## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 5 - Problems - Page 188: 53

#### Answer

(a) It takes 14.5 days for the rotor to come to rest. (b) The rotor spins through $4.0\times 10^{10}$ revolutions.

#### Work Step by Step

(a) We can find the time it takes for the rotor to come to rest: $\omega_f = \omega_0+\alpha~t$ $t = \frac{\omega_f - \omega_0}{\alpha}$ $t = \frac{0 - 5.0\times 10^5~rad/s}{-0.40~rad/s^2}$ $t = 1.25\times 10^6~s$ $t = 347.2~hours$ $t = 14.5~days$ It takes 14.5 days for the rotor to come to rest. (b) We can find the magnitude of the angular displacement: $\Delta \theta = \frac{1}{2}\alpha~t^2$ $\Delta \theta = \frac{1}{2}(0.40~rad/s^2)(1.25\times 10^6~s)^2$ $\Delta \theta = 2.5\times 10^{11}~rad$ We can express the angular displacement in units of revolutions: $\Delta \theta = 2.5\times 10^{11}~rad\times \frac{1~rev}{2\pi~rad} = 4.0\times 10^{10}~rev$ The rotor spins through $4.0\times 10^{10}$ revolutions.

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