Question

A wheel has a rotational inertia of 12 kg · m2 . Initially it is rotating with an angular velocity of 5 rad/s. A constant net torque is then applied, causing its angular velocity to increase from 5 rad/s to 6 rad/s during the time interval when the wheel turns through 5 revolutions. Which is closest to the magnitude of the net torque?

1. 0.016 m · N
2. 0.57 m · N
3. 3.6 m · N
4. 2.1 m · N
5. 0.18 m · N

Answer 1

Option 4. τ = 2.1 m.N

Step-by-step explanation:

To calculate the magnitude of the net torque (τ) we are going to use the next equation:

`\tau = I \cdot \alpha` (1)

where I: rotational inertia and α: angular acceleration

We can to find the angular acceleration from the angular velocity equation:

`\omega^(2) = \omega_(0)^(2) + 2 \alpha \theta`

`\alpha = (\omega^(2) - \omega_(0)^(2))/(2 \theta)`

`\alpha = \frac{(6 (rad)/(s))^(2) - (5 (rad)/(s))^(2)}{2 (\frac {5rev \cdot 2\pi rad}{1 rev})}`

`\alpha = 0.175 (rad)/(s^(2))` (2)

Now, introducing the angular acceleration calculated (2) in equation (1), we can find the net torque:

`\tau = I \cdot \alpha = 12 kg\cdot m^(2) \cdot 0.175 (rad)/(s^(2)) = 2.1 m \cdot N`

Hence, the correct answer is option 4: 2.1 m.N

Have a nice day!