpha\right)= m\sin\alpha\left(g+\omega^2L\sin\alpha\right).$

Step 3. Equilibrium Condition with Friction

In order for the ring to remain fixed at the given point on the rod, the friction must be able to balance the net force along the rod. Therefore, we require:

$\Bigl|\,m\omega^2L\cos^2\alpha - mg\sin\alpha\,\Bigr| \le m\sin\alpha\left(g+\omega^2L\sin\alpha\right).$

Dividing both sides by $m$ (and noting all quantities are positive) we have

$\left|\omega^2L\cos^2\alpha - g\sin\alpha\right| \le \sin\alpha\left(g+\omega^2L\sin\alpha\right).$

We must consider two cases since the net force along the rod may be upward or downward.

(a) Case 1: When $\omega^2L\cos^2\alpha \ge g\sin\alpha$

Here the net force is upward and friction must act downward. The equality limiting condition is:

$\omega^2L\cos^2\alpha - g\sin\alpha = \sin\alpha\left(g+\omega^2L\sin\alpha\right).$

Rearrange:

$\omega^2L\cos^2\alpha - \omega^2L \sin^2\alpha = 2g\sin\alpha.$

Factor out $\omega^2L$:

$\omega^2L(\cos^2\alpha-\sin^2\alpha)=2g\sin\alpha.$

Since $\cos^2\alpha-\sin^2\alpha=\cos2\alpha$, we obtain

$\omega^2L\cos2\alpha=2g\sin\alpha.$

Thus

$\omega^2=\frac{2g\sin\alpha}{L\cos2\alpha}$

or

$\omega_{\max}=\sqrt{\frac{2g\sin\alpha}{L\cos2\alpha}}.$

This is the maximum angular speed for which friction can still hold the ring.

(b) Case 2: When $\omega^2L\cos^2\alpha < g\sin\alpha$

In this case the net force is downward and friction acts upward to oppose sliding. The limiting condition becomes:

$g\sin\alpha-\omega^2L\cos^2\alpha = \sin\alpha\left(g+\omega^2L\sin\alpha\right).$

Let’s simplify:

$g\sin\alpha-\omega^2L\cos^2\alpha = g\sin\alpha+\omega^2L\sin^2\alpha.$

Subtract $g\sin\alpha$ from both sides:

$-\omega^2L\cos^2\alpha=\omega^2L\sin^2\alpha.$

This yields

$\omega^2L(-\cos^2\alpha-\sin^2\alpha)=0 \quad \Longrightarrow \quad \omega^2L(-1)=0,$

so that

$\omega=0.$

Thus, the lower limiting value is

$\omega_{\min}=0.$

Final Result

The ring remains at a constant distance $L$ on the rod (that is, is held static along the rod) provided the angular speed $\omega$ lies in the range:

$0\le \omega \le \sqrt{\frac{2g\sin\alpha}{L\cos2\alpha}}.$