Pfautz+lab+20

Rotational Inertia Demonstrator

Experiment 1:

Method 1: To find the angular acceleration of the falling mass I used the equation: α = 2θ/t^2 After substitution: =2.8π/14.23^2 =.249 rad/s^2

Method 2: Vertical Displacement: .536m Equation: a=2x/t^2 After Substitution: a=2(.536)/14.23^2 a=.00529 m/s^2

Equation: α=aT/r α=.255 rad/s^2

Experiment 2:

Method 1: Angular Acceleration of rotating apparatus: .249 rad/s^2 Linear Acceleration of falling mass: .256 m/s^2 Radius of Pulley: .0207m Falling Mass: .09kg Weight of falling mass: .882N

Equation: Ft=W-ma Substitution: Ft=(.882)-.09(.00529) Ft=.881N

Equation: I=T/α= rFt/α Substitution: I=(.0207)(.881)/.249 I=.0732 kg m^2

Method 2: I(pulley) .00058 kg m^2 Thin Rods: length .338m mass .074 kg I(thin rod) .0085 kg m^2 x4=.0339 kg m^2 Movable Mass: mass .186 kg distance from center .338 m I(movable mass) .0212 x4=.085 kg m^2 I=I(pulley)+I(rod)+I(movable masses) I=.00058+.0339+.085=.11948 kg m^2

Experiment 3:

Gravitational PE: mass: .09 kg vertical displacement: .536 m Equation: PE=mph PE=(.09)(9.8)(.536) PE=.473

Translational KE: (t1) 14.8s (t2)13.7s (t3)14.2s (t avg) 14.23s (v avg) .038 m/s Vf=2xVavg=2(.038)=.075 Equation: KE=1/2mv^2 KE=1/2(.090)(.075)^2 KE=.000255

Radius of Pulley: .0207m w=12.82 rad/s I=.1986kg m^2 Equation: KE=1/2 IV^2 KE=1/2(.0723)(3.62)^2 KE=.4737J

KE(total)=.000255(trans)+.4737(rot)=.4737J

The values were very close with a small percent error.