Physics Quiz

[Ether]

A wheel and motor assembly is connected by a rigid massless rod to a frictionless pivot. See the attached Figure. The angle "theta" between the wheel's axis and the connecting rod is fixed.

If the motor is applying a torque "tau" to the wheel, and the wheel radius is "r", what is the magnitude and direction of the floor's friction force on the bottom of the wheel? Assume that the torque is small enough so that the wheel is not moving.

....I'm just in chemistry.... However, I do look forward to the answers that others post!

[~Cory~]

Is this a super creative off season teaser?

[Ninja_Bait]

If the torque was high enough and the wheel was free to move, would it be moving in the positive y or negative?

I'm going to assume the positive case. Off the top of my head the forces on the wheel I can think of are the tension in the rod, the torque from the motor, the friction, the weight and the normal. Am I missing anything?

Also, it sounds like the wheel is in static equilibrium, so I think the friction points somewhere in the southeast direction...

I hate physics.

[ebarker]

I want to use some of that mass-less rod in our robot - saves weight.

[Ether]

Originally Posted by Ninja_Bait

If the torque was high enough and the wheel was free to move, would it be moving in the positive y or negative?

The wheel is free to move... as soon as enough torque is applied to overcome the static friction on the floor.

I forgot to clarify the direction of the torque being applied to the wheel. So here goes:

Assume that the torque being applied to the wheel is in the direction to try to make the wheel travel counter-clockwise around the pivot.

I'm going to assume the positive case.

Yes, counterclockwise.

Off the top of my head the forces on the wheel I can think of are the tension in the rod, the torque from the motor, the friction, the weight and the normal. Am I missing anything?

Do a free-body diagram of the wheel+motor assembly, and just consider all the external forces acting on that assembly in the horizontal plane. So all you need to consider are the tension in the rod and the friction force of the floor acting on the wheel.

Also, it sounds like the wheel is in static equilibrium

Correct. The problem specified that the torque was not sufficient to overcome the static friction.

so I think the friction points somewhere in the southeast direction...

A common mistake. Think free-body diagram, and make the forces balance in the horizontal plane.

I hate physics.

Say it ain't so! You're doing great so far.

[Ether]

Originally Posted by ebarker

I want to use some of that mass-less rod in our robot - saves weight.

And it's perfectly rigid too!

It's made from unobtainium.

[theprgramerdude]

When you said a torque tau is applied to the wheel, did you really mean the rod? If you meant the wheel, then wouldn't that make the angle theta irrelevant to solving the problem?

[Ether]

Originally Posted by theprgramerdude

When you said a torque tau is applied to the wheel, did you really mean the rod?

No. Torque is applied to the wheel by the motor on the wheel. This torque on the wheel then tries to drive the wheel "forward", but it can't go straight "forward" because it is constrained by the rod to travel in a circle. But it can't travel along that circle until sufficient torque is applied to break the static friction with the floor.

If you meant the wheel, then wouldn't that make the angle theta irrelevant to solving the problem?

No. The angle of the wheel affects how much friction force the floor generates in response to the torque on the wheel.

[James Critchley]

Without using any "special" observations what you really should do is draw the Free Body Diagram and sum moments about the pivot point (in the plane normal to the sketch). This gives one equation and one unknown.

Then when reconstructing the total friction force from the longitudinal (given/known/by inspection) and lateral (unknown) components on the wheel, some sines and cosines will cancel out for you. This then makes clear the "two force" member observation which Ether has stated. If you see this right away, then you can jump straight to the answer without any diagrams.

A useful analogy to this system is a static screw. The wheel angle (incline) acts similar to the pitch of a thread. Following this you should watch out when reporting any answer for theta = N*pi (for N any integer neg, 0, or positive). This should also be clear from initial observations.

[amoose136]

Hmm... cool problem.
Given that torque is F*r and force in the direction of the axle will cancel out, the magnitude of friction = (tau/r)*sin(pi/2-theta) and direction = theta-pi/2. That was quickly arrived at so I probably missed something.

****angle is given relative to the angle of the axle*********

[Ninja_Bait]

I'm still confused, but I think I might be slowly getting it.

I made an error in confusing fixed with given, with respect to theta. I thought the wheel could pivot at the origin, but it can't. :\

So on the rod there is a net torque of 0 (preventing circular motion about the pivot point) and on the wheel there is a net force of 0 (preventing translation on the xy plane), right? I think that gives me three equations, torque, force in the y and force in the x, for three unknowns, tension, magnitude of friction and direction of friction. I think I can do it...

It's still a little too early in the morning for me to solve this independently, but did amoose get the right answer?

[Ether]

Originally Posted by Ninja_Bait

I made an error in confusing fixed with given, with respect to theta. I thought the wheel could pivot at the origin, but it can't. :\

If by "origin" you mean the center of the wheel, then it cannot pivot there. The rod is rigidly connected there.

Originally Posted by Ninja_Bait

So on the rod there is a net torque of 0 (preventing circular motion about the pivot point) and on the wheel there is a net force of 0 (preventing translation on the xy plane), right?

Correct.

Originally Posted by Ninja_Bait

did amoose get the right answer?

Close, but no cigar.

[brndn]

Well since τ = rFsinθ and the wheel isn't able to move, then F equal to the force of friction. In that case F = τ/(rsinθ) opposite the direction of the wheel turning.

[John]

The sum of the moments around the pivot must be zero. The forward friction on the wheel and the sideways friction on the wheel are the only two forces with a component perpendicular to the rod. So

0 = forward friction * length * cos(theta) - sideways friction * length * sin(theta)

the forward friction would be F= tau/r, so solving for the sideways friction gets:

0 = tau/r *length * cos(theta) - sideways friction * length * sin(theta)

sideways friction = tau/r *cos(theta)/sin(theta)
sideways friction = tau/r * cotan(theta)

The total magnitude of the total friction force is just
F = sqrt ((tau/r)^2 + (tau/r * cotan(theta))^2)
F = tau/r * sqrt(1+cotan(theta)^2) = tau/r * (1/sin(theta))^2 = tau/r * cosec(theta)^2

Edit: should be F= tau/r * cosec (theta) (forgot to take square root)

The angle from the horizontal is
arctan((tau/r)/(tau/r * cotan(theta))) = arctan (theta) =theta

This makes sense because the friction force has to act exactly opposite to the tension force, or the wheel won't be in static equilibrium.