Roth+Lab+20

Lab 20 Rotational Inertia Demonstrator Two different methods were used to calculate the angular acceleration. In one, a mass hung from a string wound around the Rotational Inertia Demonstrator 4 times descended to the ground. Since the number of rotations that the rotational inertia demonstrator went through throughout the descent was 4 rotations, then the angular displacement was 8 pi radians. The average time that it took for the mass to descend to the ground from .71 meters above the ground was 18.03 seconds. To calculation alpha, angular acceleration, the following equation can be used: change in theta = initial angular velocity (t) + (1/2)alpha (t^2). In the equation, t is the average time and the initial angular velocity is zero. Therefore, when all of the numbers are plugged in you can solve for alpha. 8pi = (0 x 18.03) + (1/2)alpha (18.03^2). Alpha ends up being .155 rad/s^2. The other method involved using the equation vertical displacement = initial velocity + (1/2)a (t^2). In this case, the linear acceleration is being found. When the known values are plugged in you get the equation .71=0(18.03) + (1/2)a (18.03^2). The linear acceleration, a, ends up being .0044 m/s^2. The angular acceleration can be found by dividing the linear acceleration by the radius of the pulley, which is .028 m. Therefore, the angular acceleration by this method would equal .157 rad/s^2. The percent error would be calculated using the percent error equation. Percent Error = l (actual – experimental) / actual l x 100= l (.157-.155) / .157 l x 100 = 1.27% error. The Percent error would only be 1.27%, which is within the uncertainty of measurement.
 * Angular Acceleration Calculations:**

The first method to find moment of inertia involved calculating tension and using the equation Tension= moment of inertia (angular acceleration), which produced a moment of inertia of .0884 kg(m^2). The second method involved calculating the moments of inertia of the individual parts and adding them together, this method produced a moment of inertia of .08366 kgm^2. The percent error would be calculated using the percent error equation. Percent Error = l (actual – experimental) / actual l x 100= l (.08366-.0884) / .08366 l x 100 = 5.7% error. A percent error of 5.7% is within the uncertainty of measurement.
 * Moment of Inertia Calculations:**

The initial mechanical energy of the system is equal to the gravitational potential energy since the system begins with a velocity of zero. The gravitational potential energy was .3479 J. As the mass descends, this energy is converted into potential energy. When the mass hits the ground, the system is all kinetic energy. This kinetic energy Is divided into translational kinetic energy and rotational kinetic energy. The sum of these two types of kinetic energy would equal the total kinetic energy. The translational kinetic energy was 1.45 x 10^-4 J and the rotational kinetic energy was .327 J. The sum of these two types of kinetic energy is .3271 J. The percent error would be calculated using the percent error equation. Percent Error = l (actual – experimental) / actual l x 100= l (.3479-.3271) / .3479 l x 100 = 5.98% error. A percent error of 5.98% is within the uncertainty of measurement.
 * Energy of Rotational Motion Calculations:**