## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 5 - Problems - Page 188: 55

#### Answer

The radial acceleration component is $2.45~m/s^2$ The tangential acceleration component is $2.54~m/s^2$ The tension in the string is $11.9~N$

#### Work Step by Step

We can find the radial acceleration component: $a_c = \frac{v^2}{r}$ $a_c = \frac{(1.40~m/s)^2}{0.800~m}$ $a_c = 2.45~m/s^2$ The radial acceleration component is $2.45~m/s^2$ We can find the tangential acceleration component: $mg~sin~\theta = ma_t$ $a_t = g~sin~\theta$ $a_t = (9.80~m/s^2)~sin~15.0^{\circ}$ $a_t = 2.54~m/s^2$ The tangential acceleration component is $2.54~m/s^2$ We can consider the net force on the bob directed along the line toward the center of the circle. We can find the tension in the string: $\sum~F = \frac{mv^2}{r}$ $T - mg~cos~\theta = \frac{mv^2}{r}$ $T = mg~cos~\theta + \frac{mv^2}{r}$ $T = (1.00~kg)(9.80~m/s^2)~cos~15.0^{\circ} + \frac{(1.00~kg)(1.40~m/s)^2}{0.800~m}$ $T = 11.9~N$ The tension in the string is $11.9~N$

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