Alvarez+lab+20

rotational inertia demonstrator

Experiment #1 kinematics of rotational motion

method 1: Answer found through kinematics alpha=2(theda)/t^2 theda=2pi t=14.23

alpha= .249 rad/s^2

method 2: using a=alpha (r)

a= 2x/t^2 a= .00529 m/s^2

pulley radius: .0207 m

alpha= tangental angular acceleration/ radius alpha= .255 rad/s^2

They do agree within the uncertainty of measurement they are only .006 rad/s^2 off. The percent error is 2.35%.

Experiment #2 Determining moment of inertia

Method 1: using newton's 2nd law

F tangent= Weight of falling mass- (mass of falling mass) acceleration linearly

Weigh of falling mass: .882N Mass of falling mass: .09 kg a= .00529 m/s^2

F tangent= .881N

torque= radius (F tangent) I=(torque) alpha

I= rF/alpha

r= .0207 alpha= .249 I= .0732

Method 2: direct determination of rotation inertia from formulas

I= I(pulleys)+ I(rods)+I(movable masses) I(pulleys)=.00058 kg m^2 I(rods)=.0339 kg m^2 I(moveable masses)= .085 kg m^2 I= .119 kg m^2

This does not agree with the uncertainty because it is .0458 kg m^2 off. the percent error is 38.49%

Experiment #3: Energy of rotational motion

PE= mgh mass: .09 kg h:.536 g: 9.8 PE= .473 J

v avg= vertical displacement (h)/ time average v avg= .03767 m/s

Vf= 2xVavg Vf= .0753 m/s

Translation KE KE=(1/2) mv^2

KE= .00051 J

Rotation KE

KE=(1/2)I theda^2 KE=.48429

Rotational KE+Translation KE = total KE

Total KE= .4848 J

The total sum of KE and total PE were only .0118 J off so it does agree within the uncertainty of measurement. the percent error is 2.47%