2018 A/L Physics Past Paper · Question 34 (English Medium)

A spider landing on a rotating disc — L and ω of the disc

A disc freely rotates without friction at a certain angular speed, about a fixed vertical axis normal to the plane of the disc and passing through its centre. At time $t = 0$ , a spider vertically lowers itself with negligible speed onto the rim of the rotating disc as shown in figure, and becomes rest. Variation of the magnitudes of the angular momentum $(L)$ and the angular speed $(\omega)$ of the disc only with time $(t)$ is best represented by,

(1) Figure (1)
(2) Figure (2)
(3) Figure (3)
(4) Figure (4)
(5) Figure (5)

Answer and explanation

Answer: (3) Figure (3)

What the question asks

We want: the graph of $L$ and $\omega$ of the disc only, before and after the spider lands.

Relevant theory

The total angular momentum of a system remains constant provided no external torque acts on the system.

The angular momentum (L) of a lamina moving with an angular velocity $\omega$ about an axis passing through O is given by

Required equations

$I\omega_0 = (I + mr^2)\,\omega_1$

$\omega_1 = \frac{I}{I + mr^2}\,\omega_0 < \omega_0$

$L_{disc} = I\omega_1 < I\omega_0$

Reasoning and calculation

The flat axis is time $t$ .

The up axis is the size of $L$ and $\omega$ , for the disc only.

Before the spider ( $t < 0$ ): the disc spins alone.

Nothing outside twists it.

So $L_0 = I\omega_0$ and $\omega_0$ stay the same. Two flat lines.

At $t = 0$ the spider lands on the rim.

Very quickly it starts to turn with the disc.

To turn, the spider needs angular momentum.

It gets it from the disc. So the disc's $L$ goes down.

The disc's $I$ does not change. So its $\omega$ goes down too.

After that ( $t > 0$ ) nothing outside twists it. So both stay at their new, lower values.

$0.8\,\omega_0$

$\omega_1 = \frac{I}{1.25\,I}\,\omega_0$

$0.8\,L_0$

$L_1 = I \times 0.8\,\omega_0$

So both graphs step down at $t = 0$ , by the same ratio $\dfrac{I}{I + mr^2}$ .

Then both stay flat.

They never reach zero.

Reading the options

Only figure (3) shows both dropping at $t = 0$ and then staying flat.

Why the other options fail

This option says: graph (1): the disc's $L$ stays the same, only $\omega$ drops.

But the disc gives some $L$ to the spider. So its $L$ must drop too.

So it is wrong.

This option says: graph (2): $\omega$ goes up.

But the disc slows down when the spider lands. $\omega$ cannot go up.

This option says: graph (4): $L$ drops, but $\omega$ stays the same.

But $L = I\omega$ , and the disc's $I$ does not change. If $L$ drops, $\omega$ must drop too.

This option says: graph (5): nothing changes.

But the disc must give the spider some angular momentum. So something must drop.

Conclusion

At $t = 0$ the disc's $L$ and $\omega$ both drop by the same ratio.

Then they stay the same.

The answer is figure (3).

Saying "angular momentum stays the same" for the disc alone.

It stays the same for disc + spider together.

The question asks about the disc only. That goes down.

Explore

The disc spins at a steady angular speed, with no friction.

At $t = 0$ the spider lands on the rim and starts to turn with the disc.

The angular momentum the spider needs comes from the disc, so the disc’s $L$ falls.

The disc’s $I$ does not change, so its $\omega$ falls too, and then both stay constant: graph (3).

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