Recent content by PhysHobbiest

oof, thanks for being another set of eyes on that. I don't know how I mistaked θ with dθ there.

Thanks for the input, I believe you accidentally inverted ##\sec^2\theta## when you expanded it out. The answer should look like ##\frac {dx} {dt} = \frac {h^2 + x^2} {h} \frac {d\theta}{dt}## I'm aware of what the answer should be, I was simply looking for the flaw in my above reasoning. It...

So clearly the easiest way to relate the angular speed to the linear speed would be to start from ##\tan(θ) = x/h## and take a time derivative of both sides. However, it also shouldn't be difficult to find the angular speed geometrically. Using the diagram below one can see that: ##sin(dθ) =...

Sorry, I should have been clearer here. sin(θ) here is the length of the rod (I should have written ##lsin{\theta}## although Morin himself omits the ##l##), not for the component of the normal force perpendicular to stick, so really what I was saying is that I should have equated the torque to...

I was blind but now I see. I was equating the torque on the stick to the normal force (i.e. ##rF\sin{\theta} = N##) when I really should have been balancing torques (i.e. ##rF\sin{\theta} = Nsin{\theta}##). Silly mistake but it was driving me crazy. Thank you all for helping me correct my mistake!

What makes me think that he's using the entire weight of each stick is his equation for normal force. Which is N = (MlG) (This is the gravitational force supposedly perpendicular to the stick) x (1/2)(sin(θ)) (This being the center of gravity)

I'm getting tan3(θ) = (1/μ) as a solution. Morin's solution seems wrong to me because it seems to me that he's assuming that the entire weight of each stick is providing torque when in reality the component of the gravitational force perpendicular to each stick should be all that is providing...