Recent content by decerto

This clears it up for me I think, dL/dt != 0 in the inertial frame and there is no MA 'force'?

I understand what you are getting at but I don't understand why I need to include MA as a moment, what physical reason is there for it when the force F is clearly marked as acting at the base of the wheel. edit * ok I can see the logic if we consider MA is a fictitious force due to it being a...

The MA vector is one which acts through the CoM?, you are not talking about the F acting at the point of contact of the backwheel?

I have $$ F= \mu N=MA$$ $$N=\frac{MA}{\mu}$$ which is how I got $$A_{max} = \mu g \frac{a}{a+b}$$ in my original post. Or are you saying I missing an "MA component" from my torque equation?

So calculating the moments about the point of contact of the front wheel I got $$(a+b)N - aMg=0$$ which gives $$A_{max} = \mu g \frac{a}{a+b}$$ Is this correct? The next part of the question asks but my answer and all my formulas are independent of h and if I go back to using the...

So I just use this and the other two equations $$N_1 + N_2 = Mg$$ $$bN_1-aN_2 -h F =0$$ ? Just so I have this clear in my head, if it says "roll without slipping" this means there is no frictional force that resists the motion. Instead when a torque is applied to the wheel (like the back...

Just so I am clear, Are you saying I just have $$F=MA=\mu N_1$$?

Homework Statement http://imgur.com/a/vA0H2 Homework Equations $$F=ma$$ $$r \times F = I \alpha $$ The Attempt at a Solution Denoting the forces at the back wheel with a 1 and front wheel with a 2 and calculating the torques around the CoM I have $$F-f_1-f_2 = mA$$ $$N_1 + N_2 -mg = 0$$...

$$ma=-N$$ and $$Nd = I\dot{\omega}$$ gives $$-mad = I\dot{\omega}$$ Integrating wrt to time I have $$-mv_f d + mv_i d = I \omega$$ which is just the conservation of angular momentum about the point of contact with $$v_i= \sqrt{2gh}$$ Solving for omega $$\omega^2 = \frac{1}{I^2}\left(m^2d^2v_i^2...

Thank you all for the help. Integrating your dynamical equation just gives me back the conservation of angular momentum, is this correct? Should I end up with a quadratic in the final velocity when I combine it with the energy equation?

I don't understand how this is possible, there is forces so there are no conservation of momentum but the forces are not meaningful so I can't use Newtons laws. I clearly have some gap in my understanding here.

Honestly I am confused over each one of those sentences. What do you mean by there is no meaningful force? How is there a momentum transfer (to/from?) the rod if there is no force? If so, how do you calculate it? Surely the rod starts rotating after the collision and wouldn't that mean it...

Homework Statement [/B] http://imgur.com/a/Ssolz Homework Equations [/B] Elastic Collision so $$ mgh = \frac{1}{2}mv_i^2 = \frac{1}{2}mv_f^2 + \frac{1}{2}I\omega^2 $$ $$ F=ma $$ The Attempt at a Solution [/B] So the problem is similar to this one which allows me to work out the normal force...

It was an exam question from like 6 years ago, I was revising, that is why I wrote "Annoyingly it didn't come up in the exam" and regardless I did solve it myself.

Thanks, that was the second part of the question, here it is.