## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 5 - Problems - Page 190: 79

#### Answer

The tension in the string connecting the two blocks is $2.0~N$ The tension in the string connecting the inner block to the pole is $3.8~N$

#### Work Step by Step

The angular speed is $(1.5~rev/s)(2\pi~rad/rev) = 3\pi~rad/s$ We can find the tension $T_2$ in the string connecting the two blocks. Note that this tension provides the centripetal force to keep the outer block moving around in a circle: $T_2 = m_2~\omega^2~r_2$ $T_2 = (0.030~kg)~(3\pi~rad/s)^2~(0.75~m)$ $T_2 = 2.0~N$ The tension in the string connecting the two blocks is $2.0~N$ We can find the tension $T_1$ in the string connecting the inner block to the pole: $\sum F = m_1~\omega^2~r_1$ $T_1-T_2 = m_1~\omega^2~r_1$ $T_1 = T_2 + m_1~\omega^2~r_1$ $T_1 = 2.0~N + (0.050~kg)~(3\pi~rad/s)^2~(0.40~m)$ $T_1 = 3.8~N$ The tension in the string connecting the inner block to the pole is $3.8~N$.

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