## University Physics with Modern Physics (14th Edition)

(a) We can set up a torque equation for the wheel. $\tau = I\alpha$ $T~R = (\frac{1}{2}MR^2)(\frac{a}{R})$ $T = \frac{1}{2}Ma$ We can use this expression in the force equation for the person. Let $m$ be the mass of the person. $\sum F = ma$ $mg-T = ma$ $T = mg - ma$ $\frac{1}{2}Ma = mg - ma$ $M = \frac{2mg}{a} - 2m$ $M = \frac{2mg}{(\frac{g}{4})} - 2m$ $M = 6m$ $M = (6)(90.0~kg)$ $M = 540~kg$ (b) We can find the tension in the rope. $T = \frac{1}{2}Ma$ $T = \frac{1}{2}(540~kg)(\frac{9.80~m/s^2}{4})$ $T = 662~N$