# Chapter 10 - Dynamics of Rotational Motion - Problems - Exercises - Page 334: 10.62

(a) $a = 1.12~m/s^2$ (b) $T = 14.0~N$

#### Work Step by Step

(a) We can set up a torque equation for the flywheel. $\tau = I\alpha$ $T~R = I(\frac{a}{R})$ $T = \frac{Ia}{R^2}$ We can use this expression in the force equation for the block. Let $m$ be the mass of the block. $\sum F = ma$ $mg~sin(\theta)-F_f-T = ma$ $mg~sin(\theta)-mg~\mu_k~cos(\theta) = ma+T$ $mg~sin(\theta)-mg~\mu_k~cos(\theta) = ma+ \frac{Ia}{R^2}$ $a = \frac{mg~sin(\theta)-mg~\mu_k~cos(\theta)}{m+ \frac{I}{R^2}}$ $a = \frac{(5.00~kg)(9.80~m/s^2)~sin(36.9^{\circ})-(5.00~kg)(9.80~m/s^2)(0.25)~cos(36.9^{\circ})}{5.00~kg+ \frac{0.500~kg~m^2}{(0.200~m)^2}}$ $a = 1.12~m/s^2$ (b) $T = \frac{Ia}{R^2}$ $T = \frac{(0.500~kg~m^2)(1.12~m/s^2)}{(0.200~m)^2}$ $T = 14.0~N$

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