Object Rolling Down Inclined Plane

[sidvelu]

This isn't really a numerical question, just a conceptual questoin. I wanted to know why if you have an object rolling down an inclined plane, you can just choose to put the pivot point anywhere.

This is because I see problems where one thing is solved using F*R=I\alpha

And I also see when the formula is written as mgsin\theta R = I \alpha

I was curious about why we can do this.

 

[tiny-tim]

welcome to pf!

hi sidvelu! welcome to pf!

(have an alpha: α and a theta: θ and an omega: ω )

This isn't really a numerical question, just a conceptual questoin. I wanted to know why if you have an object rolling down an inclined plane, you can just choose to put the pivot point anywhere.

not anywhere …

you can only use the centre of mass or the centre of rotation

torque = rate of change of angular momentum is true about any point, but angular momentum = moment of inertia times angular velocity (L = Iω) is not generally true

about a general point P, L_P = mr_c.o.m. x v + I_c.o.m.ω, and that doesn't generally equal I_Pω