- #1

Hamiltonian

- 296

- 190

why can the equation ##\tau = I\alpha## only be applied about the axis passing through CoM or the IAOR?

I thought the answer to this can be found by seeing how the equation is derived I faced an issue here, I am able to derive for a point mass ##m## moving in a circular path of radii R under the application of a force F

$$\vec{R}\times \vec{F} = \vec{\tau} = \vec{R}\times M(\vec{a_t}) = \vec{R}\times M(\vec{R} \times \vec{\alpha}) = MR^2\vec{\alpha}$$

in this case ##MR^2## is the moment of inertia of the body about its CoM.

if say there is a disc fixed at its CoM and a Force F acts on its circumference we know that ##\vec{\tau} = \vec{R}\times\vec{F}##

but how exactly do we get ##\vec{\tau} = \frac{(MR^2)}{2}\vec{\alpha}## I am unable to derive ##\tau = I\alpha## in this case I don't see why we can just replace the point particle with a rigid body and just directly account for it by replacing ##MR^2## with the I of the rigid body.