Recent content by Nexus99

Yes but the frame rotates around y axis

y antiparallel to M and z perpendicular to the plane represented by the wheel

Ok so: ## \vec{L_{O'}} = I_{CM} \omega_2 \hat{z} - (\frac{I_{CM} \omega_1}{2} + m v d) \hat{y} ##

So should be: ## L_{O'} = \sqrt{(I_{CM} \omega_2)^2 + (\frac{I_{CM}}{2} \omega_1 + mvd))^2} ##

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...

Yes, i forgot to write the mass, and yes, i saw now on the book that M = 2t Nm/s.

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?

Yes, it's a physical pendulum. Sorry if it was not so clear but i had to summarise the problem

It isAnyway i don't understand what do you want to say with your previous message

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

I am sure that is easier, but i have never done a multivariable integral. Is it possible tu use the fact that: ## I_z = I_x + I_y ##?