Rolling Sphere On Incline

[Randomized10]

Homework Statement

An 8.6-cm-diameter, 320 g solid sphere is released from rest at the top of a 1.7-m-long, 17 degree incline. It rolls, without slipping, to the bottom. What is the angular velocity of the sphere at the bottom?

Relevant Equations

I=2/5MR^2
LKE=1/2Mv^2
RKE=1/2Iomega^2
mgh=LKE+RKE

I did some algebra and got the final equation mgh=1/2MR^2omega^2+1/2(2/5MR^2)omega^2, then plugged in the numbers and got an omega value of 111.09. Using 1.7 as h instead of cos(17)*1.7=1.63 I got 114.09. Both answers were wrong. Am I missing something, or did I just screw up the math?

 

[kuruman]

Homework Statement: An 8.6-cm-diameter, 320 g solid sphere is released from rest at the top of a 1.7-m-long, 17 degree incline. It rolls, without slipping, to the bottom. What is the angular velocity of the sphere at the bottom?
Relevant Equations: I=2/5MR^2
LKE=1/2Mv^2
RKE=1/2Iomega^2
mgh=LKE+RKE

I did some algebra and got the final equation mgh=1/2MR^2omega^2+1/2(2/5MR^2)omega^2, then plugged in the numbers and got an omega value of 111.09. Using 1.7 as h instead of cos(17)*1.7=1.63 I got 114.09. Both answers were wrong. Am I missing something, or did I just screw up the math?

Numbers without units are meaningless. Please repost with units this time and when you do, show more details.

Also, why do you think h = 1.7 or even cos(17)*1.7 ? Draw a diagram and see what h is.