Answer
(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.
Work Step by Step
(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.
