Ok, probably i finished the problem
## \omega_1 = \frac{v}{d} ##
## \omega_2 = \frac{v}{r} ##
## v = \frac{t^2}{2md} ##
## L_{O'} = L' + L_{CM} = I_{CM} \omega_2 + \frac{I_{CM}}{2} \omega_1 + mvd ##
where ## I_{CM} = mr^2 ##
## K = K' + K_{CM} ##
Last question can be find imposing that
## f...
I noticed it too. I only did algebra
## F = \frac{M}{d} = 2 ma ##
And considering that ## M = 2t ## you get my result. Probably the book forgot a coefficient in that law
##\frac{M}{d} = F ##
##f = ## friction
projecting the forces on the transverse direction to the circumference described by the wheel around the vertical axis:
## F - f = m a ##
Torque about z axis, perpendicular to the plane of the wheel:
## fr = I \alpha## ## \rightarrow ## ## f = ma##
So...
No, ##M## is a torque, so I thought ##\frac{M}{d}## was the force applied to the rod (or to the cm of the wheel and in opposite direction of the friction)
I had some idea to solve this problem but i can't understand where the moment M is directed and where the force/forces that has magnitude ##\frac{M}{d}## is/are directed. Can anyone help me?
I noticed in this moment that is present the solution of this problem:
"For the calculation of the moment of inertia, can be considered the moment of inertia of the elementary rod with mass dm, thickness dy and length l = l (y) indicated in the figure, where y is the distance from the apex of...
@jbriggs444 @etotheipi
thanks for explanation.
I watched some videos on youtube about double integrals and i got these results
## I_z = \frac{5 \rho h^4}{12 tan( \alpha) } = \frac{5 M h^2}{12} ##
Hope it's correct