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Chapter 10: Problem 41

A wheel turns through 2.0 revolutions while accelerating from rest at \(18 \mathrm{rpm} / \mathrm{s}\). (a) What's its final angular speed? (b) How long does it take?

Short Answer

Expert verified
The wheel's final angular speed is \(13.72 \, \text{rad/s}\) and it takes \(3.65 \, \text{s}\) to reach this speed.

Step by step solution

01

Convert angular acceleration to rad/s^2

We're given the angular acceleration in rpm/s. To work with it effectively, we want it in rad/s^2. We can use the conversion factor of \(2\pi\) rad/rev (since there are \(2\pi\) radians in one revolution) and another conversion factor of 1 min/60 s (to convert minutes to seconds), giving us \(\alpha = 18 \, \text{rpm/s} \times \frac {2\pi \, \text{rad}}{\text{rev}} \times \frac {1 \, \text{min}}{60 \, \text{s}} = 1.885 \, \text{rad/s}^2\).
02

Compute the final angular speed

We use the equation for the final angular velocity, \(\omega_f = \omega_i + \alpha t\). However, instead of time we have the number of revolutions, hence we need to convert that into radians (since 1 rev = \(2\pi\) rad). Therefore, 2.0 rev = \(4\pi\) rad. We assume the wheel starts from rest, so \(\omega_i = 0\). By rearranging the equation for displacement, \(\theta = \omega_i t + 0.5 \alpha t^2\), we get \( t = \sqrt{\frac {2 \theta}{\alpha}}\). Substituting these into the equation for \(\omega_f\), we get \(\omega_f = \alpha \cdot \sqrt{\frac {2 \theta}{\alpha}} = \sqrt{2 \alpha \theta} = \sqrt{2(1.885 \, \text{rad/s}^2)(4\pi \, \text{rad})} = 13.72 \, \text{rad/s}\).
03

Compute the time

Now we substitute the angular acceleration and displacement values into the previous formula for time, \( t = \sqrt{\frac {2 \theta}{\alpha}} = \sqrt{\frac {2(4\pi \, \text{rad})}{1.885 \, \text{rad/s}^2}} = 3.65 \, \text{s}\). So, it takes 3.65 seconds.

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