Question

A solid homogeneous sphere of mass M = 4.70 kg is released from rest at the top of an incline of height H=1.21 m and rolls without slipping to the bottom. The ramp is at an angle of θ = 27.7o to the horizontal.

a) Calculate the speed of the sphere's CM at the bottom of the incline.​

b) Determine the rotational kinetic energy of the sphere at the bottom of the incline.

Answer 1

The equation for the energy is

$m g h = \frac{1}{2} mv^2 +\frac{1}{2}I \omega^2$

$m g h = \frac{1}{2} mv^2 +\frac{1}{2} (\frac{2}{5}mr^2)\left ( \frac{v}{r} \right )^2$

$m g h = \frac{1}{2} mv^2 +\frac{1}{5} mv^2$

$m g h = \frac{7}{10} mv^2$

$v = \sqrt{\frac{10}{7}gh}=\sqrt{\frac{10}{7}(9.8)(1.21)} =4.12 m/s$

b)

The rotational kinetic energy is

$k=\frac{1}{5} mv^2 = \frac{1}{2}(4.7)(4.12)^2 = 39.81 J$