Acceleration of an object rollling down a incline.

AI Thread Summary
The discussion focuses on deriving the acceleration of a cylinder rolling down an incline. The key equation used is the torque equation, ∑τ=Iα, which incorporates the parallel-axis theorem to account for the cylinder's moment of inertia. The solution shows that the acceleration of the cylinder's center of mass is a=2/3g sinΘ. Clarification is provided regarding the necessity of the parallel-axis theorem, emphasizing that it accounts for the torque about the point of contact, which is crucial for rolling without slipping. Understanding this concept is essential for correctly applying the principles of rotational motion.
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Homework Statement


A cylinder of mass M and radius R rolls(without slipping) down an inclined plane whose incline angle with the horizontal is Θ. Determine the acceleration of the cylinder's venter of mass.(Derive the formula)


Homework Equations


I have the answer, but I just don't understand:

∑τ=Iα, use the parallel-axis theorem and a=Rα.
(Mg sin Θ)R=(1/2MR²+MR²)(a/R)
Rg sin Θ=3/2Ra
a=2/3g sinΘ

The Attempt at a Solution


Here's my question. Why do we need to use the parallel-axis theorem? isn't that the torque is just 1/2MR² ? Maybe I am confused with the concept of Torque. I don't know...Please help!
 
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When the cylinder is rolling without slipping, the relative motion of point of contact is zero.
So it acts as the axis of rotation of the center of mass. That is why you have to find the moment of inertia about this axis by using parallel-axis theorem.
 
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