b Lab Physics 211 Lab 10 Torque What You Need To Know: Please read this introductory material carefully; it covers topics you might not yet have seen in class. F (a) (b) FIGURE 1 Forces acting on an object F F Angular Systems Every lab up to this point has dealt with motion in straight lines. Even projectiles, which move on parabolas, can be understood as a combination of straight-line motion (constant velocity) in the x direction and straight-line motion (free fall) in the y direction. We have used linear quantities to describe this straight-line motion, such as the following: position, velocity, acceleration, force, and momentum. In this lab, the collection of quantities and laws of physics that deal with straight-line motion (or combinations of straight-line motions) are called the linear system, and we ll use the adjective linear to refer to quantities in that system when a distinction is important (such as linear velocity). Starting in this lab, we are going to deal with the angular system, in which objects rotate rather than move in straight lines. Every idea in the linear system will have an analogous idea in the angular system. For example, in the linear system we have velocity (which we now specify as linear velocity). Linear velocity tells you how far an object travels in a straight line every second (in units of m/s). In the angular system we have angular velocity which tells you how many times an object rotates every second (in units of revolutions/s) or how much angle an object sweeps out per second (in units of rad/s). Today, we will introduce torque, which is the rotational analog of force. Force and Newton s 2 nd Law Before we get to torque, let s jump back a bit and review some ideas in the linear system. Recall that a force is a push or a pull on an object. More specifically, if you apply a single force, F, to an object, then that object will accelerate in a straight line. See Figure 1a. According to Newton s 2 nd Law, (ΣF = ma, net force equals mass times acceleration) the object accelerates because it has a nonzero net force acting on it. (Convince yourself that if only one force acts on an object, then the net force on the object is nonzero unless the force acting has a strength of zero Newtons.) If there are two or more forces acting on an object then the object will accelerate unless the net force (i.e., the vector sum of all of the forces) is zero. This can happen with two forces if they point in the opposite directions and have the same magnitude. See Figure 1b.
r T Torque Now let s look at the rotational analog of force that is part of the angular system. The analog to force in the angular system is called torque. The net torque is the vector sum of all torques acting on an object. Here are the key ideas for force and torque: A nonzero net force causes an acceleration: the velocity changes (in magnitude, direction, or both) as time passes. If the net force is zero, then the object moves at constant velocity. A nonzero net torque causes an angular acceleration: the angular velocity changes (in magnitude, direction, or both) as time passes. If the net torque is zero, then the object rotates at a constant angular velocity. Force is the physical quantity that causes objects to change their linear motion. Torque is the physical quantity that causes objects to change their angular motion (in other words, their rotational motion). The magnitude of the torque acting on an object is defined as follows: = r? F FIGURE 2 Rope acting on a pulley Here τ is the torque acting on an object. It is determined by the force F acting on the object and by the perpendicular distance between the force and the chosen origin (which often, but not always, is chosen to be at the center of the rotational motion). From now on, we ll simplify our notation by letting r?! r so = rf, but remember, r is not a radius; it s the perpendicular distance between the place where the force is applied and the center of rotation Here s an example. Let s say you have a pulley with a rope wrapped around it. You pull on the rope which gives the rope a tension, T. This is the force F applied to the pulley. See Figure 2. Let s choose the origin to be at the center of the pulley. Then the distance, r, is shown; it is the perpendicular distance from the reference point to where the tension is acting. Perpendicular here means perpendicular to whatever direction the applied force is acting. In this case, the perpendicular distance happens to be equal to the radius of the pulley. In this lab, we will only consider torques where the force is perpendicular to the radius, although with a bit more math, we could have generalized this. See Figure 3a. The basic idea is that if you choose an origin so that the radius (the line from the origin to the place where the force is applied) is not perpendicular to the direction of the applied force, then you must break the force into components, with only the perpendicular component contributing to the torque. (Equivalently, you could treat the radius as a vector and break that into components, with only the perpendicular component contributing to the torque.)
(a) r T Torque is a vector. Besides a magnitude, it has a direction. The direction of the torque is given by a right-hand rule. If you point your right hand s fingers along the radius vector (the vector pointing from the origin to the place where the force is applied) and then curl your fingers in the direction of the applied force, the torque will point in the direction of your thumb. In Fig. 3a, the radius vector points to the right, and the tension points down. Hold your right hand fingers pointing to the right, with the radius, and then curl them downward toward the tension force. Point your right hand fingers from the origin outward along the perpendicular radius r in Fig. 3a. Curl your fingers so they now point in the direction of T. Your thumb should be pointing into the page; this is the direction of the torque caused by the force T. Question 0.a: You push open a door that opens outward, away from you, with hinges on the left. In what direction is the torque your hand exerts on the door? (Use the right hand rule.) T r + (b) r T FIGURE 3 Torques acting on an object Net Torque The equation for Newton s 2 nd Law in the angular system is very similar in form to the one for the linear system. Στ = Iα Στ is the net torque on an object (in meter-newtons, meter-newtons) I is the moment of inertia on an object (in kg m 2 ) α is angular acceleration of an object (in radians/s 2 ) All of the variables are analogous to each other; force and torque, mass and moment of inertia, and linear acceleration and angular acceleration. You simply switch out one for the other. Just as torque is the analog of force, angular acceleration is the analog of acceleration. Moment of inertia is the analog of mass: it is how much an object resists changes to its angular velocity. Note that torque has units m N. This is mathematically equivalent to N m, but we never write torque to have units of Joules, because torque is not a form of energy. It just happens to have similar units. The ideas work out the same as well. In Figure 3a there is one torque acting on the system (in this case, τ = Tr). If you apply a single torque to an object it will angularly accelerate, and according to Newton s 2 nd Law for the angular system. This is so because if there is only one torque acting on an object then there must be a net torque and therefore an angular acceleration, unless the applied torque is zero m N. If there are two torques acting on an object then the object might or might not angularly accelerate. If the two torques are equal but pointing in opposite directions then they cancel out and the net torque is zero and the object will not accelerate. See Figure 3b. Question 0.b: What direction are the torques caused by each tension force in Fig. 3b? Use the right hand rule. Are the torques in the same direction, opposite directions, or neither?
Just like in the linear system in which we have to choose a positive direction (e.g., a +x direction and a +y direction), now we will have to choose the positive direction for torque. It will typically be into or out of the page, so you can think of this as choosing the +z direction. Special note: Normally, we choose +x to point right and +y to point up. We normally choose +z so that if you point your right hand s fingers in the +x direction and curl them to point in the +y direction, then your thumb points in the +z direction. This is out of the page if +x is to the right and +y is up within your piece of paper. Throughout this lab, use this right-hand rule to find the +z direction, always choosing +x is to the right (when standing in front of the meterstick+fulcrum) and +y is up. Suppose the pulley in Fig. 3a began at rest. The torque is into the page (check this using the right-hand rule), so it is in the z direction. This torque turns out to cause the pulley to rotate clockwise, if it begins at rest. For this reason, sometimes we will call an into-the-page torque a clockwise torque. But remember that torque is a vector that points in a direction perpendicular to both the radius and the force; it isn t a curvy arrow pointing clockwise or counterclockwise. Look at Fig. 3b. Following the usual convention, we choose the direction out of the page as the +z direction (+z = out of the page). Now we can say that the tension in the rope on the right is causing a torque in the z direction (into the page), which sometimes we might call the clockwise direction. This tension tends to cause the pulley to angularly accelerate in the clockwise direction. The tension in the rope on the left in Fig. 3b is causing a torque in the +z direction, or out of the page (check this with the right-hand rule!). Therefore, the torque caused by the left rope tension tends to make the pulley angularly accelerate counterclockwise. These two torques cancel each other out and the pulley does not rotate at all. A note about the Center Of Mass and Free-body diagrams Just as with the linear system, free-body diagrams can help you add up all the torques acting on an object to get the correct net torque, with all the torques having the correct signs. In the linear system, the forces acting on the object are drawn coming from a dot. That dot actually represents the point particle the forces are acting on. We often dealt with objects that are extended, not just point particles. This works by treating the object as a point particle that exists at the location of the object s center of mass. (A reminder about the center of mass: the center of mass of an object or system is a weighted average, just like your GPA is a weighted average. Specifically, divide the object into pieces of mass m1, m2, m3, etc. The center of mass is the weighted average of the position: multiply each pieces position vector by its mass, add up the results for all the pieces, and divide by the mass. The center of mass is also the balance point of an object: if you place an object so its center of mass is directly above a spike, the object will balance.)
Now, with torque we can see that the location of the force is important. So, now you have to be more careful in which place you draw your weight vector in your free-body diagrams. The weight of an object acts at the location of the center of mass of the object. You should be drawing it at the center of mass of your object.
Here is an example of center of mass. Let s say we have a meter stick. Its mass is uniformly, evenly distributed throughout itself. Divide the ruler up into bits 1-cubicmm in size, say; each bit has the same mass. When you do the weighted average, the center of mass will turn out to be at the center of the meter stick, since each bit on the left has a bit of equal mass on the right. This is where we would place the weight vector of the meter stick. See Figure 4a. Now, let s say that we drill a hole on the right end of the meter stick. The mass of the stick is no longer distributed uniformly across the stick because of the hole. This would cause the center of mass of the stick to be shifted to the left, since there is now a bit of meter stick on the lift that is not canceled by an identical bit on the right. See Figure 4b. W (a) W (b) HOLE FIGURE 4 Centers of Mass for meter sticks In today s lab you will be dealing with a meter stick whose mass is not uniformly distributed. They have been altered specifically for this lab. So, one of the first things that you will have to do is find the center of mass of the meter stick.
The Equipment The most basic part of the lab is the fulcrum and meter stick. See Figure 5. The fulcrum is mounted to a block of wood and is used as a pivot point for the meter stick. There are also three clamps on your table. Two of them have swinging hangers on them, one doesn t. See Figure 6. You can attach clamps to the meter stick to either hold the meter stick up at the pivot point on the fulcrum or you can use them to hang extra masses. See Figure 5. CLAMP WITH SCREW POINTING UP CLAMP WITH SCREW POINTING DOWN CLAMP MEASURING EDGE MASS HANGER PIVOT POINT FULCRUM METER STICK EXTRA MASS SWINGING HANGER FIGURE 6 - Clamp FIGURE 5 - Possible Lab Setup When you use these clamps you will be taking measurements from them on their location. Use the edge on the clamp that is shown in Figure 6. You should also have two mass hangers as well as a set of masses that you set on the platform of the hanger. See Figure 5.
What You Need To Do: Part 1 - Center Of Mass In order to find all the torques on an object, you have to include the torque due to the weight of the object. Depending on the reference point that you choose, the weight of the object might have a torque. The location at which the weight of the object acts is at the center of mass of the object. So, the first thing we are going to do is find the center of mass of the meter stick. NOTE: On the last page of the lab there is a chart that you can use to fill in your data. You can answer questions on this sheet too. Use your own paper if you need more room for answers and calculations. A) Slide the clamp without the swinging hanger onto the meter stick until it gets to about the center of the stick. NOTE: Make sure the screw on the clamp is pointing down and that the metric side is facing you. B) Place the meter stick on the fulcrum with the clamp right at the pivot point. See Figure 5. Adjust the location of the clamp back and forth until the stick remains relatively horizontal. C) Once the stick is balanced, record the location of the clamp. This is the location of the center of mass of the meter stick. You will be using the value throughout the lab. Notice that, before the clamp was at the center of mass, the stick angularly accelerated from rest, rotating about the fulcrum. That means that there must have been a net torque on the stick. The weight of the stick was acting at a distance, r, away from the fulcrum and thus was causing a net torque (i.e. the stick rotated). Once you placed the clamp right at the center of mass, the weight of the stick was acting at the fulcrum. That means that the perpendicular distance r = 0 and there would not be a torque (i.e. the stick remained horizontal). Part 2 - Single Torque Leave the clamp at the center of mass of the stick and keep this setup on the fulcrum. A) While holding the meter stick horizontal, slide a clamp with a swinging hanger to about the 80 cm mark. B) Release the meter stick. Describe what happens and also explain why in terms of what has been discussed so far.
C) While holding the meter stick horizontal, remove the clamp at 80 cm and place it on the other side of the stick at 20 cm. D) Release the meter stick. Describe what happens and also explain why in terms of what has been discussed so far. For each of the instances above, in adding the clamp to one side of the stick we added a torque to the system, changing the net torque from zero. This caused the stick to angularly accelerate. One way caused a torque in the +z direction (toward you, when you re standing in front of the device), the other way caused a torque in the z direction (away from you, when you re standing in front of the device). Part 3 - Net Torque So far there has only been one torque acting on the ruler. Now you will be dealing with a system that has multiple torques acting on it. A) Using the digital scales in the back of the lab, find the mass of each hanger assembly which includes the clamp, the swinging hanger, and the mass hanger. See Figure 5 & 6. Do not assume that each assembly has the same overall mass. Find both separately. B) While the meter stick is balanced at its center of mass, place one hanger assembly at 20 cm on the stick. NOTE: Have the screw on the clamp pointing upwards for any clamp with a swinging hanger. It s easier to adjust that way. See Figure 5. C) Holding the meter stick horizontal, take the other hanger assembly and slide it on the stick from the right hand side until the stick remains horizontal when you let it go. Question 1 How many forces are acting on the meter stick? Draw a diagram of the meter stick showing each force vector acting on it. Place each force vector in the approximately correct location where that force is acting.
Question 2 If the location of the fulcrum is taken as the reference point (this will be the case for the entire lab), how many torques are acting on the meter stick? Why is this number different than the number you wrote for the answer to Question 1? Question 3 In what directions are the torques acting (+z=toward you, or z=away from you)? Question 4 What is the net torque acting on the meter stick? How do you know this by looking at the meter stick? D) Calculate the weight of each hanger assembly. (Make sure you are using units of kilograms for mass and units of Newtons for forces, such as weights.) Put these values in the chart in the row labeled Part 3, under the columns labeled as F. NOTE: In the chart there are extra columns for torques that you might not use depending on which part of the lab you are working. This is left ambiguous in order for you to determine how many torques you have, either clockwise or counterclockwise. E) Measure the distance, r, (in meters) for each torque. Remember, r is defined as the distance from your chosen reference point (in this case, the fulcrum) out to where your force is acting. Place these values in the chart.
F) Calculate the torques, τ, and put these values in the chart. Also, put these values in for the total torques since there is only one torque per side. Calculate a percent difference between these totals. If you got a value greater that 5% then you are doing something wrong. Go back and check your data. Part 4 - C of M as Torque For this section you will be moving the fulcrum away from the center of mass. This will cause the weight of the meter stick to have a torque. A) Go to the back of the lab room and measure the mass of the meter stick WITHOUT the fulcrum clamp on it. Reattach the clamp and place it at 60 cm. B) Put the meter stick back on the fulcrum. Add an extra mass of 100 g to a hanger assembly. Place this hanger assembly at a point on the meter stick so that the meter stick is balanced. Question 5 How did you know where to place the hanger to balance the stick? Explain in detail using the formula for the magnitude of torque. C) Draw a diagram of the meter stick showing each force vector acting on it. Place each vector in the approximately correct location as well. Describe the torques acting on the stick. D) Calculate the weights that are causing the torques on the stick and place these values in the appropriate places in the row labeled Part 4. Make sure you now include the 100 g mass for the hanger assembly. E) Measure the distance, r, for each torque. Place these values in the chart.
F) Calculate your torques and total torque. Calculate a percent difference between your total torque values. If you got a value greater that 5% then you might have made a mistake; please go back and double-check. Part 5 - Multiple Torques on a Side Remove the hanger assembly currently on the meter stick. Keep the fulcrum at 60 cm. A) Add an extra mass of 200 g to one of the hanger assemblies and place it at 90 cm. B) Add an extra mass of 100 g to the other hanger assembly and place it on the meter stick to balance the system. C) Draw a diagram of the meter stick showing each force vector acting on it. Place each vector in the approximately correct location as well. Describe the torques acting on the stick. D) Calculate the weights for each torque acting on the stick and place these values in the appropriate places in the row labeled Part 5. E) Measure the distance, r, for each torque. Place these values in the chart. F) Calculate your torques and total torque. Calculate a percent difference between your total torque values. If you got a value greater that 5% then you are doing something wrong. Go back and check your data. Part 6 - Mixin It Up Remove all hanger assemblies. A) Place the fulcrum at 75 cm. B) Take one hanger assembly and place it at 84 cm. C) Take the other hanger assembly and add 20 g to it. By either adding mass (NOTE: add no more than 150 g) to the first hanger assembly AND/OR adjusting the location of the second hanger assembly, balance the system. (You can place the second hanger anywhere on the meter stick.) D) Draw a diagram of the meter stick showing each force vector acting on it. Place each vector in the approximately correct location as well. Describe the torques acting on the stick. E) Appropriately fill out the row labeled Part 6 in your chart as you have been.
Part 7 Calculating Tension So far the only torques you ve been dealing with have been caused by weights. Now, a tension is going to cause a torque. A) Open the file FORCE PROBE. Push the CONNECT button on the extra window that opens. With nothing hanging from the force sensor, push the button to the left of the COLLECT button to zero the sensor. B) Place the fulcrum at 20 cm. C) On the table there is a piece of string with one end tied in a big loop and the other in a small loop. Slide the big loop around the ruler until it is at 70 cm. Have the small loop hang on the hook on the Force Sensor which is hanging from a stand. Adjust the height of the Force Sensor until the meter stick is horizontal. See Figure 8. FULCRUM FORCE SENSOR HOOK LOOPED STRING FIGURE 8 - Lab Setup for Tension D) Take one hanger assembly and place it at 8 cm. Put an extra mass of 150 g on the mass hanger. E) Take the other hanger assembly and place it at 83 cm. Put an extra mass of 100 g on the mass hanger. F) Draw a diagram of the meter stick showing each force vector acting on it. Place each vector in the approximately correct location as well. Describe the torques acting on the stick. So far you have been verifying that the clockwise torques equal the counterclockwise torques. Based on this idea, make an equation that will allow you to calculate the tension in the string. Make sure you use units of kilograms and meters so that you will get Newtons for tension. Calculate the tension. G) On the screen there will be a small window that tells you the tension in the string. Record this value. H) Calculate a percent difference between the computer value and your calculated value. If you got a value greater than 10% then you are doing something wrong. Go back and check your data and calculations. I) Remove all clamps from the meter stick. Thanks!
What You Need To Turn In: This lab report Use your own paper for additional pages. Make sure you show your work for each section in the corresponding section in your report. Answer all of the questions. Torque Lab Report P a r t Torques in +z direction torques in z direction 1 st Torque 2 nd Torque Total 1 st Torque 2 nd Torque F r τ F r τ Torque F r τ F r τ Total Torque % Diff. 3 4 5 6