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

2.5 s

Step-by-step explanation:

We use the equation of motion for rotational motion:

`\theta = \omega_it + (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}) + (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}`

Answer 2

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

Explanation: Please see the attachments below