A cylinder rolling down an incline

[arunma]

So here's a pretty straightforward question. Given a cylinder with some specific mass and radius rolling down an inclined plane with a specific angle of inclination, what is the cylinder's acceleration?

I can figure out the answer pretty easily by finding the Lagrangian of the cylinder (I worked it out to \dfrac{2}{3} g sin( \theta ), someone let me know if that's wrong). But what's bugging me is that I can't remember how to do this problem using regular Newtonian mechanics. Can anyone help me out of this brain fart? Thanks.

 

[Doc Al]

The usual suspects:
(1) Identify the forces acting on the cylinder (don't forget friction)
(2) Write Newton's 2nd law for translation
(3) Write Newton's 2nd law for rotation
(4) Include the constraint of rolling without slipping
(5) Solve!

 

[arunma]

It's number 1 that's eluding me at the moment (which is really bugging me, since this is supposedly something that a freshman can do). There is a static frictional force at the base of the cylinder. But since the cylinder isn't at the point of slipping, how can this be computed?

 

[arunma]

Wow, I just realized the source of my brain fart. I had forgotten that the speed of the cylinder is equal to its radius time the angular velocity. That clears everything up. Thanks!

 

[Doc Al]

It's number 1 that's eluding me at the moment (which is really bugging me, since this is supposedly something that a freshman can do). There is a static frictional force at the base of the cylinder. But since the cylinder isn't at the point of slipping, how can this be computed?

I'm sure you've figured it out for yourself by now, but just for the record: You calculate the friction force by solving the force equations. The amount of static friction depends upon the angle (as you realize, you certainly cannot assume that friction equals \mu N).

 

[arunma]

I'm sure you've figured it out for yourself by now, but just for the record: You calculate the friction force by solving the force equations. The amount of static friction depends upon the angle (as you realize, you certainly cannot assume that friction equals \mu N).

Yup, it turns out I need to use Newton's Second Law for rotating bodies. Anyway, I figured it out. And it's always satisfying to see that the Newtonian and Lagrangian methods agree!