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Physics
Classical Physics
Mechanics
A doubt in the rotational analogue of F=ma
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[QUOTE="Hamiltonian, post: 6432078, member: 680682"] If we have a cylinder rolling on the ground ##\tau = I\alpha## can only be applied about the point in contact with the surface(Instantaneous axis of rotation) and its CoM I don't see why this should be the case. 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. [/QUOTE]
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Physics
Classical Physics
Mechanics
A doubt in the rotational analogue of F=ma
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