## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 5 - Problems - Page 186: 22

#### Answer

(a) $T = \frac{mg}{cos~\phi}$ (b) The period of the pendulum is $2\pi~\sqrt{\frac{L~cos~\phi}{g}}$

#### Work Step by Step

(a) The vertical component of the string's tension is equal in magnitude to the bob's weight. We can find an expression for the tension $T$: $T~cos~\phi = mg$ $T = \frac{mg}{cos~\phi}$ (b) The horizontal component of the tension provides the centripetal force to keep the ball moving around in a circle. We can find the ball's angular speed: $T~sin~\phi =m~\omega^2~r$ $(\frac{mg}{cos~\phi})~sin~\phi =m~\omega^2~L~sin~\phi$ $\frac{mg}{cos~\phi} =m~\omega^2~L$ $\frac{g}{L~cos~\phi} =\omega^2$ $\omega = \sqrt{\frac{g}{L~cos~\phi}}$ We can find the period $P$ of the pendulum: $P = \frac{2\pi}{\omega}$ $P = \frac{2\pi}{\sqrt{\frac{g}{L~cos~\phi}}}$ $P = 2\pi~\sqrt{\frac{L~cos~\phi}{g}}$ The period of the pendulum is $2\pi~\sqrt{\frac{L~cos~\phi}{g}}$

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