Answer
(a) At the bottom of the hill, the wheel is moving at a speed of 29.3 m/s.
(b) $K = 1930~J$
Work Step by Step
(a) We can use conservation of energy to solve this question. The potential energy and the initial kinetic energy at the top will be equal to the total kinetic energy at the bottom.
$\frac{1}{2}mv_2^2+\frac{1}{2}I\omega_2^2 = mgh+\frac{1}{2}mv_1^2+\frac{1}{2}I\omega_1^2$
$\frac{1}{2}mv_2^2+\frac{1}{2}(mR^2)(\frac{v_2}{R})^2 = mgh+\frac{1}{2}mv_1^2+\frac{1}{2}(mR^2)(\frac{v_1}{R})^2$
$\frac{1}{2}v_2^2+\frac{1}{2}v_2^2 = gh+\frac{1}{2}v_1^2+\frac{1}{2}v_1^2$
$v_2^2 = gh+v_1^2$
$v_2 = \sqrt{gh+v_1^2}$
$v_2 = \sqrt{(9.80~m/s^2)(75.0~m)+(11.0~m/s)^2}$
$v_2 = 29.3~m/s$
At the bottom of the hill, the wheel is moving at a speed of 29.3 m/s.
(b) We can find the total kinetic energy at the bottom of the hill.
$K = \frac{1}{2}mv_2^2+\frac{1}{2}I\omega_2^2$
$K = \frac{1}{2}mv_2^2+\frac{1}{2}(mR^2)(\frac{v_2}{R})^2$
$K = \frac{1}{2}mv_2^2+\frac{1}{2}mv_2^2$
$K = mv_2^2$
$K = (2.25~kg)(29.3~m/s)^2$
$K = 1930~J$
