A spinning wheel is slowed down by a brake, giving it a constant angular acceleration of 25.60 rad/s2. During a 4.20-s time inte

Question

rewona

9 days ago

6

A spinning wheel is slowed down by a brake, giving it a constant angular acceleration of 25.60 rad/s2. During a 4.20-s time inte

rval, the wheel rotates through 62.4 rad. What is the angular speed of the wheel at the end of the 4.20-s interval?

Physics

1 answer:

Yuliya22 [1.1K] · Answer 1 · 2026-02-24T21:39:04+03:00

We will use the equations of rotational kinematics,

$\theta =\theta _{0} + \omega_{0} t+ \frac{1}{2}\alpha t^2$ (A)

$\omega^2= \omega^2_{0} +2\alpha\theta$ (B)

Here, $\theta$ and $\theta _{0}$ denote the final and initial angular displacements, respectively, whereas $\omega$ and $\omega_{0}$ represent final and initial angular velocities, and $\alpha$ is the angular acceleration.

We are provided with $\alpha = - 25.60 \ rad/s^2, \theta = 62.4 \ rad$ and $t = 4.20 \ s$ .

By substituting these values into equation (A), we have

$62.4 \ rad = 0 + \omega_{0} 4.20 \ s + \frac{1}{2} (- 25.60 \ rad/s^2) ( 4.20)^2 \\\\ \omega_{0} = \frac{220.5+ 62.4 }{4.20} =67.4 \ rad/s$

Now, using equation (B),

$\omega^2=(67.4 \ rad/s)^2 + 2 (- 25.60 \ rad/s^2)62.4 \ rad \\\\\ \omega = 36.7 \ rad/s$

This indicates that the wheel's angular speed at the 4.20-second mark is 36.7 rad/s.