Lab+20+-+Rotational+Inertia+Demonstrator

Lab 20 - Rotational Inertia Demonstrator

Angular acceleration of the falling mass (rotational kinematics) ∆θ = ωot + 1/2αt^2... since ωot = 0 due to 0 initial angular velocity... ∆θ = 1/2αt^2... solving for α yields... α = 2(∆θ)/(t^2)... substituting found values in for the appropriate variables yields... α = 2(10π)/(9.5^2)
 * Experiment #1**
 * α ≈ 0.696 rad/s^2**

Angular acceleration of the falling mass (linear kinematics) ∆x = vot + 1/2at^2... since vot = 0 due to 0 linear velocity... ∆x = 1/at^2... solving for a yields... a = 2(∆x)/(t^2)... substituting found values in for the appropriate variables yields... a = 2(.92)/(9.5^2)
 * a ≈ 0.0204 m/s^2**

Since the relationship between α = a/r... substituting the found values for the appropriate variables yields... α = 0.0204/0.3 = **0.68 rad/s^2**

To find the percent difference between the two values, the following formula is used: %difference = | value 1 - value 2 | / ((value 1 + value 2)/2) x 100... substituting the found values for the appropriate variables yields... %difference = | 0.696 - 0.68 | / ((0.696 + 0.68)/2) x 100
 * %difference ≈ 2.325%**

Given the percent difference between two values, it could be said that these two values do agree within a reasonable amount of uncertainty.

Method 1: Using Newton's 2nd Law Mass = 0.23 kg Tension = mg = (0.23)(9.81) = **2.256 N**
 * Experiment #2**

(Ft x r) = τ = Iα I = τ /α = (2.256 x 0.03)/0.696 **≈ 0.0972 kgm^2**

Thin rod length: 0.39 m Thin rod mass: 0.0748 kg I(thin rod) = 1/3ML^2 = 1/3(0.0748)(0.39^2) = 0.00379 x 4 rods **≈ 0.0152 kgm^2**

Movable mass mass: 0.018 kg Movable mass radius: 0.38 m I(point mass) = MR^2 = (0.018)(0.33^2) = 0.00196 x 4 disks **≈ 7.84 x 10^-3 kgm^2**

I(pulley) **≈ 5.8 x 10^-4 kgm^2**

∑I = 5.8 x 10^-4 + 0.0152 + 7.84 x 10^-3 **≈ 0.0236 kgm^2**

%difference = | value 1 - value 2 | / ((value 1 + value 2)/2) x 100... substituting the found values for the appropriate variables yields... %difference = | 0.0972 - 0.0236 | / ((0.0972 + 0.0236)/2) x 100 **≈ 121.85% difference**

Given the percent difference between two values, it could be said that these two values do not agree within a reasonable amount of uncertainty.

__Gravitational Potential Energy__ Mass = 0.23 kg Vertical Displacement = 0.9 m
 * Experiment #3**

PE = mgh = 0.23 x 9.81 x 0.9 = **2.0307 J**

__Translational Kinetic Energy__ t1 = 9.4 s t2 = 9.4 s t3 = 9.8 s tavg = 9.53 s vavg = 0.094 m/s vf = 0.189 m/s

KE = 1/2mv^2 = (0.5)(0.23)(0.189^2) = **0.004 J**

__Rotational Kinetic Energy__ Radius of pulley = 0.03 m ωf = 6.3rad/s I = 0.972 kgm^2 Rotational KE = 1/2Iω^2 = 1/2(0.0972)(6.3^2) =**1.929 J**

KE Total = 0.004 + 1.929 = **1.933 J** %difference = | 2.0307 - 1.933 | / ((2.0307 + 1.933)/2) x 100 **≈ 4.930% difference**

The kinetic energy of the pulley system was conserved to a reasonable measurement. However, this was using the moment of inertia that was calculated via the analytical way and not the experimental way since the math does not lie but there could have been a huge error in measuring distances, masses, etc in the calculations for the moment of inertia of the entire pulley system.