Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 8 - Dynamics II: Motion in a Plane - Conceptual Questions: 5

Answer

(a) The tension in string A is greater than the tension in string B. (b) The tension in string A is less than the tension in string B.

Work Step by Step

The centripetal force $F_c$ is the net force which keeps objects moving around in a circle. In general: $F_c = \frac{mv^2}{r} = m\omega^2~r$ (a) We can write an expression for $T_A$: $\sum F = \frac{mv^2}{r}$ $T_A+mg = \frac{mv^2}{r}$ $T_A = \frac{mv^2}{r} - mg$ We can write an expression for $T_B$: $\sum F = \frac{mv^2}{2r}$ $T_B+mg = \frac{mv^2}{2r}$ $T_B = \frac{mv^2}{2r} - mg$ Since $\frac{mv^2}{r}$ is greater than $\frac{mv^2}{2r}$, then $T_A$ is greater than $T_B$. The tension in string A is greater than the tension in string B. (b) We can write an expression for $T_A$: $\sum F = m\omega^2~r$ $T_A+mg = m\omega^2~r$ $T_A = m\omega^2~r - mg$ We can write an expression for $T_B$: $\sum F = m\omega^2~(2r)$ $T_B+mg = 2m\omega^2~r$ $T_B = 2m\omega^2~r - mg$ Since $m\omega^2~r$ is less than $2m\omega^2~r$, then $T_A$ is less than $T_B$. The tension in string A is less than the tension in string B.
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