Rolling Sphere On Incline

AI Thread Summary
The discussion centers on calculating the angular velocity of a solid sphere rolling down an incline. The user derived the equation mgh=1/2MR^2ω^2+1/2(2/5MR^2)ω^2 but obtained incorrect omega values of 111.09 and 114.09. Participants pointed out the necessity of including units in calculations and suggested verifying the height (h) used in the equations. They recommended drawing a diagram to clarify the correct height measurement. The conversation emphasizes the importance of accuracy in both calculations and understanding the physical setup.
Randomized10
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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?
 
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Randomized10 said:
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.
 
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