Angular and CoM Velocities of a Solid Sphere

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SUMMARY

The discussion focuses on calculating the angular velocity (ω2) and center of mass velocity (vcm) of a solid sphere rolling down a curved rail from height h1 to height h2. The relevant equations include the moment of inertia for a solid sphere, I = (2/5)mR², and the kinetic energy equation KE = (1/2)mv² + (1/2)Iω². The final expressions derived are ω = √((10/7)g(h1-h2))/R and v = √((10/7)g(h1-h2)). The relationship v = ωR confirms the consistency of the calculations.

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Homework Statement


A solid sphere of mass M and radius R is rolling,without slipping, down a curved rail. The sphere is initially at rest at a height of h1. Find the angular velocity ω2 and the center of mass velocity of the sphere vcm at the end of the rail of height h2. You may assume that no vibration and heat are generated as the sphere rolls along the rail.

Homework Equations


solid sphere, I = \frac{2}{5}mR^2


The Attempt at a Solution


I'm not sure if I began with the correct equation.

KE = \frac{1}{2} m v^2 + \frac{1}{2} I ω^2
= \frac{1}{2} m(ωR)^2 + \frac{1}{2} (\frac{2}{5}mR^2) ω^2
mg(h_1 -h_2) = \frac{7}{10}m ω^2 R^2
ω = √(\frac{10}{7}g(h_1-h_2)) / R

KE = \frac{1}{2} m v^2 + \frac{1}{2} I ω^2
= \frac{1}{2} m v^2 + \frac{1}{2} (\frac{2}{5}mR^2) (\frac{v}{R})^2
mg(h_1 -h_2) = \frac{7}{10}mv^2
v = √(\frac{10}{7}g(h_1-h_2))

Thanks in advance!
 
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Looks good to me. (Of course there was no need to solve it twice, since v = ωR.)
 

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