Physics Quiz 3
 

This is a continuation of the Physics Quiz posted here yesterday:


The problem is changed as follows:

- The wheel is rotating about its axis at a constant rate omega_w

- The (constant) torque exerted on the wheel by the motor is tau_k

- The wheel+motor assembly is revolving CCW around the pivot at a constant rate omega_p

- The length of the connecting rod is L

- The weight of the wheel+motor assembly is W

- The coefficient of kinetic friction between wheel and floor is mu_k


What is the magnitude and direction of the floor's friction force on the wheel?

Re: Physics Quiz 3
 

Is the majority of information given not needed?
f = μN = μmg = (mu_k)(W/g)(g) = (mu_k)(W)

Re: Physics Quiz 3
 

And what direction is that force directed?

I would say the majority of the information given is needed. It just might not be needed for a particular part of the problem.

Re: Physics Quiz 3
 

Well isn't friction always opposite the applied force?

Re: Physics Quiz 3
 

So what's the "applied force" ?

Caution: think this through.

Re: Physics Quiz 3
 

I wanted to think about this kind of extension after I figured out part one, but I had to learn how to drive instead.

I'm going to go very slowly... and try to not be too confused...

1. Centripetal acceleration is constant so:
Centripetal force on the wheel = (WL/g)(omega_p)^2 = Tension in the rod - Friction parallel to the rod

2. Tangential acceleration is zero, so net force perpendicular to the rod is zero.

3. And I think brndn was right in saying F=(mu_k)(W), yes?

4. And then friction parallel plus friction perpendicular would still equal tau_k/(rsintheta) if there was no wheel slip. However, there is wheel slip, which is why point 2 is true.

So, the only forces are tension and friction parallel, so friction is anti-parallel to the rod. At least I think so...?

Is there another way to get the magnitude of friction using the difference in wheel speed and rod speed?

Re: Physics Quiz 3
 

The magnitude of the centripetal acceleration is constant, but the centripetal acceleration vector is not constant because its direction is changing.


The tension on the rod is not the only force acting on the wheel+motor assembly, so you can't set it equal to (WL/g)(omega_p)^2


Angular acceleration is zero.

Tangential acceleration is the rate at which the tangential velocity is changing. Since the tangential velocity is not constant, the tangential acceleration is not zero.

Re: Physics Quiz 3
 

I said that centripetal force = tension minus friction parallel to the rod. Is there another force in addition to those two?

I'm a little confused by this statement. Let me see if my understanding of circular motion is right, or if my physics teachers have been teaching me falsehoods. An object moving in a circle is always accelerating. That acceleration has two components, one that is tangential and one that is centripetal. The centripetal component is (mv^2)/r and the tangential component is the derivative of ds/dt, where s is the arc length. ds/dt is equal to d(theta)/dt times radius and tangential acceleration equals angular acceleration times radius.

In this problem the radius is constant and d(theta)/dt is constant as well. That means that angular acceleration is zero and tangential acceleration is also equal to zero.

(Maybe the problem is my axes. I created another system where one axis is always perpendicular to the rod and the other is always perpendicular. Then the centripetal acceleration is actually constant in both direction and magnitude, and I think tangential velocity is also actually constant)

Re: Physics Quiz 3
 

I'm thinking one of us needs to build this (massless rod might be tough to find...) and put a piece of paper under the wheel and see which way the paper flies.

Re: Physics Quiz 3
 

@Ninja_Bait:

My apologies, I misunderstood your earlier post. I see you edited it. I had the original version still on the screen when I replied to it (got interrupted).

The net force acting on the motor+wheel assembly points directly to the pivot. This must be the case, since the angular acceleration is zero (the problem stated that the angular velocity is constant).

The net force is equal in magnitude to the tension minus the friction. Since these are the only two external forces acting on the motor+wheel assembly, the friction must be pointing directly away (radially) from the pivot.

As posted earlier, the magnitude of the friction force is simply mu_k*W (if we assume the Coulomb friction model).

Re: Physics Quiz 3
 

Since the tangential acceleration is zero, the frictional force must act in the direction of the rod (so it must be angle theta from the horizontal).

The forward friction force on the wheel must be equal to tau_k/r or the wheel would not be spinning at constant speed. The sideways friction force for the friction to act in the correct direction must be equal to
forward friction * tan(90 - theta)
so the total friction force would be
tau_k/r * sqrt(1 + tan(90-theta)^2) = tau_k/r * sqrt(sec(90-theta)^2)
= tau_k/r * sec(90-theta) = tau_k/r * cosec(theta) = tau_k/(r*sin(theta))

Re: Physics Quiz 3
 

Yes.

But according to the Coulomb friction model, the total friction force must be mu_k*W.

So that means tau_k must equal mu_k*W*r*sinθ.

So, what happens if the voltage to the motor is increased?

Re: Physics Quiz 3
 

I did a quick model in LEGO, but it's kind of crummy, since obviously I don't have a frictionless pivot or massless rod. I might film it and post it, though.

@ Ether: Sorry to confuse you; I accidentally hit submit instead of preview and had to go back to finish editing. :\ Lesson learned in clarity of answer as well. However, I'm glad I was right about direction. That makes everything better.

I'm very interested by John's last answer. It seems that friction is always tau/(rsin(theta)) and parallel to the rod, pointing away from the pivot, no matter the conditions. Is that the end result?

Re: Physics Quiz 3
 

See this post:

http://www.chiefdelphi.com/forums/sh...0&postcount=12

Re: Physics Quiz 3
 

You were just a minute faster than me.

I think I saw the effect you mention in my model. Does it speed up but not gain torque (that is, omega_w increases but omega_p would not because friction does not increase)? Because tau_k is equal to a constant based on your reply to John

My physics knowledge doesn't extend this far yet. I think I have been told that speed and torque on a motor aren't necessarily linked in voltage control... but I don't understand the science behind that.

Re: Physics Quiz 3
 

Would omega_w increase until the point on the motor curve where
tau_k/r = mu_k * W * sin(theta)
So the wheel would start spinning faster along its axis, but the rod would rotate at the same speed as before?

Re: Physics Quiz 3
 

Say you have a small DC brush motor mounted on a workbench and you connect it to a fixed voltage. The motor will spin up to a certain speed and then stay there. (That's called the "free speed" or the "no load speed").

The reason the motor reaches a certain speed and then stops increasing is because as the motor speeds up it generates more and more "back emf". This back emf is a voltage that opposes the applied voltage. At free speed, the back emf is almost equal to the applied voltage. The small difference in voltage creates just enough current in the motor to keep it spinning at the free speed (some current is required at free speed to supply the torque necessary to balance the friction and other losses in the motor). This current is called the "free current" or the "no load current".


DON'T DO THE FOLLOWING, YOU MIGHT HURT YOUR HAND, IT IS A THOUGHT EXPERIMENT: Let's say Superman walks in and holds the motor shaft and gently squeezes it. What happens? The motor slows down. It slows down because it has a load on it. It also draws more current in order to generate the torque necessary to balance that load. If Superman squeezes harder, the motor slows down more. If he squeezes hard enough, the motor will stop turning. This is called "stall". The torque at stall is called the "stall torque". The current is called the "stall current".

The important take-away from all this is that, for a fixed applied voltage, the torque output and the speed of the motor are inversely related. Put more load on the motor and it slows down. Put less load on the motor and it speeds up.

Now consider what happens if you have a fixed load on the motor. If you increase the voltage applied to the motor, the motor will speed up and reach a new equilibrium speed. If you decrease the voltage applied to the motor, the motor will slow down and reach a new equilibrium speed. At the equilibrium speed, the applied voltage minus the back emf provides just enough voltage to push the current through the motor required to balance the applied load.

If you want to run some "virtual" experiments to explore these concepts, I have written a small motor calculator which is available for download here. Grab the most recent revision, at the bottom. It has all the FRC 2011 motors in it. You can change voltage, load, speed and see how the motor reacts.

Re: Physics Quiz 3
 

Yes.

It gets kind of messy. mu_k is not really constant in real life.

Typically, omega_p would increase as omega_w increases.

Under certain conditions, I think the system tries to minimize the relative speed between the floor and the wheel. I believe this occurs when omega_p = (r/L)*cosθ*omega_w

Re: Physics Quiz 3
 

Actually the model contains the answer to this and it is a simple matter of kinematics.

All of the analyses performed are in equilibrium and part of that is the wheel slipping on the floor. In order for the force of friction to be pointed in the radial direction, the velocity of the point on the bottom of the wheel must be in that direction (for this model). Once you accelerate the wheel you will introduce a wheel slip velocity relative to the ground which is not aligned and thus results in an angular acceleration changing omega_p. You will reach a new equilibrium point if you specify a new constant wheel speed.

Quiz 4: Given omega_w determine the steady state omega_p.



Okay... got you between edits :)


Quiz 5: Qualitatively, what happens to the torque requirements, friction force, and reaction force at the pivot, if the motor and wheel are instead a thin disk of mass m (i.e. with rotational inertia) AND the pivot connection only resists moments about the radial axis (e.g. is a Hooke's joint)? The equations get tough, so we need only discuss the character and origin of the effects.

Answer the following:
Does the answer to quiz 4 still apply? Why or why not?
Is the applied torque requirement constant? Why or why not?
Is the friction force constant magnitude and radially oriented? Why or why not?
Is the reaction force at the pivot different in any meaningful way? How so?

Re: Physics Quiz 3
 

This doesn't sound right to me. The torque applied by the motor should not be changing as theta changes. I think the problem is that in order to find the friction equal to tau_k/(r*sin(theta)), you must assume the wheel is gripping. This does make sense because it is able to rotate and drive around the pivot but is inconsistent. I feel like one of the boundary conditions is actually overconstraining the model.

Re: Physics Quiz 3
 

tau_k is the motor torque necessary to be in equilibrium at a given angle theta, given the assumptions in the model (mu_k constant, frictionless pivot, frictionless wheel bearing, no wind resistance, no rolling friction, no friction due to rotation of the wheel about the Z axis).

What the equation is saying is that if you reduce the angle theta, it takes less tau_k to be in equilibrium. If you increase theta, it takes more.

Sustaining equilibrium when theta is very small takes very little torque. In fact, when theta is zero it takes zero torque to sustain equilibrium since the pivot is frictionless and we are ignoring wind resistance, rolling friction, friction in the wheel bearings, and friction due to rotation of the wheel about the Z-axis.

So what happens if you start with the system in equilibrium, and you then increase tau_k to tau_k' ?

Re: Physics Quiz 3
 

After thinking about it more, the error I made was getting wrapped up in angular motion, thinking that theta was the wheel's angular position about the pivot as the system rotates. What it really represents, though is a constant in this problem, the angle between the wheel's axis of rotation and the pivot arm.

I think that if you increase the torque, nothing will change because the wheel is already slipping. The friction cannot be higher than W*mu so no more torque is able to be useful.

Re: Physics Quiz 3
 

If you increase tau_k, the system will no longer be in equilibrium. That means something must change.

Re: Physics Quiz 3
 

Can normal force increase to balance the increase in tau_k? I don't think that actually makes sense but it seems to be the only variable that can change in this case.

Re: Physics Quiz 3
 

No.

If the system is in equilibrium, and then you increase the motor torque that's driving the wheel, what happens?

Think torque=momentOfInertia*angularAcceleration. Then follow the causes and effects as they ripple through the system.

Re: Physics Quiz 3
 

Ninja_Bait: Ether has already answered this question in an earlier post... I think the issue is the notion of supplying a larger torque indefinitely. Perhaps it will help to say it another way?

The system will accelerate to a speed where you are no longer able to supply torque greater than tau_k. Physical systems are power limited and power = torque * angular rate. The system will accelerate to the limits of your power supply or motor torque vs speed curve (which ever comes first). At that point the torque supplied by the motor will again be tau_k and you will have constant speeds.

If you could somehow supply a constant torque independent of speed, the system as modeled would properly accelerate forever. Other effects would eventually dominate and the model would become invalid so this wouldn't actually happen, but you would still want to patent that motor and power supply! :)

Re: Physics Quiz 3
 

Ah, I see. Thank you.