Physics Quiz 3

[Ether]

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?

[brndn]

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

[EricH]

Originally Posted by brndn

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

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.

[brndn]

Well isn't friction always opposite the applied force?

[Ether]

Originally Posted by brndn

Well isn't friction always opposite the applied force?

So what's the "applied force" ?

Caution: think this through.

[Ninja_Bait]

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?

[Ether]

Originally Posted by Ninja_Bait

1. Centripetal acceleration is constant

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

Centripetal force on the wheel = (WL/g)(omega_p)^2 = Tension in the rod

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

Tangential acceleration is zero

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.

[Ninja_Bait]

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

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

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.

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)

[SenorZ]

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.

[Ether]

@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).

[John]

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))

[Ether]

Originally Posted by John

so the total friction force would be ... tau_k/(r*sin(theta))

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?

[Ninja_Bait]

Originally Posted by SenorZ

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.

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?

[Ether]

Originally Posted by Ninja_Bait

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?

See this post:

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

[Ninja_Bait]

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.