# Rotational Dynamics : An Investigation

#### Rotational Dynamics : An Investigation

Rotational Dynamics : An Investigation Abstract: Every theorem in rotational dynamics can be derived from corresponding linear dynamics equations if variables are properly defined. While this report doesn’t derive everything from scratch,  we started with several definitions and verified that the theoretical framework indeed describes physical reality by conducting experiments.

We have verified several key concepts in rotational  dynamics: conservation of angular momentum, calculation of moment of inertia, and parallel axis  theorem. In the first experiment we applied a torque on an aluminium disk. By recording its angular displacement with respect to time we experimentally determined its moment of inertia, and then we  compared this value with its theoretical value given by I=MR^2.

Then we repeated the procedure by  putting the ring above the disc to obtain the moment of inertia for the ring. From the experiment, the  moment of inertia for the ring is 4.9+/- 0.1×10^-4 kg m^2, which matches well our theoretical  value 5.00±0.03e-4kgm^2. In the second experiment we dropped the ring on the disc. Using the  moment of inertia previously determined to can calculate the total angular momentum before and after  the collision, and they are found to be J-before 2.56+/-0.01e-3 kgm^2s^-1, and 2.54+/-0.01e-3 kgm^2s^-1 respectively for one set of data. In the third experiment we used parallel and perpendicular  axis theorems to derive the moment of inertia of a rod connected with two bells.

Then we plotted an inertia vs radius diagram and verified the theoretical prediction. For all three experiments our  theoretical values compare well with quantities experimentally determined as corresponding quantities differ by no more than 2 standard deviations.

#### Introduction

Theory [1]

Consider a rigid body to consist of a large number N of small mass elements of mass mi whose
relative positions are fixed and whose absolute positions measured with respect to an origin fixed in
the laboratory are ri. If we consider the origin to lie somewhere along the axis of rotation, the angular
momentum J is defined by:
J = ∑miri ×vi

(where vi is the velocity of each element) and its rate of change is given by:

dJ/dt= ri × Fi Γ

(where Γ is the torque resulting from the applied forces Fi) – the rotational equivalent of Newton’s
second law. Thus, J is conserved in the absence of applied torques.

For an isolated rigid body rotating at angular velocity ω rad/s about a fixed axis the z-component of may be written:

Jz = miR2i ω = Izω

where Iz =integral [ ρ(x^2 +y^2 )dxdydz]

Using the above definitions, we can derive other formulas needed.

We can integrate Iz =integral [ ρ(x^2 +y^2 )dxdydz] and plug in ρ=constant for a uniform disk, we get  Iz =1/2 MR^2

Derivation of parallel axis theorem and perpendicular axis theorem from Wikipedia [2] [3] Experiments

The aim of experiments is to investigate a number of different aspects of rotational dynamics using an  apparatus consisting of a set of rotating masses, a low friction bearing with an optical encoder  connected to a PC, and a pulley system to accelerate the rotating masses with known torques.

The experiment is in three sections: the first section entails measuring the moment of inertia of a  metal disc, comparing the theoretical value with that experimentally observed. The second section involves investigation of the conservation of angular momentum, in which we drop a metal ring on an  aluminium disc. In the final section we examine factors which contribute to the moment of inertia. We  applied parallel and perpendicular axis theorems to get the moment of inertia theoretically and then  compared theoretical value with that experimentally observed.

Lab script Question: How do figure skaters vary their moment of inertia, and why? Show that, if the  angular momentum of the figure skater is conserved, varying the moment of inertia requires changing  their kinetic energy. Where does this energy come from?

When a skater spreads her arms, rotational inertia increases while angular momentum is held  constant. Therefore, angular velocity must decrease as a result. Since KE is proportional to (angular  velocity) ^2, and angular momentum is proportional to (angular velocity), KE decreases as her  muscles do negative work. The opposite is true if the skater contracts arms.

Experimental apparatus used

1. The blue rotational motion sensor.

2. The clear plastic pulley set.

3. The black low friction pulley.

4. The thin aluminium plate.

5. The heavy black aluminium ring.

6. The pair of brass weights on a long thin rod.

7. The hanger mass set.

Section 1: comparing the theoretical value of aluminium disc and ring with that experimentally  observed

Experimental Arrangement

Obtain precise values for the diameters of the torque pulleys by measuring them using the callipers supplied. Check the extent to which the pulleys are circular and use an average radius. An  approximately constant torque can be applied by attaching the torque pulley to the hanger set using  the fine cotton thread supplied. Suspend the hanging mass over the edge of the bench using the low  friction pulley. Raise the hanging mass to just below the motion sensor by rotating the thin aluminium  plate, wrapping the thread around the thinnest diameter on the clear plastic torque pulley.

Figure 1: The apparatus assembled to measure the moment of inertia of the thin aluminium plate.

To determine the rotational inertia of the ring, we repeat some of the measurements as did previously with the combined ring and plate assembly in order to measure the combined moment of inertia. It is  very important to make sure that the centre of the ring and disc coincide; otherwise, we would have to  use the parallel axis theorem to calculate the total moment of inertia.

Figure 2: Ring and plate assembled for moment of inertia measurements.

Measurements

We record the radius of the torque pulleys as r and the mass of the hanger as m.

The angular velocity vs time graph is plotted on the PC. We perform a linear fit to the data so that the  slope represents angular acceleration of the disc, known as α

Figure 3: Example dataset recorded from the freely rotating thin aluminium disc, showing gradual  deceleration due to friction. The highlighting tools in figure 5 are used to select a region of interest in  the data.

Methods

By knowing the mass and effective radius, we can calculate the torque applied to the thin aluminium  plate.

Γ ≈ mgr (1)

By equating

Γ==Idw/dt =Id2θ/ dt^2

We can get the experimental value for moment of inertia of the aluminium disc through its COM and  orthogonal to its plane

I≈ mgr/α. (2)

The theoretical expression for the moment of inertia of a uniform disk of mass M, calculated about the  axis of symmetry, is:

Iz =1/2 Ma2 (3)

Itotal = Iring + Iplate

Subtract the moment of inertia due to the thin aluminium plate found previously to find the moment of inertia of the ring itself.

The theoretical moment of inertia of a ring is given by:

I = ½(Rinner^2+Router^2)Mring (4)

where Mring is the ring mass, and Rinner and Router are the inner and outer radii of the ring.

Compare the moment of inertia found from your angular acceleration measurements with that derived from the equation.

Analysis and Results

Error calculation

Error is a measure of standard deviation of measured sample

Absolute error=Measured value*percentage error (5)

Propagation of errors:

Percentage error from one variable is multiplied by the power being raised

If F=kV^n

Ferror=n*Verror (6)

Total Error=(error1^2+error2^2) ^1/2 (7)

The torque formula is an approximation because when there is an angular acceleration of the disk, the  mass attached experience a corresponding linear acceleration, a=αR, which means some weight of the  mass is acted on the mass itself to provide linear acceleration. So real torque provided=rm(g-a). This effect is small given small hanging mass and big moment of inertia. In the experiment, this αR term is  on the order of magnitude of 0.1ms^-2, which is way less than g and translates into a fractional error of  1. Therefore error because of friction dominates

Using equation (3), plug in mass and radius measured.

Add in quadrature for mass and diameter fractional uncertainty to obtain fractional uncertainty of I Percentage uncertainty=0.43%

I=1.35±0.01e-4kgm^2

From PC output we can use equation (2) to obtain experimental rotational inertia.

I=1.41±0.06e-4kgm^2

They are within 1 standard deviation from each other.

For the ring moment of inertia determination, we have

Moment of Inertia = 6.3 +/- 0.1×10^-4 kgm^2

so Iring=Itot-Idisk= 4.9+/- 0.1×10^-4kgm^2

Using formula (4)

I=5.00±0.03e-4kgm^2

Section 2: Conservation of angular momentum

Experimental Arrangement

The conservation of angular momentum can be illustrated and verified by the inelastic  collision/coalescence of two rotating objects. To explore this, we drop the non-rotating heavy black  aluminium ring onto the spinning thin aluminium plate, recording the angular velocities of the plate  before and after the collision. It’s important to note that in order for us to be able to use eq. 14, we  would have to drop the black aluminium ring so that it lands exactly centred on the aluminium. If the  ring is dropped off-centre onto the plate, then the moment of inertia of the final state will be greater  than suggested and has to be calculated using the parallel axis theorem.

To perform this experiment:

1. Remove the cotton thread and hanger set, so the aluminium plate can rotate freely.  2. Start the PC taking data by clicking Record.

3. Start the thin aluminium plate spinning by hand.

4. Hold the black aluminium ring centred over the spinning aluminium plate.

5. Drop the black aluminium ring onto the spinning plate.

6. Record a few seconds of data after the collision before clicking

Measurements

We read the output from the PC as before and fit lines to determine angular velocities before and after  the collision occurs.

Method

Since we have already determined moment of inertia of the ring and disc, we can use

J=Iw (8)

for the angular momentum in each case.

Finally, we calculate total angular momentum as below

Jbefore total = Jbefore ring + Jbefore plate

Jafter total = Jafter ring + Jafter plate

Analysis and Results

We get slopes of fitted lines as before and plug the angular velocity into equation 8 to get the angular  momentums.

J-before = 2.56+/-0.01e-3 kgm^2s^-1

J-after = 2.54+/-0.01e-3 kgm^2s^-1

They pretty much agree.

By getting the derivative of the slope of the disc and ring rotation, we can implicitly calculate the magnitude of friction torque. Since it takes a small amount of time for the 2 objects to coalesce,

J-after= J-before-t* torque (9)

The t* torque term roughly equals to 0.01e-3 kgm^2s^-1, which reduces the error further.

Section 3: determine moment of inertia of different shapes

Experimental Arrangement

Build the variable-radius mass assembly on the rotational motion encoder as shown in figure 7.

Figure 7: The variable-radius mass apparatus.

Let Ic be the moment of inertia of a cylindrical mass about a line through its centre of mass and  perpendicular to its axis. Use the adaptor to mount one of the brass cylinders on the rotary motion  sensor.

Measurements

Measure Ic and compare your value with that given by the theoretical expression for a hollow cylinder  on its side:

In the experiment, we take the slope of the angular velocity vs time diagram. In this case it is  important to note that air resistance plays a significant role as angular velocity increases. Therefore, we take the data of the first second and fit the line.

Then we move the brass cylinders along the radius and plot moment of inertia vs radius.

Analysis and Results

mass of light rod = (26.4+/-0.1) g. length of light rod = (340+/-10) mm. inner diameter of light rod =  (5.3+/-0.1) mm. outer diameter of light rod = (7.8+/-0.1)mm.

Theoretical inertia of bare rod = (0.25 +/- 0.02) x10^-4 kgm^2

Using parallel axis theorem and perpendicular axis theorem, we can get the rotational inertia of the  (10)

theorem where Mc is the cylinder mass, l is its length, and about and bin are, respectively, its outer and  inner radii.

Experimental Moment of Inertia = (0.29 +/- 0.02) x10^-4 kgm^2

The air resistance and resistance of other types still play a significant role, but they would agree if we  account for that.

inertia vs radius diagram should be in the form I = A + B*r^2

The fitted parabola approximates the relationship pretty well.

Fitted parameters A = 4.6e-4 kgm^2 B = 0.149 kg

Theoretically, I = (Ic + Irod) +2*Mc*r^2

B should be twice the mass of the cylinders, 0.151kg, and compares well with actual mass 0.149kg.

Conclusions

With these three experiments, we have verified several key concepts in rotational dynamics, including  conservation of angular momentum, calculation of moment of inertia, and parallel axis theorem. For  all three experiments our theoretical predictions compare well with quantities experimentally  determined as corresponding quantities differ by no more than 2 standard deviations. Furthermore,  we can reduce the difference between experimental and theoretical values further if we take  resistance into account.

Appendices

A. Measurement of components

mass of Aluminium disk =(118.90+/-0.1)g.

diameter of Aluminium disk =(95.2+/-0.2)mm.

mass of black steel ring =(459.7+/-0.1)g.

inner diameter of black ring =(53.4+/-0.2)mm.

outer diameter of black ring =(76.5+/-0.2)mm.

mass of light rod = (26.4+/-0.1)g.

length of light rod = (340+/-10)mm.

inner diameter of light rod =(5.3+/-0.1)mm.

outer diameter of light rod =(7.8+/-0.1)mm.

mass of fixation screw = (3.4+/-0.1)g.

diameter of fixation screw =(8.1+/-0.1)mm.

mass of brass cylinders = (75.5+/-0.5)g.

height of brass cylinders =(20.3+/-0.1)mm.

inner diameter of brass cylinders =(8.0+/-0.1)mm. outer diameter of brass cylinders =(25.1+/-0.1)mm. mass of triangle = (449.1+/-0.1)g.

B. Example of Experimental data for aluminium and ring

C. selection of Python code used for data analysis import numpy as np

import matplotlib.pyplot as plt

from scipy import optimize

import pandas as pd

def linear(t, m, c):

return m*t+c

fileID=’AluBaseOnly’

#fileID=’AluBaseBlackRing’

import numpy as np

#fileID=’BareRod’

#fileID=’BrassAt40mm’

#fileID=’BrassAt80mm’

#fileID=’BrassAt120mm’

#fileID=’TriangleHypotenuseVertical’

#fileID=’TriangleHypotenuse30degToVertical’  #fileID=’TriangleHypotenuse60degToVertical’

dataframes={}

torques=[]

accels=[]

for df in dataframes:

torque=df/1000*9.81*0.014

torques.append(torque)

t_data=data[‘Time (s)’]

fig, ax = plt.subplots(1, 1, figsize=(15,5))

ax.scatter(t_data,v_data)

ax.set_xlabel(‘Time [s]’)

timeMax = data.loc[data[‘Angular Velocity (rad/s)’].idxmax()][‘Time (s)’]-0.1  t_data_fit = t_data[(t_data<timeMax)&(t_data>1.0)]

v_data_fit = v_data[(t_data<timeMax)&(t_data>1.0)]

params, params_covariance = optimize.curve_fit(linear, t_data_fit, v_data_fit, p0=[  plt.plot(t_data_fit, linear(t_data_fit, params[0], params[1]),label=’Straight line’  angAcc=params[0]

Bibliography

[1] GP08 Rotation dynamics lab script

[2] “Parallel axis theorem”, Wikipedia, https://en.wikipedia.org/wiki/Parallel_axis_theorem [3] “Perpendicular axis theorem”, Wikipedia, https://en.wikipedia.org/wiki/Perpendicular_axis_theorem