A circular disc of moment of inertia about a perpendicular axis through the centre is 8 kg m². It is smoothly pivoted from the centre and initially rotates at constant angular speed of 40 rad s⁻¹. When a constant torque is applied for 10 s, the angular speed of the disc reduces to 20 rad s⁻¹. What is the magnitude of the applied torque?

2024 A/L Physics Past Paper · Question 08 (English Medium)

The torque that slows a spinning disc

A circular disc of moment of inertia about a perpendicular axis through the centre is $8\ \text{kg m}^2$ . It is smoothly pivoted from the centre and initially rotates at constant angular speed of $40\ \text{rad s}^{-1}$ . When a constant torque is applied for $10\ \text{s}$ , the angular speed of the disc reduces to $20\ \text{rad s}^{-1}$ . What is the magnitude of the applied torque?

(1) $8\ \text{N m}$
(2) $16\ \text{N m}$
(3) $32\ \text{N m}$
(4) $40\ \text{N m}$
(5) $80\ \text{N m}$

Answer and explanation

Answer: (2) $16\ \text{N m}$

Given information

What we know:

The moment of inertia of the disc is $I = 8\ \text{kg m}^2$ .

The angular speed falls from $40\ \text{rad s}^{-1}$ to $20\ \text{rad s}^{-1}$ .

This takes $10\ \text{s}$ , under a constant torque.

What the question asks

We want: the magnitude of the torque $\tau$ .

What each letter means:

$I$ ( $\text{kg m}^2$ ): the moment of inertia of the disc.

$\omega_1$ , $\omega_2$ ( $\text{rad s}^{-1}$ ): the angular velocity at the start and at the end.

$\alpha$ ( $\text{rad s}^{-2}$ ): the angular acceleration.

$\tau$ ( $\text{N m}$ ): the torque.

Relevant theory

The rate of change of angular velocity of a moving body at a moment is the angular accederation at that moment.

From the relation $L = I\omega$ $\frac{\Delta L}{\Delta t} = \frac{\Delta I\omega}{\Delta t}$ , I does not change with time $\therefore \frac{\Delta L}{\Delta t} = I\frac{\Delta\omega}{\Delta t}$ $\tau = I\alpha$

The first quote says that the angular acceleration is the rate of change of the angular velocity, $\alpha = \dfrac{\Delta\omega}{\Delta t}$ . The second quote gives the rotational form of Newton's second law, $\tau = I\alpha$ .

For example, a potter's wheel is spinning fast. The potter presses a hand on its rim and it slows down steadily.

A heavier wheel, or a faster slow-down, needs a harder press.

Required equations

$\alpha = \dfrac{\omega_2 - \omega_1}{t}$

$\tau = I\alpha$

Applying it to this question

$-2$

$\alpha = \dfrac{20 - 40}{10}$

$2$

$|\alpha|$

$16$

$\tau = 8 \times 2$

Reasoning and calculation

$I * (w1 - w2) / t$

Now match it with the five options.

Option (2) is $16\ \text{N m}$ . It matches.

The others would make the disc slow down by $10$ , $40$ , $50$ or $100\ \text{rad s}^{-1}$ in $10\ \text{s}$ , not by $20$ .

Checks

Check the units. $\text{kg m}^2 \times \text{rad s}^{-2}$ is $\text{kg m}^2\,\text{s}^{-2}$ , which is $\text{N m}$ .

Does it make sense?

Each second the speed drops by $2\ \text{rad s}^{-1}$ : $40$ , $38$ , $36$ , … and after $10$ seconds it is $20$ . That fits.

Conclusion

$\tau = 16\ \text{N m}$

The torque is $\tau = 16\ \text{N m}$ .

The answer is (2).

Turn the change in angular speed into $\alpha$ first, then use $\tau = I\alpha$ .

It works just like $F = ma$ for straight-line motion.

Exam tip

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