Rotational Energy, Rolling downhill

[Elysian]

Homework Statement

A solid steel sphere of radius 10 cm and a mass of 1.5 kg rolls down a 1.25 m incline that makes an angle of 20 degrees with the horizontal. Calculate the linear velocity of the following points relative to the ground, when it reaches the bottom of the incline.
a) The center of the sphere
b) A point at the top edge of the sphere
c) A point at the bottom of the sphere

Homework Equations

mgh = .5Iw²+.5mv²

I = 2/5(mr^2)

The Attempt at a Solution

Ok so using the equation above, energy, I substituted in the values, using wr = v1 for tangential velocities, putting W = v1/r for the angular velocity, But the velocity (v2) of the .5 mv², is different than the tangential, as the velocity of the kinetic energy (normal) is V2, V2 = .5*V1, which is derived from the fact that a point on a sphere goes twice the distance compared to the center of mass, the point on the spheres velocity is V1, v2 is the speed of the Center of mass.

Substituting we get

mgh = .5(.4*m*r^2)(.5V1/r)² + .5*m*V1²

which reduces down into

gh = .1(v1^2/4) + .5v1^2, and I get 2.70 m/s for the linear speed of the center of mass... is this right?

Also would the answers for b and c be the same?

 

[Delphi51]

I'm not running on all cylinders tonight, but I'm pretty sure you are off the mark on your assumption that v1 and v2 are different. Please take a look at this explanation:
http://teacher.pas.rochester.edu/phy121/lecturenotes/Chapter12/Chapter12.html
It should clear up any difficulty with parts b and c as well.