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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 final angular speed is \(13.84 rad/s\) and it takes \(7.34 s\).

Step by step solution

01

Convert Revolutions to Radians

Convert the total angular displacement from revolutions to radians by using the factor \(1 rev = 2 \pi rad\). Therefore, \(2.0 revs = 2.0 * 2 \pi rad = 4 \pi rad\).
02

Convert Angular Acceleration to rad/s虏

Angular acceleration given is in \(\mathrm{rpm}/s\), so convert that to \(\mathrm{rad}/s虏\) using the factor \(1 min=60s\) and \(1 rev = 2 \pi rad\). Thus, \(\alpha = 18 rpm/s = 18 * \((2 \pi rad)/60s\)= 1.88496 \mathrm{rad/s}^2\).
03

Find the Final Angular Speed

Because the wheel starts from rest, \(\omega_i=0\). But we know that \(\omega_f = \omega_i + \(\alpha t\)\). As we know that the motion endured two full revolutions, we can equate that with the equation for rotational motion under constant angular acceleration \(\theta=\(\omega_i t + 0.5 \alpha t^2\)\). Solving for angular velocity using \(\theta = 4 \pi\).\n Therefore, \(\omega_f = \sqrt{2 \alpha \theta}\) = \(\sqrt{2 * 1.88496 rad/s^2 * 4 \pi rad} = 13.84 \mathrm{rad/s}\)
04

Determine the time

Use the equation: \(t = (\omega_f- \omega_i) / \alpha\), where \(\omega_i = 0\), \(\omega_f = 13.84 rad/s\) and \(\alpha = 1.88496 rad/s^2\). Therefore, the time \(t = (13.84 rad/s) / (1.88496 rad/s^2) = 7.34 s\)

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