1. Find and record the position of the center of gravity of the meter stick by adjusting the pivot clamp until the meter stick balances with no other clamps or weights on the stick. 2. Mass the clamps from which the mass hangers will hang. 3. a) Determine the mass of your meter stick in a manner similar to the method used in Part 2 of your preparation sheer. b) Remove the pivot clamp and mass the meter stick using the electronic balance. c) Calculate the percent difference between the results of a) and b). 4. Return the pivot clamp to the position of the center of mass, then do the following: a) Place a mass (include the mass of the clamp and hanger) of approximately 250 g at the 0.010 m mark Place another mass at some position between the 0.500 m and 1.000 m marks and adjust the position until balance is achieved. Repeat this for ve additional masses at different positions on the right. b) Calculate the counterclockwise torque and the six clockwise torques with respect to the pivot point. c) lCompare the above results by calculating the percent difference between the two Net Torques, clockwise and counter-clockwise, for each of the six cases. Note: You only want to compare the magnitudes of the torques, so remove the negative sign in front of the clockwise torques when making the comparison. Revised 12-19-08 Experiment 3 - Page 3 All Rights Reserved 5. Move the pivot to the 0.400 m position. Place a clamp and mass hanger at about 0.3 m and another clamp and hanger at 0.150 In. Place a clamp, mass hanger and enough mass to make a total of 100 g at position 0.500 m. Adjust the position andfor mass of the hanger near 0.3 m such that the whole system is balanced. What you now have is a model of a balance which can be used for massing. The hanger at the 0.150 m mark is the "pan" of the balance and the mass at 0.500 m is the sliding weight on the arm of the balance. However, we still need to calibrate the scale. Calibrate the scale by placing a 50 g mass on the "pan" and then moving the sliding mass on the right until balance is obtained. Use additional 20 gram masses on the "pan" until there is no more room on the end of the meter stick for the sliding mass. Use your data to produce a calibration graph that relates the mass on the "pan" to the distance scale on the arm. In Procedure 1, a meterstic is placed in a pivot clamp that allows the stick to be placed in a stand. It is found that the stick balances in equilibrium when the clamp is located at the 49.8 cm mark. This is the location of the center of mass of the stick. (It is not unusual for the center of mass to precisely at the 50.0 cm mark.) In Procedure 3, a 19.6-gram clamp is placed at the 10.0 centimeter mark and a mass of 100.3 grams is suspended from the clamp. The pivot clamp that supports the meterstick must be placed at the 34.2 cm mark in order for the the stick to be balanced. Calculate the magnitude of the torque (in Newton-meters) on the meterstick about the pivot clamp by the weight of the clamp and added weight. on 2 0 out of10 points The weight of the meterstick acts on a line passing through the center of mass. Use the fact that the torque on the meterstick about the pivot clamp is equal to the torque from the clamp and added mass to calcuate the weight (in newtons) of the meterstick. 0" 3 0 out of10 points Given your answer from the preceding question, calculate the mass of the meterstick. on 4 10 out of10 points In Procedure 4, the pivot clamp is moved back to the center of mass of the meterstick. A clamp is placed at the 0.050 meter mark supporting a total mass of 253.0 grams (including the clamp itself.) A second clamp, supporting a total mass of 0.289 kilograms is placed at 0.893 meters. Calculate the weight of the 253 gram mass in newtons. on 5 10 out of10 points In Procedure 4, the pivot clamp is moved back to the center of mass of the meterstick. A clamp is placed at the 0.050 meter mark supporting a total mass of 253.0 grams (including the clamp itself.) A second clamp, supporting a total mass of 0.289 kilograms is placed at 0.893 meters. Calculate the lever arm of the 253 gram mass (in meters) about the center of mass. on 6 10 out of10 points In Procedure 4, the pivot clamp is moved back to the center of mass of the meterstick. A clamp is placed at the 0.050 meter mark supporting a total mass of 253.0 grams (including the clamp itself.) A second clamp, supporting a total mass of 0.289 kilograms is placed at 0.893 meters. Calculate the magnitude of the torque from the 253 gram mass (in newton-meters) about the center of mass. on 7 10 out of10 points In Procedure 4, the pivot clamp is moved back to the center of mass of the meterstick. A clamp is placed at the 0.050 meter mark supporting a total mass of 253.0 grams (including the clamp itself.) A second clamp, supporting a total mass of 0.289 kilograms is placed at 0.893 meters. Calculate the weight of the 0.289 kilogram mass in newtons. 0" 8 0 out of10 points In Procedure 4, the pivot clamp is moved back to the center of mass of the meterstick. A clamp is placed at the 0.050 meter mark supporting a total mass of 253.0 grams (including the clamp itself.) A second clamp, supporting a total mass of 0.289 kilograms is placed at 0.893 meters. Calculate the lever arm of the 0.289 gram mass (in meters) about the center of mass. on 9 10 out of10 points In Procedure 4, the pivot clamp is moved back to the center of mass of the meterstick. A clamp is placed at the 0.050 meter mark supporting a total mass of 253.0 grams (including the clamp itself.) A second clamp, supporting a total mass of 0.289 kilograms is placed at 0.893 meters. Calculate the magnitude of the torque from the 0.289 gram mass (in newton-meters) about the center of mass. on 10 Needs Grading Given your answers for the two torques in procedure 4, can we say that the torques on the meterstick are equal; in other words, does the data support the claim that the stick in in equilibrium? Be sure to explain why or why not