Kinetic Energy of a Rotating Hoop - Physics Forum

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The discussion centers on calculating the kinetic energy of a rotating hoop using the Lagrangian approach. The initial formula presented for kinetic energy, T = (1/2) M a^2 w^2, is corrected by considering the moment of inertia. The correct expression for kinetic energy is T = (1/2) I w^2, where I is the moment of inertia for the hoop. The final agreement confirms the kinetic energy formula as T = (1/2) w^2 (1/2) M a^2. This highlights the importance of incorporating the moment of inertia in rotational dynamics calculations.
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Hey PF, this is just a quick question:

If there's a hoop of mass M and radius a rotating around its vertical axis (see pic) and I want to write the kinetic energy for the Lagrangian is it just T = \frac{1}{2} M a^2 w^2 ? Considering w the angular velocity
 

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Not quite. What is the moment of inertia for this set up?
 
T =\frac{1}{2} w^2 \frac{1}{2} M a^2 ? is it more close to the answer now?
 
Yeah it's right now
 
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