O cmll 12 14 5 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 40 uTorque and Equilibrium Lab goals To correctly complete the Lab on center of mass. To understand the concepts of equilibrium and balance of torques. To find a relation between a given mass and an unknown mass. Gain experience in researching and setting up an experiment by themselves. Learn to be independent and self-reliant by performing an experiment by themselves. Theory The conditions for mechanical equilibrium of a rigid body are: (1) the sum of the forces must be equal to zero, and (2) the sum of the torques must be equal to zero. This ensures that there is no linear acceleration and no rotational acceleration, respectively. This may be written mathematically as: EF =0 (1) Er = 0 (2) The first condition ensures that the object is at a particular location (not moving linearly) or that it is moving with constant linear velocity (Newton's First Law of Motion). In this experiment, the rigid body (the ruler) is restricted from linear motion and 2 7 = is automatically satisfied. To be in static equilibrium, a rigid body must also be in rotational static equilibrium. Although the sum of the forces on the body may be zero and it is not moving linearly, it is possible that it may be rotating about some fixed axis of rotation. However, if the sum of the torques is zero, the object is in rotational equilibrium, and either it does not rotate, or it rotates with a constant angular velocity. Forces produce linear motion and torques produce rotational motion. A torque results from the application of force F acting at a distance r from an axis of rotation. If the angle between the vector F and the vector " is , the magnitude of the torque is equal to t = rF sin(9) (3) The unit of torque is the meter -Newton (mN). Relative to an axis of rotation, a rigid body can rotate in only two senses, clockwise and counterclockwise. Torques may be referred to as clockwise torques or counterclockwise torques, that is torques that may produce clockwise rotations and torques that may produce counterclockwise rotations. Since clockwise and counterclockwise torques work to rotate an object in opposite directions, the condition for equilibrium can be expressed as:EtcCH - Etc =0 (4) In this lab, the angle between the vector Fand the vector r is 90 and the Eq. (4) becomes: paper paper coincoin Where r paper and r expressed in centimeters are the respective distances of the objects measured from the center of the ruler. Canceling the acceleration of gravity on both sides, we have: "paper paper coin coin (5) Material Ruler, one page of printed paper, scotch tape, one coin, string (Dental floss, sewing thread, etc.) See picture with my setup RULER COIN PAPER Activity Finding a relation between the paper mass and the coin mass. Write all your results on a Google sheet shared with the instructor. You need to use the school Google drive. 1. Using the scotch tape and a string secure the ruler hanging free from a table. You want to suspend it through its center. 7.2.cm 15.1 cm 2 23.8 cm y=(15.1-7.2) = 7.9 cm = .079 m My y X = (23.8-15.1)=8.7 cm = .087 m2. Cut two equal pieces of a string and attach them to the coin and to the paper page. I made two loops that I hung on the ruler. 3. Place the coin and the paper on the opposite side of the ruler center until you find the perfect equilibrium. 4. Comparing the left torque with the right torque, find a relation between the paper mass and the coin mass. Use Eq. (5) as a guide. Use the Google sheet formulae for your calculations. paper mass .087/.079 = 1.101265823 m = 1.1 m coin mass 5. Use this result to correct the lab on the center of mass. You want to include the mass of the paper as a fourth object in your theoretical calculations for the center of mass coordinates. Open the report for the previous lab and add the correction showing all the calculations. Tip: If you have four discrete objects and you want to find the x-coordinate of their center of mass you would use the following equation: m x +m x, +m,x , +mix . (2.5*5.4)+(2.5*1 1.2)+(2.5*21.4)+(1.6*0.087)/(2.5*3)+1.6 = 10.63 = x X CM m +m , +m , +m . (2.5*13.7)+(2.5*1.9)+(2.5*6.8)+(1.6*0.79)/(2.5*3)+1.6 = 2. = x GRADING RUBRIC as posted on Canvas (For grading, if you are not in class, you want to add a picture of your carefully balanced ruler with the paper and the coin attached. No picture, no grade.) Ipt for writing the lever arms for the paper r and the coin r on your report. 1 pt for complete and correct calculations of the paper mass following the handout's formula. Ipt for using the same google sheet you used on the Lab Center of mass part 2 and adding on the same page the new data according to the tables below. (Must use the google formulae for the calculations). Ipt for adjusting the center of mass lab value including the paper mass and showing all the calculations on your report. 1pt for re- calculating the percent error between the experimental center of mass and the value obtained with the new math formula corrected with four objects. 3Previous experiment Mass (leave m for the x-coordinate (cm) y-coordinate (cm) coins) (g) Data and results Coin 1 m 5.4 13.7 Coin 2 m 11.2 1.9 Coin 3 m 21.4 6.8 CM with the equation (leave this cell blank) 12.7 7.5 CM from the (leave this cell blank) experiment 12.9 7.6 Percent error (leave this cell blank) 1.84% 1.79% Torque experiment Mass of the page as a CM x-coordinate (cm) CM y-coordinate (cm) function of m Just the Paper information New center of mass (leave this cell blank) using the corrected formula including the page's mass New percent error with (leave this cell blank) the CM from the experiment