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Rotational Equilibrium and Center of Gravity
Jomar Anthony J. Biñan [email protected] I. Introduction In Newton’s first law we are taught that if the net force of an object is equals to zero then it is in the state of translational equilibrium. There is also what we call torque. This is the rotation of an object because of different forces exerting on it. For example, a fan could rotate on a point but its not moving at all in terms of distance. So, if the net force is zero there is still a possibility that the net torque is not zero. If the object is not rotating or if the net torque is zero then this is what we call rotational equilibrium. Figure 1: A visualization of torque. It shows the rotation of the wrench when there is an application of a force on one side.
Torque is the measure of force that causes an object to rotate on an axis (KhanAcademy, n.d.). As torque requires directions for it to work properly then it is a vector quantity. Before torque is determine the center of gravity must be located. Its formula is given as,
τ = Fl
the net torque equation is given as,
τnet = F 1 l 1 + F 2 l 2 + F 3 l 3 +…+ Fnln
In this experiment, the center of mass of objects with different shapes and sizes would be determined, find the tipping angle of an object with varying mass distribution, the torques would be computed by the forces that makes the object rotates, and determine the unknown mass with the concept of rotational equilibrium. II. Methodology The center of gravity of objects are determined by using the plumb bob method.
Part C. Systems in rotational equilibrium Part 1C. Axis of Rotation (pivot point) at the center of the object The third part of the experiment was about the rotational equilibrium. The setup was assembled just like in figure 4. The meter stick was mounted in a way that the 50cm mark was the pivot point. The balance was checked and was adjusted b adding masking tape to the end of the meter stick where it is lighter to have it balanced. The mass of the meter stick was then measured using the digital balance. The value was recorded on Table 2. A 40.0-g load at the 40.0-cm mark of the meter stick and a 80.0-g load was placed on the other side of the pivot point. The 80.0g load’s position was adjusted to have the setup balanced. When the setup has been balanced, the lever arms were determined for each loads and recorded values in Table 2. The 40.0g load to the 30.00 cm mark of the meter stick and found the new position of the 80.0g load that maintained the balance. The lever arms were then measured and recorded the values. The process was then repeated when the 40.0g load is at the 20.00cm- mark and at the 10.00 cm mark. Torques were then calculated. Figure 4: Rotational Equilibrium setup. This shows the proper setup to check the net force to be zero.
Part 2C. Axis of rotation (pivot point) away from the center of the object The pivot point was moved from the center to the 40-cm mark. The 40.0g load was used to balance the meter stick and the corresponding lever arm was recorded. Repeat the steps using the 80.0g, 130.0g, and 180.0g loads. Lastly, the torques were calculated. Part 3C. Unknown Mass The meter stick was mounted at the 40.00-cm mark and a load of unknown mass was used as a counter balance bringing it to rotational equilibrium. The unknown mass was then determined. III. Results and Discussions Part A. Locating the center of gravity
30-cm mark 0.2000 0.0784 0.1 0. 20-cm mark 0.3000 0.1176 0.15 0. 10-cm mark 0.4000 0.1568 0.20 0. After the meter stick has been balanced at the 50.0-cm mark the 40.0g load has been place at different cm marks and was counter balanced by the 80.0g load. Their net torque resulted to zero making it in rotational equilibrium. Also, the results show that the lever arm of the 80.0g load is half of the 40.0g load and that is because to allow the torques to cancel each other out to bring balance. Part 2C. Axis of rotation (pivot point) away from the center of the object mmeter stick = 0.11521 kg l- = 0.1m wmeter stick = 1.1 N t- = 0.11 Nm load mass (kg) weight (N) l+ t+ 40.0g 0.04 0.392 0.31 0. 80.0g 0.08 0.784 0.15 0. 130.0g 0.13 1.274 0.09 0. 180.0g 0.18 1,764 0.07 0. The meter stick has been positioned with its pivot point is located at the 40.0-cm mark. The meter stick was then balanced using different loads. As seen on the data, it shows that heavier the load is the nearer it becomes to the pivot point. This shows that torque is dependent on the lever arm and its weights. This results to the net torque of the system equal to zero. Part 3C. Unknown Mass m (meter stick) = 0.11521 kg m (unknown) = 0.054 kg w (meter stick) = 1.10 N w (unknown) = 0.5292 N l- = 0.1 m l+ = 0.2150 m t- = 0.11 Nm t+ = 0. To determine the value of our unknown mass:
tm = tu where: tm = torque of the meter stick, tu = torque of the unknown mass
Fm(lm) = Fu(lu)
(mmg)(lm)=(mug)( lu)
mu = mmlm / lu
mu = (0.11521kg)(0.1m) / (0.2150m) = 0.054 kg
The torque of the two sides should cancel out to zero because the system is in rotational equilibrium. To determine the unknown mass, the unknown mass is mu is solved from the equation because the other quantities of the torques are already been recorded. IV. Conclusions This lab experiment showed how the center of gravity of many objects are determined by using the plumb bob method. The rotational equilibrium was then investigated by using balances of different weights to get the same torque in the system. As the rotational equilibrium have been met unknown mass could be determined using the equation of the rotational equilibrium. The center of gravity is where all the distributed mass of an object is equal to zero. If the center of gravity then is determined by balancing an object in a way that intersects with each other. Then that intersection is deemed to be the center of gravity. In the results, torques is affected by the its weight and the distance from the pivot point (lever arm) of the certain object. That is there is varying distances and weights. This experiment made the visualization and practical use of rotational equilibrium. It made the concept easier to grasp by providing a good application experiment that showcases the various characteristics of torque. The experiment could be improved by providing more figures to show how to setup the systems and allow the user to understand more of what the procedures are trying to convey. V. Acknowledgements I would like to thank my partner, Eric Sumarago, for helping me in the experiment. He has assisted and made the calculations that allowed us to determine the correct torques and equilibrium.
