#### Answer

(a) $v = 8.37~m/s$
$\omega = 24.3~rad/s$
(b) The ratio of the translational to rotational kinetic energy at the bottom is 2.5
(c) The translational speed and the ratio of kinetic energy does not depend on mass or radius. However, the rotational speed at the bottom does depend on the radius. A sphere with a smaller radius will have a faster rotational speed. A sphere with a larger radius will have a slower rotational speed.

#### Work Step by Step

(a) We can find the vertical height $h$ that the sphere drops is it rolls along the incline:
$h = d~sin(\theta) = (10.0~m)~sin(30.0^{\circ}) = 5.0~m$
As the sphere rolls down the incline, the initial potential energy is converted into translational and rotational kinetic energy. We can use conservation of energy to solve this question.
$KE_t + KE_r = PE$
$\frac{1}{2}mv^2+ \frac{1}{2}I\omega^2 = mgh$
$\frac{1}{2}mv^2+ \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = mgh$
$\frac{1}{2}mv^2+ \frac{1}{5}mv^2 = mgh$
$\frac{7}{10}v^2 = gh$
$v = \sqrt{\frac{10gh}{7}} = \sqrt{\frac{(10)(9.80~m/s^)(5.0~m)}{7}}$
$v = 8.37~m/s$
The translational speed at the bottom of the incline is 8.37 m/s. We can use the translational speed to find the rotational speed.
$\omega = \frac{v}{r} = \frac{8.37~m/s}{0.345~m} = 24.3~rad/s$
(b) We can find the ratio of the translational to rotational kinetic energy at the bottom.
$\frac{KE_t}{KE_r} = \frac{\frac{1}{2}mv^2}{\frac{1}{5}mv^2} = 2.5$
(c) The translational speed and the ratio of kinetic energy does not depend on mass or radius. However, the rotational speed at the bottom does depend on the radius. A sphere with a smaller radius will have a faster rotational speed. A sphere with a larger radius will have a slower rotational speed.