# Homework Help: Motored wheel

1. Dec 4, 2008

### saltine

1. The problem statement, all variables and given/known data
A motor with wheel is attached to the end of a pole. The device is initially vertical and at rest when the motor is switch on and supplies torque τ to the axle. As the wheel begins to roll clockwise, does the top of the pole moves to the left or to the right?

3. The attempt at a solution
I am confused by this problem, because as the motor exert the torque to the wheel, doesn't the wheel exert the same torque to the motor? So if I take off the wheel and look at only the free body diagram of the rod&motor, then torque acting on it would be counter-clockwise, so the rod would move to the left. But the wheel is turning clockwise, so it should be moving to the right also (?).

How should I consider the problem?

- Thanks

2. Dec 4, 2008

### LowlyPion

The motor is attached to the pole. So shouldn't the pole reflect the rotational motion of the motor in producing the torque. (That action/reaction thing?)

3. Dec 4, 2008

### saltine

So the pole will fall to the left with respect to the axle, while the axle will be translating to the right?

4. Dec 5, 2008

### LowlyPion

That's what it looks like to me.

5. Dec 5, 2008

### saltine

To calculate the translation acceleration at the axle, I would related the clockwise τ to α by

α = τ/I ... (1)

Then if the wheel rolls without slipping, then a = r α. In (1), what I should I be using? Is it just the moment of inertia of the wheel? Does the mass of the pole have an effect? I am thinking that since τ is given, whatever effect the mass has must be taken care of by τ, so I would be that of the wheel only.

atransx = r τ/I = 2 τ/mr

Then I use the counterclockwise τ to get the acceleration at the top of the rod by using the moment of inertial of the rod.

atangx = -r τ/I = - 3τ/m(2d) = -1.5τ/md ... (3)

Summing them up:

atopx = ( 2/r - 1.5/d ) τ/m ... (4)

This means that whether the rod fall to the left or the right depends on r and d. In particular when:

(2d - 1.5r)/(rd) > 0
d > 0.75 r

The top of the rod would be moving forward. According to the drawing, since d is longer than r (not to mention 2 times d, which is the length of the rod), the rod will be moving to the right (while falling counter-clockwise).

Is this interpretation correct?

But now I have another question:

Suppose the rod is shorter than r, how do I know that the wheel would turn at all? How do I know that the motor is not just spinning itself about the axle, while the wheel does not move?