# Cylinder over a step -- Is the normal force radial?

1. Feb 9, 2015

### Sol_Solved

1. The problem statement, all variables and given/known data

A horizontal force, $F$, is applied to a cylinder of mass $M$ and radius $R$ (the line of action of $F$ goes through the center) in order for the cylinder to climb over a step of height $R/2$.
1) Find the torque around $A$ and find force $F$ as a function of the normal force $N$ with the floor for the system to be in equilibrium. (Note that the minimum force needed to climb up the step is obtained when $N$=0).
2) Write down the forces once the cylinder has rotated an angle $\theta$ and has left the floor. Which forces do work?
3) The cylinder finishes its ascent once it is completely over the height of the step. At that moment, force $F$ stops being applied. What is the rotational velocity of the cylinder? What is the velocity of the center of mass? (both as functions of $F$). Consider $I_{CenterOfMass}$ given.

2. Relevant equations
Choosing axes from A: x positive towards the right, y positive upwards, rotation positive counterclockwise. Calling the normal force from the step $A$ and the normal force from the floor $N$.
If I draw a right triangle between the center of the cylinder and A, the hypothenuse is R, the shorter leg is R/2, that makes the longer leg $R\ \sqrt{3}/2$. Let's call that $d$.

1)
$\sum\ \ F_x = F-A_x = m \ddot x$
$\sum\ \ F_y = A_y - mg + N = m \ddot y$
$\sum\ \tau_A = -FR/2 + mgd - Nd = I_A \dot \omega = (I_{CM} + mR^2) \dot \omega$

2)
$\sum\ \hat \theta = - mg \ cos \ \theta + F \ sin \ \theta + A \ cos \ \theta = m r \ddot \theta$
$\sum\ \hat r = - mg \ sin \ \theta - F \ cos \ \theta + A \sin \ \theta= -m r \dot \theta^2$
$W= \vec F \cdot \vec s$

3. The attempt at a solution
1) When the system is at equilibrium, $\ddot x = 0, \ddot y = 0, \dot \omega = 0$
$\vec F=(mg-N)d/R \ \ \hat x =>$ $\hat F = (mg-N)\ \sqrt{3} \ \hat x$

2)
$W= \vec F \cdot \vec s$ The weight and F do work.

I'm still working on the third part, but I am confused about the direction of force A. If I calculate it from the first part of the exercise, when the cylinder is still touching the ground:

$A_x = (mg-N)\ \sqrt{3}$ and $A_y = (mg-N)$ therefore the direction of the force is $tan \alpha = 1/ \sqrt{3}$, $\alpha= \pi/6$

Is there anything that should be telling me that the force A becomes purely radial after the cylinder leaves the floor or should I carry on with $\hat \theta$ and $\hat r$ ?

Thanks!

2. Feb 9, 2015

### Sol_Solved

$\hat x$ is the direction of $\vec F$, since it is a horizontal force.

In 2), the question just mentions writing down the forces involved. $A$ is the force of contact between the step and the cylinder. I do intend to solve for $A$, but I can't seem to move forward because I can't figure out if $A$ is purely radial or if it has a tangential component (hence my question). Maybe I'm just too fixated on that and should just proceed assuming it has both components.

3. Feb 9, 2015

### haruspex

Yes, sorry, I posted in too much haste earlier. I later figured out the answers to my questions and hoped to delete my post before anyone read it! Will try to provide a more useful response later.

4. Feb 9, 2015

### Staff: Mentor

To solve this problem as posed, you don't need to know the force at A. That's because the problem can be solved soley using the moment balance. Let me help you.

The line from point A to the center of the cylinder initially makes an angle of 30 degrees with the negative x axis, correct? If the cylinder rotates clockwise (not counterclockwise) through an angle of θ, what is the new angle that the line from point A to the center of the cylinder makes with the negative x axis. If θ is the rotation angle at time t, what are the x and y coordinates of the center of the cylinder at time t? Since as the cylinder rotates, the force F is always pointing horizontally at the center of the cylinder, what is the moment balance about point A at time t?

After we get squared away with the answers to these questions, I'll help you with the next steps. (There are only a couple of steps remaining to arrive at the final answer).

Chet

5. Feb 17, 2015

### Sol_Solved

Thank you! I'm sorry it took me so long to reply. I'm having a hard time with the problem. Let's see if I can get this right:

After rotating $\theta$:
The new angle between A and the center of the cylinder would be $\pi/6 + \theta$.
The coordinates of the center of the cylinder at time t are: $(\pi/6 + \theta) \ R \ \hat x$, $R/2 * (\pi/6 + \theta) / \pi/3 \ \hat y$
The moment balance about A is $-R \ sin (\pi/6 + \theta) \ \vec F + R \ cos (\pi/6 + \theta) m\ \vec g$

6. Feb 17, 2015

### Staff: Mentor

No. The x and y coordinates of the center of the cylinder are:

$x=-rcos(\pi/6 + \theta)$

$y=rsin(\pi/6 + \theta)$

Yes, that's the net counterclockwise moment about A, but the force F and the gravitational constant g in this equation should not be indicated as vectors. They should be the scalar magnitude of F and the scalar magnitude of g. What is the clockwise moment about A?

I asked for the moment balance about A. All you have given is one side of the equation. What is the other side of the equation (in terms of time derivatives of θ)?

Chet