Sliding Yardstick

Photo Courtesy of Wikimedia Commons
Original Image by RJFJR

Support a yardstick (or meter stick) horizontally atop your open palms, with one hand near each end. (Or you can rest it on two fingers, or the backs of your hands).

If you slowly slide your hands towards each other, the yardstick will have to respond in some fashion. Generally it will try to stay in the same place on your hand, restrained by friction, until your hands have moved too far and it is forced to slip. Then it will slip on one side or the other, or sometimes both. Usually it will slip on one side briefly, then on the other, then back to the first, so that when your hands finally meet, they are very nearly in the center of the yardstick.

What's interesting is that your hands will end up at the center of the yardstick even if they don't start out at approximately equal distances from the center. If one hand is slightly closer to the center, the stick will slip more against the other hand. You can even start with one hand very nearly at the center, and at the end both hands will end up at the center of the stick.

Why is this? What determines where the stick ends up relative to your hands? And how can you change the outcome so that your hands end up, say 1/3 of the way from one end?

Part A

What are the forces acting on the Yardstick before the hands start moving?

Solution

System:

The Yardstick is a rigid body subject to torque (single-axis).

Interactions:

External forces due to gravity (near-earth) and the two hands.

Model:

Single-Axis Rotation of a Rigid Body and Point Particle Dynamics.

Approach:

Diagrammatic Representation

A Force Diagram of the yardstick looks like this. gravity (interaction) pulls downward at the center of mass with force mg (where m is the yardstick's mass) and is resisted by the normal forces F₁ and F₂ of the hands

Mathematical Representation

We now write the equations of Newton's 2nd Law for the center of mass and of torque balance for the system. Since the system is at rest, we can put the axis wherever we choose. The simplest point is at the center of mass , since that removes one term from the expression for torque (single-axis). The torque must be zero, since the yardstick is not rotating.

Force balance in the vertical direction gives:

Using the first equation to eliminate one of the normal forces and substituting into the second gives, after a little algebra:

The closer one hand is to the center of mass, the greater the normal force on it.

Part B

Now imagine that the hands start moving slowly toward the center. There is frictional force preventing easy sliding of the yardstick on the hands, with coefficient of friction μ , but eventually the stick must slide against the hands. What happens?

The yardstick will try to remain stationary, with the skin on the hands moving against the underlying tissue until it can go no farther. We know that the force of friction will prevent motion until the force exceeds F_max given by

where N is the normal force.

Solution

System:

The Yardstick is a rigid body subject to torque (single-axis) and friction.

Interactions:

External forces due to gravity (near-earth) and both frictional and normal forces from the two hands.

Model:

Single-Axis Rotation of a Rigid Body and Point Particle Dynamics.

Approach:

Diagrammatic Representation

We again sketch the situation, this time adding friction.

Mathematical Representation

The magnitude of the sum of the horizontal forces is

Substituting from Part A for the normal forces F₁ and F₂ gives

If a > b , then the normal force F₂ is greater than the normal force F₁, and hence the frictional force due to the hand a distance b from the center of mass is greater. Therefore that hand will remain fixed relative to the yardstick and the other hand will slide, bringing it closer to the center of mass of the yardstick.

As it does so, the distance a will change, and the distribution of normal forces will change – that on the left hand will begin to increase, and that on the right hand will decrease by the same amount. The two normal forces will become equal when the new distance between the left hand and the center of mass equals b. At this point, the normal forces will be equal, and so will the frictional forces.

As you continue to press your hands towards each other, the forces will eventually cause the yardstick to slip against your hand. In an ideal world, with the same normal force on each hand and the same coefficient of friction, the forces on both hands will be equal and you would expect both hands to move toward the center. Inevitably, however, all things will not be equal – the coefficient of friction will vary from place to place on the yardstick, for instance, and one side will move first. But, by the same logic as above, the hand closer to the center of mass will have more normal force on it, and hence the frictional force will increase and will eventually stop the hand from moving.

We then have the re-appearance of the initial state, with one hand closer to the center of mass than the other, and consequently having more normal force, so it is the hand farther from the center of mass that now moves until the hands are again equidistant from the center of mass.

The hands and the yardstick proceed in this way, with one hand moving a little bit closer, then the stick stopping and the alternate hand catching up, until the hands meet, at which point the center of mass will be very nearly overhead, to within the range of one of these motions.

Part C

How can we make the hands meet at some other point than the center of the yardstick?

Solution

System:

The Yardstick is a rigid body subject to torque (single-axis) and friction.

Interactions:

External forces due to gravity (near-earth) and both frictional and normal forces from the two hands.

Model:

Single-Axis Rotation of a Rigid Body and Point Particle Dynamics.

Approach:

Since the hands slip or stick based upon their distance from the center of mass, and they therefore end up under that point, the obvious way to make the hands finish at a different location than the center of the stick is to move the center of mass. One can use a stick with a mass that varies with position to achieve this. The easiest way, however, is simply to put a large mass at one end of the yardstick. If you're doing this as a demonstration, you can tape a mass at one end, or stick a ball of clay there. If the mass added is M at the extreme end of the stick, then the new center of mass will be at a distance D from the end without the mass, where

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\begin

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[D = \frac{m {\frac

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{2}}+ML}

Unknown macro: {m + M}

= \left( \frac

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\right) \left( \frac

Unknown macro: {m + 2M}

Unknown macro: {m + M}

\right) ]\end

As M increases, the center of mass moves from the center of the yardstick towards the far end.

Question

The center of mass is the point which behaves as if all the mass is concentrated at – if you were to throw the yardstick, the center of mass would execute a parabolic path, just as if I threw a point mass. But we have performed the calculations above on the assumption that the torques due to gravity cancel out. Is it only coincidence that the point your hands end up at is also the center of mass?

For a one-dimensional object, the center of mass of a stick having density ρ is defined as

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\begin

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[CoM = \frac

Unknown macro: {int x rho (x) dx}

Unknown macro: {int rho (x) dx}

= \frac

{M_{total}} ]\end

The condition that the torques due to gravity cancel out is the integral of torques on one side plus the integral of torques on the other side sum up to zero

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\begin

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[ gM_

Unknown macro: {total}

\int (x - CoM) \rho dx = 0 ]\end

We can divide by gM_total without changing the result (since it's equal to zero) and substitute from the above expression to get

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\begin

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[ CoM M_

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-CoM M_

= 0]\end

So the center of mass is not only the point that acts as if all the mass is concentrated there, but is also the point about which the torques due to gravity cancel out. You can suspend an object from its center of mass and it won't rotate.

Child pages

Sliding Yardstick

Part A

Solution

Part B

Solution

Part C

Solution