2021 A/L Physics Past Paper · Question 38 (English Medium)

Answer: (3) $25\ \text{rad s}^{-2}$

Given information

We know:

The moment of inertia of the flywheel is $I = 0.4\ \text{kg m}^2$.

It turns at a steady angular speed $\omega = 10\ \text{rad s}^{-1}$.

The motor has power $P = 100\ \text{W}$.

What the question asks

We want: the angular deceleration $\alpha$ after the motor is switched off.

The plan is in two steps.

First, find the torque that the motor gives. It balances the friction torque.

Second, let that torque alone slow the wheel.

Relevant theory

The rate at which a body is doing work or the rate at which it releases energy is called power

power = force × velocity

In translatory motion, force is defined as the rate of change of linear momentum. Likewise in rotational motion, torque or moment of the force is defined as the rate of change of angular momentum.

Required equations

$P = Fv = F r\omega = \tau\omega$

$\tau = I\alpha$

Applying it to this question

The wheel turns at a steady speed.

So its angular acceleration is zero, and the net torque is zero.

The motor torque is equal to the friction torque.

$100$

$P_f = P = 100$

$P_f = \tau\omega$

$10$

$\tau = \dfrac{P_f}{\omega} = \dfrac{100}{10}$

$I\alpha = \tau$

$25$

$\alpha = \dfrac{\tau}{I} = \dfrac{10}{0.4}$

All quantities are in SI units.

Power in W, speed in rad per s, torque in N m, and $I$ in kg m$^2$.

So $\alpha$ comes out in rad s$^{-2}$.

Reasoning and calculation

The torque from the motor is $10\ \text{N m}$.

With the motor off, this friction torque slows the wheel.

The angular deceleration is $\dfrac{10}{0.4} = 25\ \text{rad s}^{-2}$.

Option (3).

Checks

$P / (I * w)$

Does it make sense?

At $25\ \text{rad s}^{-2}$ the wheel stops in $\dfrac{10}{25} = 0.4\ \text{s}$.

That is quick for a small wheel with a strong friction torque of $10\ \text{N m}$.

The friction power at $10\ \text{rad s}^{-1}$ is $10 \times 10 = 100\ \text{W}$. It matches the motor.

Conclusion

$\alpha = \dfrac{P}{I\omega} = 25\ \text{rad s}^{-2}$

The friction torque is $\tau = \dfrac{P}{\omega} = 10\ \text{N m}$.

The angular deceleration is $\alpha = \dfrac{\tau}{I} = 25\ \text{rad s}^{-2}$.

The answer is (3).

At a steady angular speed, the motor torque equals the friction torque.

Find the torque from $P = \tau\omega$. Then use $\tau = I\alpha$ for the slowing.

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