AWeaver+Rotational+Inertia

Rotational Inertia:

__**Experiment 1: Kinematics of Rotational Motion:**__ __**Experiment 2: Determining Moment of Inertia**__: To calculate torque, we measured the radius of the pulley to be .0207 meters. To calculate tension, we used the equation w-ma. We found the force of tension to be .882 Newtons. Using the equation I=rtf/a we found that I=.0732 kg/m/s^2. Using the other method of calculating the moment of inertia for the individual pieces (the 4 thin rods, the pulley, and the 4 moveable masses) we found that the I=.11948. These two values should ideally have been closer. The percent error was rather large for the system. The error could be due to friction, inaccurate measurments, or miscalculations.
 * In order to solve for angular accelation in part 1 we used the equation** θ = θ0 + ω0t + .5αt^2. Using this equation we found that a=.249 radians/ sec^2. We also used the vertical displacement 's' which was .536 meters and average time which was 14.23 seconds to calculate the value which we found to be .255 rad/sec^2 These two acceleration values were very similar and agree within the uncertainty of the measurements taken.

__**Experiment 3: Energy of Rotational Motion:**__ We found the potential energy with the equation mgh. (.090)(.536)(9.8)=.4728j We then found the total kinetic energy by adding the translational and rotational kinetic energies. We used .5mv^2 for translational. (.5)(.09)(.075)^2=.000255 For rotational kinetic energy we used the equation .5IV^2 (.5)(.0723)(3.62)^2=.4737 So, the total potential energy lost was 4728 Joules while the total kinetic energy was .4737 Joules. These two values are very close, and agree within the uncertainty of measurment.