Moment of Inertia Nov. 4- 2014

Moment of Inertia Lab

Objective:

Calculate the object's mass and predict the rotational acceleration if a cart was attached to it on a string going down a frictionless slope.

Set up:

First we are given a large metal disk on a central shaft(it has a known overall mass) and a caliper to measure its volume. Second, we have a track and a cart on a light string tied to the shaft of the apparatus.

The Approach:

  We approached this problem assuming that the disk has a uniform distribution of density, and that the shaft has negligible friction. From there find the volume of the apparatus by separating it into the sections (two small cylinders that stick out of a large disk) as seen in the figure below. Once we find their volumes, we find what ratio the large disk is compared to the small cylinders since density is evenly distributed by the same ratio.

M = V               M,V= values for large disk

m     v                m,v= values for small cylinders (each)

Since V= 13.18v, then compared to 2v, V is 6.59 times larger in comparison to the sum of the cylinders. Thus, the disk's mass is 6.59 times larger resulting to mass of 3.915 kg from the total 4.615 kg.

 Now that we have the mass values, we calculate each of their inertia and add them together for a system inertia of:

(Big) + 2(small)= (.02 kg*m^2)+ 2(3.99*10^-5 kg*m^2)= 2.008*10^-2 kg*m^2

Figures 1A and B. A picture of the rolling disk apparatus and Lab work for each part including a small diagram of the
second set up for this experiment.

 With the total inertia known, there was a need to derive the predicted angular acceleration  for a cart of a mass (m) moving down a frictionless track while attached to the shaft of the rolling disk. We start off as treating this problem as a dynamic problem or unequal torque is the tension of the string times the force on the cart parallel to the track which turns out to be the sine value of the force of gravity on the cart. From there we had the tangential acceleration become terms of angular acceleration and radius of the turn shaft. The result is the following:

α= mgsinθ*r       Where m is the mass of the cart, r is radius of the shaft, θ is the angle of the slope,
  (m(r^2)-Itot)         and Itot is the total inertia of the apparatus.

We set our surface at 25 degrees with a .4 kg cart attached to the shaft. Before beginning the trial we predicted with out derived formula that the α of the cart would be -1.253 rads/s^2 and our actual came to be -1.266 rads/s^2, only 1.04% off. 
Conclusion:
It seems our derived equation is accurate enough to predict the angular acceleration of our cart set up. Any apparent source of error would source from the friction of the disk, friction of the surface and friction of the axle.