1. A solid sphere with uniform density and a mass of 0.0669 kg rolls down a track with two rails, so that the sphere touches the track at two points instead of one, and the lowest part of the sphere is slightly below the points of contact, as shown in the figure below. a. Given that the speed of the sphere when it leaves the track is v = 1.54 m/s, the diameter of the sphere is D = 2.50 cm, and the vertical distance from one of the points of contact to the top of the sphere is L = 2.4 cm, determine 1) the angular speed of the sphere just before it leaves the track, and 2) the total kinetic energy of the sphere just before it leaves the track. Divide this total kinetic energy by the mass and compare the result with the number obtained in class.

b. If the sphere had rolled down a plane with a flat surface instead of two rails, would the final speed of the sphere have been the same, less than, or greater than 1.54 m/s? Assume the sphere to be released from the same height in both cases.

The sphere rotates about the contact points. It means the radius of gyration of the center is

$R= L-D/2 =2.4 -2.5/2 =1.15 cm$

(it is like when for an instant the sphere rotates around the plane that section the sphere and contains both points of contact) See figure below

$v = \omega*R$

$\omega = V/R =1.54/0.0115 =13.39 rad/sec$

Momentum of inertia is

$I =(2/5)*m*(D/2)^2 =(2/5)*0.0669*(0.025/2)^2 =4.18*10^{-6} kg*m^2$

$E_c = I*\omega^2/2 + m*V^2/2 = 3.75*10^{-4} + 7.93*10^{-2} =7.97*10^{-4} Joule$

b) when the sphere rolls with one contact point, the radius of gyration is greater, it means it will acquire more rotational kinetic energy, it means less translational kinetic energy, it means less speed than 1.54 m/s

(rotational kinetic energy + translational kinetic energy $= M*g*H$