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

(a) The gravitational potential energy decreases by 0.882 J. (b) $\omega = 5.42~rad/s$ (c) $v = 5.42~m/s$ (d) $v = 4.43 ~m/s$ The speed of a point on the end of the meter stick is greater than an object falling freely from one meter.
(a) The center of mass falls a distance of 0.500 meters. $PE = mgh = (0.180~kg)(9.80~m/s^2)(-0.500~m)$ $PE = -0.882~J$ The gravitational potential energy decreases by 0.882 J. (b) We can use conservation of energy to find the angular speed. $KE = PE$ $\frac{1}{2}I\omega^2 = 0.882~J$ $\frac{1}{2}(\frac{1}{3}ML^2)\omega^2 = 0.882~J$ $\omega = \sqrt{\frac{(6)(0.882~J)}{ML^2}}$ $\omega = \sqrt{\frac{(6)(0.882~J)}{(0.180~kg)(1.00~m)^2}}$ $\omega = 5.42~rad/s$ (c) $v = \omega ~R = (5.42~rad/s)(1.00~m)$ $v = 5.42~m/s$ (d) $v = \sqrt{2gh} = \sqrt{(2)(9.80~m/s^2)(1.00~m)}$ $v = 4.43 ~m/s$ The speed of a point on the end of the meter stick is greater than an object falling freely from one meter.