Cylinder inside a cylindrical track

[Nexus99]

Homework Statement

A homogeneous cylinder of mass m and radius r rolls into a rough cylindrical rigid surface of radius R. The static friction coefficient between cylinder and surface is $\mu_s$.
Knowing that the cylinder is allowed to start at zero speed from the position shown in the figure at an angle $\theta_0$ respect to the vertical line. Calculate:
1) Assuming perfect rolling, the linear speed of the cylinder when it passes through the lowest point of the surface.
2) Assuming perfect rolling, the normal reaction to the surface when the cylinder passes through the lowest point of the surface
3) The minimum static friction coefficient for perfect rolling during the whole motion

Relevant Equations

pure rolling condition, conservation of energy, centripetal acceleration

1) Conservation of energy
$mg(R-r)(1-cos \theta_0) = \frac{1}{2}mv^2 + \frac{1}{2} I \omega^2$
because of pure rolling $\omega = \frac{v}{r}$
So i got:
$v = \sqrt{\frac{4}{3} g (R-r) (1-cos(\theta_0))}$

this is how i got normal force:
2) $N - mg = m \frac{v^2}{R-r}$
where v is the velocity that i found in point 1
i got:
$N = mg(\frac{7}{4} - \frac{4}{3}cos(\theta_0))$

3) this point is quite difficult, I'm not sure of my solution
minimum $\mu_s$ can be obtained by the relation
$| F_r | ≤ \mu_s N$ [ ... ]
I thought that the point where there was more possibility of slipping was the one with the greatest tangential acceleration, that is the starting point
in that point:
$N = mgcos( \theta_0)$
and:
$F_r r = I \alpha = I\frac{a_{CM}}{r}$
$-F_r + mgsin( \theta_0) = ma_t = m a_{CM}$
where $a_t$ is the tangential acceleration and $a_{CM}$ is the acceleration of the the center of mass of the cylinder
combining these 2 equations i got
$F_r = \frac{mg sin( \theta_0 )}{3}$
Inserting that value in the equation [ ... ], i found:
$\mu_{s_{min}} = \frac{tan{\theta_0}}{3}$
is this correct?

 

[haruspex]

For 2, did you mean 7/4 or is that a typo?
For 3, sanity check you equations by considering R infinite.

 

[Nexus99]

For 2, did you mean 7/4 or is that a typo?
For 3, sanity check you equations by considering R infinite.

There was something wrong, i modified my previous post

 

[haruspex]

Your answer to 2 only needed one digit corrected, but now it is dimensionally wrong. You have a g² term.

Edit: but I see you changed (1) as well... give me a moment.

Ok, why did you add the mgr at the end? This is already taken care of by the "-r)(1 " part of the LHS.

 

[Nexus99]

Your answer to 2 only needed one digit corrected, but now it is dimensionally wrong. You have a g² term.

Edit: but I see you changed (1) as well... give me a moment.

Now should be fine, it was a mistake

 

[haruspex]

Now should be fine, it was a mistake

I still don’t see how you get 7/4.

 

[haruspex]

Part 3 looks ok except for a typo in your working: F_rR instead of F_rr.

 

[Nexus99]

this is how i got $\frac{7}{4}$
$N - mg = m \frac{v^2}{R-r}$
$N = mg + m\frac{4}{3} g (1-cos(\theta_0)) = mg(1 + \frac{4}{3} - \frac{4}{3}cos(\theta_0) ) = mg (\frac{7}{4} - \frac{4}{3}cos(\theta_0) )$
for part 3 you are right, i made a mistake it should be $F_r r$

 

[haruspex]

this is how i got $\frac{7}{4}$
$N - mg = m \frac{v^2}{R-r}$
$N = mg + m\frac{4}{3} g (1-cos(\theta_0)) = mg(1 + \frac{4}{3} - \frac{4}{3}cos(\theta_0) ) = mg (\frac{7}{4} - \frac{4}{3}cos(\theta_0) )$
for part 3 you are right, i made a mistake it should be $F_r r$

$1+\frac 43=?$

 

[Nexus99]

Ok, it's $\frac{7}{3}$ ...

 

[Nexus99]

$1+\frac 43=?$

I have a question, when i solved point 3 i thought that $a_{cm} = a_t$ because i visualized the vector $\vec{a}_{cm}$ and i believed that it can't be radial, if that's true, what about centripetal acceleration? isn't connected with $a_{cm}$ ? and the angular acceleration due to circular motion of the cylinder is different from the angular acceleration due to the rotation of the center of mass around the point of contact?

 

[haruspex]

what about centripetal acceleration?

At the point of release, the velocity is zero, so no centripetal acceleration.

the angular acceleration due to circular motion of the cylinder is different from the angular acceleration due to the rotation of the center of mass around the point of contact?

The rate of rotation of a rigid body wrt the axes of a coordinate system is the same regardless of the choice of origin. If you consider a line through two points of the body it turns through a certain angle in a certain time, independently of the choice of the two points.
But don’t confuse that with the question of angular momentum of a body about an axis not at the body’s mass centre , since that gets a contribution from the linear motion of the mass.