Need help understanding KE rotation

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The discussion revolves around the understanding of kinetic energy (KE) in rotational motion, specifically using the formula KE = (1/2)Iw^2 and its translation to KE = (1/2)v^2/r^2. The variable 'v' represents the tangential velocity of a point on the rotating body, which is related to the angular velocity by the equation v = ωr. When considering a rolling object down an incline from rest, the velocity in question is indeed the velocity of the center of mass. In the case of a wheel rolling without slipping, the speed of the center of mass equals the tangential speed of the wheel's rim. This highlights the relationship between rotational motion and translational motion in rolling objects.
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i used the formula for KE rotation KE=(1/2)Iw^2 and i translated it to (1/2) v^2/r^2, but i wanted to know what does the v in the equation stand for. is it initial velocity or velocity of center of mass. if its is the velocity of center of mass can u tell me how to solve for it
 
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Whether you know it or not, you made use of the relation v = \omega r, where v is the tangential velocity of some point on the rotating body and r is its distance from the axis.
 
ok thanks a lot my other question is what if the rotating object was rolling down a incline and it started at rest. would the velocity be the velocity of center mass of the object that is being rolled.
 
Let's take an example: A wheel, with radius r, rolls without slipping along some surface. The speed of its center with respect to the surface will be equal to the tangential speed of its rim with respect to its center.
 

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