Rotational Equilibrium - Hendrix College



Names:

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.)

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, 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,

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where m is the mass of the object, a is its acceleration, and F are the various forces acting 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 acceleration is zero and the object will happily continue to do whatever it has been doing without any change.

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”?

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 it on your finger? Can you draw any general conclusions about how the stick had 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?

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 to accelerate we can replace the stick with a point at the COM with the same mass and pretend that the force is acting there. The caveat to this is that the actual force must either act equally on all parts of the meter stick or act directly at the COM. In other words, the force cannot act in such a way that it tends to cause the object to rotate. This type of situation is actually the rare exception, not the norm, but 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.

Is there anything about your result that would lead you to believe that you need to do more than just have equal amounts of mass on either side of the rotation point? If so what?

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.

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. 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.

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 product is called torque, or ( (the greek letter tau). Actually it torque is not the product of mass and distance that is torque but the product of force and distance:

[pic]

where F is the amount of force and r is the distance of the force from the rotation point. (The force in our case has been the gravitational force, mg, on the masses. 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:

[pic].

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.

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.

[pic].

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.

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?

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 forget about the non-zero masses of the hangers and remember that the COM is not at the rotation point.)

|Mass (g) |Position (cm) |

|Meter stick (mMS) |xCOM=40 |

|50 |5 |

|300 |30 |

|200 |70 |

|100 |?? |

Guidebook Entry I.4: The Scales of Justice

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.

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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