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1. Feb 21, 2016

### SoNiiC

1. The problem statement, all variables and given/known data
The following problem is part of a physics lab with unclear directions. The lab prompt says: Find the angular acceleration of a meterstick as it slips down a wall. Here is my interpretation of the lab: A meterstick of mass 0.15kg is leaning against the wall. At an angle of 30 degrees, it begins to slip until it hits the floor. During this time, what is the angular acceleration of the center of the meterstick?
2. Relevant equations
Sum of Torques = Moment of Inertia * Angular acceleration.
Moment of Inertia of center of meterstick = 1/12 ML^2

3. The attempt at a solution
I started with a free body diagram and drew the forces for the Normal force of the wall, mass of meterstick, normal force of the floor, and static friction. My plan of action was to sum the torques of the ladder and then divide them by the moment of inertia, giving me the angular acceleration. However, I am having some trouble with the torques as I am not certain on how to take static and kinetic friction into account. What I do know is that at the moment of equilibrium, the normal force of the wall would be equal to the force of static friction. However, as soon as normal force of the wall (mg/tan(theta)) is greater than the force of kinetic friction (uk*normal force of the floor), the friction changes to static. By setting these two equations equal I can solve for the force of static friction since theta is known to be 30. However, what I really need in order to sum the torques is the force from kinetic friction. Is there any way for me to do so?

2. Feb 21, 2016

### haruspex

Not quite clear - are you saying you do not know the coefficient of kinetic friction, so you are looking for some way to calculate it, then from that calculate the acceleration?
If so, there is no way. The only way to find the kinetic coefficient will be to find the acceleration first some other way, e.g. from the time to hit the ground.

3. Feb 21, 2016

### SoNiiC

Would it be possible for me to do this problem if I just let the static and kinetic friction be the same thing?

4. Feb 21, 2016

### haruspex

Yes, as a function of position. If you want it as a function of time that may be a problem. With the two coefficients the same, and starting from only just slipping (an infinitesimal net torque) you may find the time to reach a given point is theoretically infinite.

5. Feb 21, 2016

### SoNiiC

Assuming no wall friction:
I found the value of the friction coefficient by setting the sum of the torques equal to 0 (since i knew the mass and angle) (Torque = 0 = mg(L/2)costheta - mg(friction coefficient)(L)sin theta.) However, this dosent help with the rest of the problem at all (I was planning on finding the torque and then diving it by the moment of inertia to find angular acceleration).

Assuming wall friction:
Another thing i could do would be top set the wall friction and the floor friction equal to each other, cancelling friction out of the equation completely. Torque = mg(L/2)sin theta. However, I don't think this is correct either.

The only other way that I can think of doing this problem is by just setting a = g; perpendicular acceleration = gcostheta and dividing by the radius to find the angular acceleration. However, this would be inaccurate due to friction on the bottom part of the ladder.

Any ideas?

6. Feb 21, 2016

### haruspex

Your problem statement says you are to find the angular acceleration "as it slips down the wall". It will not be constant, so you need to find it either as a function of position, or of angle, or of time. If the torque starts off only infinitesimally exceeding frictional torque, the initial acceleration will be zero.
Try to find the torque at some angle theta.