Irodov 1.145: A Ring Obeys the Conical Pendulum – Amit & Deepali Sinha Foundation

Problem Irodov 1.145

A closed chain, bent into a ring, has total mass $m = 0.36$ kg and hangs from a single thread tied to one point of the ring. The other end of the thread is fixed to a vertical shaft, which spins the chain steadily about the shaft’s vertical axis at constant angular velocity $\omega = 35$ rad/s. In this steady state the thread makes a fixed angle $\theta = 45^\circ$ with the vertical. Find (a) the distance $r$ from the rotation axis to the chain’s center of gravity, and (b) the tension $T$ in the thread.

The closed chain A hangs as a horizontal ring from a thread tied to one point of its rim; the thread's upper end is fixed to the vertical shaft, which turns at ω, and it makes the angle θ with the vertical axis. — The closed chain $A$ hangs as a horizontal ring from a thread tied to one point of its rim; the thread’s upper end is fixed to the vertical shaft, which turns at $\omega$ , and it makes the angle $\theta$ with the vertical axis.

Given

$m=0.36\text{ kg}$
$\omega=35\text{ rad/s}$
$\theta=45^\circ$

Find

r

, the center of gravity’s distance from the axis;

T

, the thread tension

Solution

Step 1Only the center of mass motion enters Newton’s second law

Point-mass equivalent: the chain's mass m at its center of gravity travels the gray horizontal circle of radius r about the shaft's axis, under the tension T at angle θ to the vertical and the weight m g; the loop marks ω. — Point-mass equivalent: the chain’s mass $m$ at its center of gravity travels the gray horizontal circle of radius $r$ about the shaft’s axis, under the tension $\vect T$ at angle $\theta$ to the vertical and the weight $m\vect g$ ; the loop marks $\omega$ .

The chain is an extended, possibly rotating body, but the only external forces on it are the thread’s tension $\vect T$ (acting where the thread ties on) and its weight $m\vect g$ (acting at every element, with resultant $m\vect g$ through the center of gravity). For any system, however it is shaped or how it spins about its own structure, the sum of the external forces equals the total mass times the acceleration of the center of mass.

\vect T + m\vect g = m\,\vect a_C

In the steady state the center of gravity travels in a horizontal circle of radius $r$ at constant angular velocity $\omega$ , so $\vect a_C$ is purely centripetal, of magnitude $\omega^2 r$ directed horizontally toward the axis, with no vertical component.

Step 2Resolve the thread’s tension along the shaft and the horizontal

Along the vertical, $\vect T$ balances the weight; along the horizontal, $\vect T$ supplies the centripetal force.

\begin{align*}T\cos\theta - mg &= 0\\ T\sin\theta &= m\omega^2 r\end{align*}

Step 3Solve the two equations for

r

and

T

, then evaluate

Dividing the horizontal equation by the vertical one eliminates $T$ and leaves $r$ ; the vertical equation alone then gives $T$ . Substituting $\theta=45^\circ$ (so $\tan\theta=1$ , $\cos\theta=1/\sqrt2$ ), $g=9.8\text{ m/s}^2$ , $\omega=35\text{ rad/s}$ , $m=0.36\text{ kg}$ carries both straight to numbers.

\begin{align*}\frac{T\sin\theta}{T\cos\theta} &= \frac{m\omega^2 r}{mg}\\ \tan\theta &= \frac{\omega^2 r}{g}\\ r &= \frac{g\tan\theta}{\omega^2} = \frac{(9.8\text{ m/s}^2)(1)}{(35\text{ rad/s})^2}\\ T &= \frac{mg}{\cos\theta} = (0.36\text{ kg})(9.8\text{ m/s}^2)\sqrt2\end{align*}

r = \dfrac{g\tan\theta}{\omega^2} = 0.8\text{ cm}

(1)

T = \dfrac{mg}{\cos\theta} \approx 5.0\text{ N}

(2)

Check

Dimensionally $[g][\omega^{-2}]=(\text{m/s}^2)(\text{s}^2)=\text{m}$ , a length, and $[m][g]=\text{N}$ , a force, as required. As $\theta\to0$ the thread hangs vertical, $r\to0$ and $T\to mg$ : a chain not being spun collapses onto the axis and simply hangs, which is the expected limit.

Remark

The result did not need to know the chain’s radius, how the thread attaches to the ring, or how the ring itself spins about its own rim: the center-of-mass theorem strips all of that away and leaves exactly the ordinary conical pendulum equations for a fictitious point mass $m$ sitting at the chain’s center of gravity. Any rigid or floppy body on a single thread, spun steadily about a vertical axis, reduces to the same two equations.