Question

A wheel has a constant angular acceleration of 3.5 rad/s2. During a certain 9.0 s interval, it turns through an angle of 220 rad. Assuming that the wheel started from rest, how long had it been in motion before the start of the 9.0 s interval

Answer 1

Answer: It had been 2.48 s before the start of the 9.0 s interval

Explanation: Please see the attachments below

Answer 2

Answer:

2.5 s

Explanation:

We use the equation of motion for rotational motion:

$\theta = \omega_it + \frac{1}{2}\alpha t^2$

where

• $\theta$ is angular displacement,
• $\omega_i$ is initial angular velocity
• $\alpha$ is the angular acceleration
• $t$ is the time

Substituting values,

$220 = \omega_i(9.0 \text{s}) + \frac{1}{2}(3.5\ \text{rad/s}^2) (9.0\ \text{s})^2$

$\omega_i = 8.69\ \text{rad/s}$

Before the 9.0 s interval, this value is the final angular velocity. The initial angular velocity is 0 rad/s since it started from rest.

We use the equation:

$\omega_f = \omega_i+\alpha t$

Note that $\omega_f$ is the value we got earlier and $\omega_i$ is the initial angular velocity and is equal to 0 rad/s.

$8.694 \ \text{rad/s}= (0\ \text{rad/s})+(3.5\ \text{rad/s}^2)(t)$

$t = 2.5\ \text{s}$