# Rotational Equilibrium

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```					                         PHYS 210/230 General Physics I Lab

Names:                                 Formatted: Tab stops: 3.5", Left

Problem of the Day
I am at the playground with my two year old son who loves nothing more than to swing.
So, of course, I have to put him in the swing and get him going. I do this by pulling
straight back on the swing and then letting go. Before I let him go he is hanging there
motionless on the swing. Assume that he weights 15 kg and that I pull on him with a
force of 100 N parallel to the ground. The chains of the swing exert a force on him as
well. Determine the magnitude and direction of the force that the chains must exert to
keep him hanging there motionless. (Hint: It would be wise to draw a free-body diagram
of this system and label the angle you are using to indicate the direction of the force.)

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PHYS 210/230 General Physics I Lab

Rotational Equilibrium

Introduction

So far this semester we have only studied so-called translational motion of objects. The
orientation of an object has always been constrained relative to the coordinate axes but
the object itself has been allowed to move in any direction. For instance, when we use
the dynamics carts on the tracks the cart is always constrained to point in one direction,       Comment [TMT1]: Do you mean the cart’s
motion is constrained to point along the track?
along the track. It never changes its orientation relative to the track even though it is free
to move back and forth along the track. For this type of motion we use Newton’s laws to
determine how and under what circumstances an object moves. If the object is initially
motionless it will remain motionless unless acted on by a net force. Similarly if the
object is in motion it will move at a constant velocity (in the same direction at the same
speed) unless acted on by a net force. The ultimate motion of the object is determined by
Newton’s second law,

ma   F

where m is the mass of the object, a is its acceleration, and F are the various forces acting    Comment [TMT2]: I’d italicize m, a, and F so
that they appear as variables.
on it. When we say an object is in equilibrium we mean that there is no net force acting
   on it (the right hand side of the equation is equal to zero). The result is that the object’s    Formatted: Font: Italic

acceleration is zero and the object will happily continue to do whatever it has been doing       Formatted: Font: Italic
without any change.                                                                              Formatted: Font: Italic
Formatted: Font: Italic
In this lab we will begin exploring a new type of motion: rotational motion. As the name
suggests, we will relax the constraint that the object always remain in the same
orientation relative to the coordinate axes and allow it to rotate around. Many of the
concepts and quantities that describe rotational motion have analogues in translational
motion. The intuition you have developed about translational motion over the course of
the semester will serve you well in discussing rotational motion once we have defined
those analogous quantities. This lab will focus on rotational equilibrium and determining
the conditions necessary to achieve it.

Session I

Guidebook Entry I.1: Rotational Equilibrium Defined

What do you suppose is meant by the term “rotational equilibrium”? What would
it mean if we prefixed the term with “static” or “kinetic”?

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PHYS 210/230 General Physics I Lab

Now it is time to demonstrate rotational equilibrium. Try to balance a meter stick
on one of your fingers so that it (and you!) is motionless. When you have it
balanced motionless on your finger you have achieved rotational (and
translational) equilibrium. Congratulations! Now describe what you had to do to
achieve this state. How was the meter stick oriented? Where did you have to put
to be arranged to get it to balance?

Let’s repeat this exercise in a more formal way. At your station there should be a
little blue stand and a collection of metal clamps, some with hangers and at least
one without. Slide the clamp without hangers onto the meter stick and line it up
so the arrow on the clamp points to the location on the meter stick where you
expect it to balance. Suspend the meter stick on the stand using the little metal
pieces that project from the clamp. Does it balance out? If not, adjust the
location of the clamp until it balances and record the final position of the clamp
on the meter stick. What did you have to do to make it balance?

In an ideal world, clamping the meter stick at the 50 cm mark should have
worked. In all likelihood you found that you had to clamp it at some other point
to make it balance. Why do you think that was the case?

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PHYS 210/230 General Physics I Lab

The point about which the meter stick balances itself is called the “center of
mass”, or “COM” for short. The nice thing about the COM is that we can pretend
that all the mass of associated with the meter stick itself is concentrated at that
point. If we want to figure out how a given force will cause the entire meter stick       Comment [TMT3]: I don’t really like this
description of center of mass. I think it is a fine
to accelerate we can replace the stick with a point at the COM with the same mass         statement for translational motion, but we know that
and pretend that the force is acting there. The caveat to this is that the actual force   when we consider rotational motion, how mass is
distributed becomes relevant, e.g. moment of inertia.
must either act equally on all parts of the meter stick or act directly at the COM.       I read what follows, and I totally get what you are
In other words, the force cannot act in such a way that it tends to cause the object      saying, but if this is their transition into rotational
motion, they’re about to find out these statements
to rotate. This type of situation is actually the rare exception, not the norm, but       (like your caveat) are rare exceptions not the rules.
let’s keep it simple for now.

Guidebook Entry 1.2: Rotational Analogue to Force

At this point we have demonstrated one type of rotational equilibrium but have
not really attempted to characterize what is was necessary to achieve that
stateequilibrium. You probably have a pretty good idea about what was necessary
to balance the meter stick: you needed to get equal amounts of mass on each side
of the point about which the stick wants to rotate. This is true for a perfectly
straight, uniform meter stick but it is not the in general. requirement. Let’s see if
we can figure out what that the general requirement is.

Take one of the clamps with a hanger and place it 20 cm from the COM of the
meter stick. Hang a 200 g mass on it. Now, where do you suppose you will need
to hang another 200 g mass to get the meter stick to balance again?

Try it and record your reout. If you couldn’t get the stick to balance exactly
where you expected adjust the position of the mass until it is balanced and record
your result.sult.                                                                         Comment [TMT4]: What is their result?
Whether it rotates and falls or not? That is, when
you say “try it,” are you asking them to balance it
and record whether they were correct, or are you
asking them to put it where they guessed, and record
what happens?

do more than just have equal amounts of mass on either side of the rotation point?
If so what?                                                                               Comment [TMT5]: What are you expecting for

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PHYS 210/230 General Physics I Lab

Now try displacing one of the 200 g masses you just put on by 1 cm. Does the
meter stick still balance? What does that do to this idea about equal masses?

You should (hopefully!) have found that displacing one of the masses by a
centimeter completely spoils the rotational equilibrium. So apparently the amount
of mass is not the sole deciding factor. T but the location of theat mass along the
meter stick is also importantappears to play a role, as well. Or, to put it more
bluntly, the distance of the mass from the rotation point is important. Let’s try to
make that relationship clearer.

Adjust the position of the Start with the 200 g mass you just moved untiland put it
at the point where the meter stick is balanced. Record the mass and the distance
from the rotation point in the following table. Increase and decrease the mass on
one of the hangers by 50 g units and record where you have to put it to keep the
stick balanced. Do not forget to include the mass of the hangers themselves in
this and subsequent exercises! Try to get values for at least five different
masses.

Mass (g)     Distance (cm)

Put this data into Excel and make a plot of distance vs. mass. What does that
graph look like? (Print out a copy of the graph and include it with your
worksheet.) Can you come up with an equation that describes the relationship
between mass and distance? If so, wWrite it here.                                        Comment [TMT6]: Graphically, is it important
that it is hard to see the difference between x^-1 and
x^-n? Are they supposed to make this guess off they
graph (unlikely) or their table (more likely) or some
previous experience with the subject? Is a log/log
graph in appropriate?

You should have found that putting smaller masses further out kept the meter
stick balanced. However, it looks like the relationship is not linear. It is a curve
of some kind but it is not clear exactly what the relationship is. (Is it parabolic?
Is it exponential? Is it something else?) It is very difficult to tell just by looking
at a curved line what kind of curve it is. Let’s try a different approach. Add
another column to your Excel spreadsheet that is the product of mass and distance
and plot that column vs. mass. If you feel so inclined you can also plot it vs.

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PHYS 210/230 General Physics I Lab

distance. Doeses oneeither of those plots make the relationship between mass and
distance clearer than the other? (Print them out and include with the worksheet.)
Write down an equation that relates mass and distance and check with the
instructor before moving on.

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PHYS 210/230 General Physics I Lab

You should have found that the product of mass and the distance from the rotation
point was a constant (if not you had better check in with the instructor). This        Comment [TMT7]: How does it work in the
product is called torque, or  (the greek letter tau). Actually it torque is not the
discovery approach when you tell them what they
were supposed to get? Do most students not go back
product of mass and distance that is torque but the product of force and distance:     and change their answers? How do you get
discrimination between groups? Or is that not
important in this method?
  rF

where F is the amount of force and r is the distance of the force from the rotation    Formatted: Font: Italic
point. (The force in our case has been the gravitational force, mg, on the masses.     Formatted: Font: Italic

We have been neglecting g because it cancels when you look at both masses.)

This equation gives us the magnitude of the torque but not the direction. It should
be fairly clear that this is a quantity that should have a direction. Masses hanging
to the right of the rotation point tend to cause the meter stick to rotate in a
clockwise direction whereas those on the left tend to rotate in a counter-clockwise
direction. This is because the forces that the masses exert due to the gravitational
force are always in the same direction, down, but are acting on opposite sides of
the rotation point. To perform calculations we assume that torques that cause the
object to rotate counter-clockwise are positive and those that cause clockwise
rotation are negative.

This may strike you as an odd choice but it is due to the fact that torques are
actually vector quantities:

  r  F .

The multiplication is called a cross product and is a way of combining two vectors
in such a way that you get a third vector. (It comes from vector calculus so don’t
   be concerned if you don’t recognize it.) We use the right-hand rule to figure out
which way the torque vector points. With your right hand point your fingers in
the direction of r (from the rotation point to the point where the force acts). Then
curl them in the direction that the force acts. If you stick your thumb out it will
point in the direction of the torque vector. If the torque points “into the page”
(rotations in a clockwise direction), then it is considered to be negative.

To make things even more complicated, the magnitude of the torque is really
rFsin() where  is the angle between the vector pointing from the rotation point
to the point where the force acts and the force vector. In the situations we have
considered so far those vectors have been perpendicular. As a result the sin()
has always equaled one. We will continue to work with systems with similar
constraints in this lab.                                                               Comment [TMT8]: Is there a way for them to
“discover” this?

The condition for rotational equilibrium is essentially the same as the one for
translational equilibrium: the sum of all the torques acting on a system must
equal zero.

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PHYS 210/230 General Physics I Lab

0  .

In this case, all torques that would cause the system to rotate in a counter-
clockwise direction are considered to be positive whereas those that cause it to
   rotate clockwise are negative.

Guidebook Entry I.3: Applying Torques to Different Systems

Let’s get some practice in dealing with torques on different physical systems.
Remove the masses and hangers from the meter stick and make sure it balances
on its own. Record the location of the COM if it is different from before. Hang a
5200 g mass at the 20 cm mark on the meter stick. Use a spring scale to exert a
force downward on the meter stick such that the system balances. Record the
location and amount of force. Calculate the total torque in the positive and
negative directions and calculate the percent difference. Do your results make
sense? Why or why not? Do this at a couple of different points along the meter
stick.

Repeat this exercise but this time useusing the spring scale to exert a an upward
force upward.

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PHYS 210/230 General Physics I Lab

Place a 100 g mass at the 10 cm mark on the meter stick and a 200 g mass at the
75 cm mark. Experimentally determine where you have to hang a third mass of
50 g to balance the system and record the position.

Calculate the total torque in the positive and negative directions and calculate the
percent difference. Do your results make sense? Why or why not?

Now let’s see if we can determine the mass of the meter stick without weighing it
on a scale. This can be done with a single known mass, say 200 g, and the known
location of the COM. How would you go about doing this? Write down the
appropriate equation (or equations) and solve them for the mass of the meter stick,
mMS. Remember, we can assume that all the mass of the meter stick is
concentrated at the COM.

Use your procedure to determine the mass of the meter stick. Compare the result
of your technique with the result you get from weighing it on the scale. Do the
results agree?

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PHYS 210/230 General Physics I Lab

You should be getting pretty good at manipulating torques, theoretically and
experimentally, to achieve rotational equilibrium for a given system of masses.
So let’s try a slightly harder system, one involving four masses plus the meter
stick. Adjust the clamp on the meter stick so that it is at the 40 cm mark; we want
the meter stick to rotate about the 40 cm point. The following table lists the
masses and positions to be used in this system. Your job is to determine the
position, theoretically and experimentally, for the fourth mass. Start by finding
the position of the fourth mass theoretically then place the mass there and see if it
works. Adjust the mass’ position until the system balances and calculate a
percent difference between your theoretical and experimental results. (Don’t            Comment [TMT9]: Don’t you need to state from
which point these positions are measured from? And
forget about the non-zero masses of the hangers and remember that the COM is            when you say Xcom, don’t you mean the position of
not at the rotation point.)                                                             the hanger? I’m just a little confused.

Mass (g)            Position (cm)
Meter stick (mMS)        xCOM=40
50                    5
300                   30
200                   70
100                   ??

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PHYS 210/230 General Physics I Lab

Guidebook Entry I.4: The Scales of Justice                                      Formatted: Font: Bold

As a final exercise let’s use this system to determine the mass of an unknown
object. Hang an unknown object of your choosing on the meter stick. Use two
known masses placed at different points on the meter stick to make the system
balance. Determine the unknown mass using this information.

Formatted: Indent: Left: 0"

Weight the unknown mass on the scale and compare the two results. Comment
on the accuracy and precision of this approach to weighing objects.

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