Purpose: To determine the moment of inertia of a right triangular thin plate around its center of mass, for two perpendicular orientations of the triangle.
Model: Iparallel axis=Icm + Mdparallel axis displacement^2 and I=(mgr)/α-mr^2
m is the mass of hanging mass, α is the angular acceleration of the disk, and r is the radius of the pulley.
Process: We will collect three group of data in different set up. First, we do not mount the triangle on a holder and disk. Second, we mount the triangle on a holder and disk vertically. Third, we mount the triangle on a holder and disk horizontally. A string is wrapped around a pulley on top of and attached to the disk and goes over a freely-rotation "frictionless" pulley to a hanging mass. The tension in the string exerts a torque on the pulley-disk combination. By measuring the angular acceleration of the system, we can determine the moment of inertia of the system.
Then, you can mount the triangle onto the disk-pulley-holder system and measure α again and determine the I of the new system. The moment of inertia of the triangle is the difference between the moment of inertia from these two measurements.
We determine the mass of the hanging mass and the radius of the pulley, and then we use the equation that we get from last experiment: I=(mgr)/α-mr^2 to find the moment of inertia.
Iempty=0.0027021 kg*m^2
Ivertical=0.0029899 kg*m^2
Ihorizontal=0.0033051 kg*m^2
We already know the theory moment of inertia:
I1=(MB^2)/18=0.000246275 kg*m^2
I2=(MH^2)/18=0.000567324 kg*m^2
According to the parallel axis theorem:
Ivertical-Iempty=0.0002878 kg*m^2 ≈ 0.000246275 kg*m^2 =I1
Ihorizontal-Iempty=0.000603 kg*m^2 ≈ 0.000567324 kg*m^2 =I2
The the theoty moment of inertia approximately equal to the experiment moment of inertia.
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