Find the sphere's total kinetic energy...?

A solid sphere of mass 4.0 kg and radius 0.12 m is at rest at the top of a ramp inclined 15(degrees). It rolls to the bottom without slipping. The upper end of the ramp is 1.2 m higher than the lower end. Find the sphere's total kinetic energy when it reaches the bottom.





So I know I need to use the equation





KE = 1/2 I*w+1/2Mv^2





and I know how to find the "I" and the "v", but I don't know how to find the w, I know w=(change in degrees)/(change in t), but I don't know what I would plug into that equation.





Just help me find the w, and I can do the rest! And show me equations and numbers, not words and sentences! Thanks.

Find the sphere's total kinetic energy...?
The sphere's total kinetic energy at the bottom has to equal the gravitational energy at the top, so it simply becomes mgh = KE.





Plug in the m=4 kg, g = 9.8 m/s^2, and h=1.2 and you're done.





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If you want more detail, you can find much more below:





First, I would make one change to your equation above - don't forget that w is squared.





KE = 1/2 I*w^2+1/2Mv^2





And here's the part you were missing v = rw





That tells you that linear velocity (v) equals the angular velocity times the radius.





since v = rw, then v^2 = r^2w^2


rearranging for w^2 gives





w^2 = v^2/r^2





I'm not sure how you're going to get the v at the bottom. I would do it this way:





mgh = 1/2 I*w^2+1/2Mv^2





and then substitute I=2/5mr^2 and w^2 in and solve for v.





mgh = 1/2 (2/5 mr^2)v^2/r^2 + 1/2 mv^2





gh = 1/5v^2 +1/2v^2


gh = 7/10 v^2


gh 10/7 = v^2


sqroot(gh 10/7) = v





For your numbers





v = sqroot(9.8 x 1.2 x 10 / 7) = 4.1 m/s





Then fill this into the equation from the top





KE = 1/2 I*w^2+1/2Mv^2


KE = 1/2 (2/5 mr^2)v^2/r^2 + 1/2mv^2





again the r^2 cancels out leaving





KE = 1/5mv^2 + 1/2mv^2





Just fill in the mass of 4 kg and the v of 4.1 m/s and solve