Irodov 1.280: Turning the Spin Turns the Platform – Amit & Deepali Sinha Foundation

Problem Irodov 1.280

A horizontal platform $P$ stands on a bearing that lets it turn freely about a fixed vertical axis $OO'$ . It carries a motor $M$ on one side and a balance weight $N$ on the other, and the moment of inertia of the platform together with the motor and the weight, about $OO'$ , is $I$ . A light frame driven by the motor holds a uniform sphere $A$ whose center sits on $OO'$ ; the sphere turns freely, on frictionless bearings, about its own shaft $BB'$ , and its moment of inertia about that shaft is $I_0$ . At the start the platform is at rest, $BB'$ lies along $OO'$ , and the sphere is spinning at angular velocity $\omega_0$ . The motor then swings the shaft $BB'$ , carrying the sphere with it, about an axis through the sphere’s center. Find:

the work the motor performs in turning $BB'$ through $90^\circ$ , and the work in turning it through $180^\circ$
the moment $N$ of the external forces that hold the axis of the whole arrangement vertical once $BB'$ has been turned through $90^\circ$

The starting arrangement. The platform P rests on its bearing, with the motor M and the balance weight N riding on it; the sphere A spins at ω_0 about the shaft BB', which lies along the vertical axis OO'. — The starting arrangement. The platform $P$ rests on its bearing, with the motor $M$ and the balance weight $N$ riding on it; the sphere $A$ spins at $\vect\omega_0$ about the shaft $BB'$ , which lies along the vertical axis $OO'$ .

Given

$I$
$I_0$
$\omega_0$

Find

(a)

A_{90^\circ}

and

A_{180^\circ}

, (b)

N

Solution

Step 1Two facts about the sphere, and one about the whole machine

Nothing outside the machine has a moment about the vertical: gravity pulls along it, the bearing’s reaction is applied on it, and the motor is internal. So the vertical component of the total angular momentum keeps the value it had at the start, when only the sphere was moving.

L_z = I_0\omega_0 \quad\text{at all times}

The sphere itself needs two more statements. First, it is uniform, so its moment of inertia is $I_0$ about every axis through its center and its angular momentum follows its angular velocity in direction as well as in size. Write $\vect e$ for the unit vector along the shaft $BB'$ and $\vect\omega_A$ for the sphere’s own angular velocity.

\vect L_A = I_0\,\vect\omega_A

Second, its own bearings are frictionless, so nothing exerts a moment on it about $BB'$ . That alone would not settle the matter, because $BB'$ is itself being swung, and differentiating the component of the angular momentum along the shaft picks up a second term from the turning of the shaft’s own direction.

\frac{d}{dt}\bigl(\vect L_A\cdot\vect e\bigr) = \frac{d\vect L_A}{dt}\cdot\vect e + \vect L_A\cdot\frac{d\vect e}{dt}

The frictionless bearings kill the first term outright.

\frac{d\vect L_A}{dt}\cdot\vect e = 0

The second term needs the first fact. The shaft direction is turned by the frame, so $d\vect e/dt$ is the frame’s angular velocity crossed into $\vect e$ and is therefore perpendicular to both of those; the sphere’s angular velocity is built out of the frame’s rotation and $\vect e$ alone, and $\vect L_A$ is parallel to it.

\vect L_A\cdot\frac{d\vect e}{dt} = 0

Both terms vanish, so the component of the sphere’s angular velocity along $BB'$ stays at the value it had at the start, while the components across $BB'$ are dragged along by the frame.

\vect\omega_A\cdot\vect e = \omega_0

Writing $\vect\omega$ for the platform’s angular velocity, the sphere’s own angular velocity is therefore the frame’s motion with its $BB'$ part replaced by the free spin.

\vect\omega_A = \vect\omega - (\vect\omega\cdot\vect e)\,\vect e + \omega_0\,\vect e

(1)

Step 2After the quarter turn the sphere is carried round bodily

After the motor has swung BB' through 90°: the shaft is horizontal, the sphere still spins at ω_0 about it, and the platform has picked up ω about OO'. The vertical member below the sphere is the light frame carrying the shaft. — After the motor has swung $BB'$ through $90^\circ$ : the shaft is horizontal, the sphere still spins at $\vect\omega_0$ about it, and the platform has picked up $\vect\omega$ about $OO'$ . The vertical member below the sphere is the light frame carrying the shaft.

With $BB'$ horizontal, $\vect e$ is perpendicular to the vertical, so $\vect\omega\cdot\vect e$ vanishes in (1) and the sphere turns with the platform as if it were bolted to it, its spin merely added on sideways. Its vertical angular momentum is then $I_0\omega$ , and the conserved $L_z$ reads

I_0\omega_0 = I\omega + I_0\omega

Solving for the rate the platform has picked up,

\omega = \frac{I_0\omega_0}{I+I_0}

(2)

The two parts of $\vect\omega_A$ are at right angles, so the sphere’s kinetic energy uses $\omega^2+\omega_0^2$ , and the whole store of energy afterwards is the platform’s plus the sphere’s.

\begin{align*}K_{90^\circ} &= \frac{1}{2}I\omega^2 + \frac{1}{2}I_0\left(\omega^2+\omega_0^2\right) \\ &= \frac{1}{2}(I+I_0)\,\omega^2 + \frac{1}{2}I_0\omega_0^2\end{align*}

The last term is exactly the energy the machine started with, so subtracting it leaves the motor’s work.

A_{90^\circ} = K_{90^\circ} - \frac{1}{2}I_0\omega_0^2 = \frac{1}{2}(I+I_0)\,\omega^2

Substituting (2) cancels one power of the combined moment of inertia.

A_{90^\circ} = \frac{1}{2}(I+I_0)\,\frac{I_0^2\omega_0^2}{(I+I_0)^2}

A_{90^\circ} = \dfrac{I_0^2\omega_0^2}{2(I+I_0)}

(3)

Step 3After the half turn the sphere is not carried at all

Turn the shaft the rest of the way and $\vect e$ becomes the downward vertical, so the platform’s rotation is now entirely along the sphere’s free axis. Putting $\vect e = -\vect k$ into (1), with $\vect k$ the upward vertical, the platform’s contribution cancels itself.

\begin{align*}\vect\omega_A &= \omega\vect k - \left(-\omega\right)\left(-\vect k\right) + \omega_0\left(-\vect k\right) \\ &= -\omega_0\vect k\end{align*}

The sphere therefore contributes $-I_0\omega_0$ to the vertical angular momentum, and the platform has to make up twice its original share.

I_0\omega_0 = I\omega - I_0\omega_0

\omega = \frac{2I_0\omega_0}{I}

(4)

The sphere’s kinetic energy is $\frac{1}{2}I_0\omega_0^2$ again, unchanged from the start, so every bit of the new energy sits in the platform.

A_{180^\circ} = \frac{1}{2}I\omega^2

Substituting (4) leaves one power of $I$ in the denominator.

A_{180^\circ} = \frac{1}{2}I\cdot\frac{4I_0^2\omega_0^2}{I^2}

A_{180^\circ} = \dfrac{2I_0^2\omega_0^2}{I}

(5)

Step 4The moment that keeps the axis upright

The machine from above after the quarter turn, the axis OO' end on. The sphere's spin angular momentum L_0 lies horizontal along BB'; over the turn dφ it swings to the dashed position, the closing side d L having length L_0 dφ. — The machine from above after the quarter turn, the axis $OO'$ end on. The sphere’s spin angular momentum $\vect L_0$ lies horizontal along $BB'$ ; over the turn $d\varphi$ it swings to the dashed position, the closing side $d\vect L$ having length $L_0\,d\varphi$ .

Of the machine’s total angular momentum, the vertical parts are constant, since $L_z$ is conserved and the platform turns steadily. What does change is the sphere’s spin angular momentum $L_0=I_0\omega_0$ , which is horizontal and is swept round by the platform. The figure gives its rate of change as a magnitude times an angular rate, and that rate must be supplied by external forces at the bearing.

\left|\frac{d\vect L}{dt}\right| = L_0\,\frac{d\varphi}{dt} = I_0\omega_0\,\omega

The platform turns at the rate (2) found for the quarter turn, so the moment the bearing must supply is

N = I_0\omega_0\cdot\frac{I_0\omega_0}{I+I_0}

N = \dfrac{I_0^2\omega_0^2}{I+I_0}

(6)

Check

Audit the $180^\circ$ state against the conservation law it was built from: the platform carries $I\cdot 2I_0\omega_0/I = 2I_0\omega_0$ and the sphere carries $-I_0\omega_0$ , and the two add to $I_0\omega_0$ , the value fixed at the start. Now let the platform be enormously heavy, $I\gg I_0$ . Both works fall to zero, and that is right: a platform that cannot be spun up leaves the sphere with the same $\frac{1}{2}I_0\omega_0^2$ it started with at every stage, so the motor has nothing to pay for. The moment $N$ vanishes in the same limit, and $I_0^2\omega_0^2/(I+I_0)$ does carry the units of a moment, since $I\omega^2$ is an energy and an energy has the units of a torque.

Remark

This machine is a control moment gyroscope, and the two answers say why spacecraft are built with them. No external moment about the vertical is applied anywhere; the motor only tilts a spinning rotor, and the platform turns anyway, because the vertical angular momentum has to stay put and tilting the rotor is the one way to move it between the two bodies. What the pedestal must then withstand, part (b), is a purely horizontal moment perpendicular to the rotor’s axis, and it does no work at all, being at right angles to the platform’s own rotation. That is the price of the trick: no propellant is thrown away to turn the platform, only the motor’s own work, but the mounting has to carry a gyroscopic moment that grows with the square of the rotor’s angular momentum.