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

Answer:

Option 4. τ = 2.1 m.N

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:            

$\omega^{2} = \omega_{0}^{2} + 2 \alpha \theta$

$\alpha = \frac{\omega^{2} - \omega_{0}^{2}}{2 \theta}$  

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

$\alpha = 0.175 \frac{rad}{s^{2}}$ (2)          

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

$\tau = I \cdot \alpha = 12 kg\cdot m^{2} \cdot 0.175 \frac{rad}{s^{2}} = 2.1 m \cdot N$  

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

Have a nice day!