# Chapter 8 - Problems - Page 312: 27

(a) A length of 3.14 meters of rope unwinds in one revolution. (b) The rope does $15.7~J$ of work on the wheel. (c) The torque on the wheel due to the rope is $2.50~N \cdot m$ (d) In 1.00 revolution, the angular displacement is $6.28~rad$ (e) The product $\tau~\Delta \theta$ is $15.7~J$, which is equal to the work done on the wheel by the rope, as was calculated in part (b).

#### Work Step by Step

(a) We can find $d$, the length of rope that unwinds in one revolution: $d = 2\pi~R = (2\pi)(0.50~m) = 3.14~m$ A length of 3.14 meters of rope unwinds in one revolution. (b) We can find the work done by the rope: $Work = F~d = (5.00~N)(3.14~m) = 15.7~J$ The rope does $15.7~J$ of work on the wheel. (c) We can find the torque on the wheel due to the rope: $\tau = R~F = (0.50~m)(5.00~N) = 2.50~N \cdot M$ The torque on the wheel due to the rope is $2.50~N \cdot m$ (d) In 1.00 revolution, the angular displacement is $2\pi~rad$ which is $6.28~rad$ (e) We can find the product $\tau~\Delta \theta$: $\tau~\Delta \theta = (2.50~N \cdot m)(6.28~rad) = 15.7~J$ The product $\tau~\Delta \theta$ is $15.7~J$, which is equal to the work done on the wheel by the rope, as was calculated in part (b).

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