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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

2 Answers

1 vote

Answer:

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}

User HolyBlackCat
by
6.8k points
4 votes

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

Explanation: Please see the attachments below

A wheel has a constant angular acceleration of 3.5 rad/s2. During a certain 9.0 s-example-1
A wheel has a constant angular acceleration of 3.5 rad/s2. During a certain 9.0 s-example-2
User Rimil Dey
by
6.9k points