Kinetic Energy of a Rotating Hoop - Physics Forum

[rmfw]

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

 

[BOYLANATOR]

Not quite. What is the moment of inertia for this set up?

 

[rmfw]

$T =\frac{1}{2} w^2 \frac{1}{2} M a^2$ ? is it more close to the answer now?

 

[BOYLANATOR]

Yeah it's right now

 

[HalfLostAndSearching]

Hi there,

Yes, you are correct in your calculation of the kinetic energy for a rotating hoop. The kinetic energy of a rotating object can be calculated using the formula T = 1/2 I w^2, where I is the moment of inertia and w is the angular velocity. In the case of a hoop rotating around its vertical axis, the moment of inertia can be calculated as I = 1/2 M a^2, where M is the mass and a is the radius of the hoop. Substituting this into the formula for kinetic energy, we get T = 1/2 (1/2 M a^2) w^2, which simplifies to T = 1/4 M a^2 w^2. So, your calculation is correct.

It's important to note that this formula assumes that the hoop is a solid, uniform object. If the hoop has a different shape or is not uniform, the moment of inertia may be different and the calculation for kinetic energy would also be different.

I hope this helps answer your question. Keep up the good work in your studies of physics!