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SIMULATIONS RELATED TO MOTION OFPARTICLES IN ROTATING SYSTEMS
Birla Institute of Technology and Science, Pilani
K.K. Bir la Goa Campus
18 Apr il 2011
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A
Repor t onSIMULATIONS RELATED TO MOTION OF
PARTICLES IN ROTATING SYSTEMS
Submitted in partial fulfillment for the completion of the course
PHY C231 Physics Project Lab
Prepar ed by: Under the supervision of:
Akshat Agha 2009A4PS380G Dr. Toby Joseph
Sahil Gupta 2009 A4PS393 G
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Acknowledgement
We would like to take this opportunity to thank our instructor Dr. Toby Joseph
for giving us an opportunity to undertake this study. We are thankful to him for
constantly supporting and guiding us throughout this project. Also, we are
thankful to our friends who helped us in making the code and rectifying it.
Thank You!
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Table of Contents
Acknowledgements 3
Abstract 4
1. Introduction 62.
Rotating Disc 7
3. Rotating Cone 104. The MATLAB Code- Rotating disc 125. Observations and Findings 14
5.1.CASE 15.1.1. Projection Point at R=1m 15
5.1.2. Trajectory traced by the body at different speeds 16
5.1.3. Variation of angle of projection with escape velocity 18
5.2. CASE 2
5.2.1. Projection Point at R=5m 19
5.2.2. Variation of angle of projection with escape velocity 20
5.3. CASE 3
5.3.1. Projection Point outside at R=10.6m 215.3.2. Trajectory traced by the body at different speeds 22
6. Results and Conclusions 24
6.1. Inferences 25
7. References and Bibliography 27
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Rotating Disc
The first system we studied is a bullet entering a disc just sliding along itssurface with initial velocity V, angular velocity of disc being and frictioncoefficient .
3D View:
Bullet at radius
Top View:
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The system involves friction between particle and disc. As, the disc rotates,
centripetal force is provided to particle by friction force. As a result particle
tends to follow a curved path. The forces acting on the particle are friction and
centrifugal force.
The quantities we require for solving the problem are:
= Friction coefficient of disc
= Angular velocity of the diskVni = Velocity w.r.t non-inertial frame
ni = Angle of bullet w.r.t non-inertial frame
= Position of particle
The other symbols used are:
Fi = force in inertial frame
Vi = velocity w.r.t inertial frame
i = angle of bullet w.r.t inertial frame
Now, the velocity of particle in inertial frame is:
= [() + + 2 cos2 ] = [ cos() ] = cos() = sin() = =
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The friction force acts on the particle in the direction opposite to the
instantaneous relative velocity of the particle w.r.t disc i.e. (
r
-
) So, the
magnitude of acceleration is given by:
= (( ) + )
= + = ( ) =
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Rotating Cone
The second system we studied is a particle projected in an arbitrary direction ina cone rotating at an angular velocity and friction coefficient .
Equations of motion in spherical coordinate system:
=
+
+
0(Since particle moving along the cone, = 0 ) = + +
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The MATLAB Code-Rotating Disc
%[output] = tablerot(v0,th0,mu,omg)%DIALOGUE BOX GENERATION
prompt={'Coefficient of friction:','Angular speed of disc in rps:','Initial Speed of
bullet in m/s:','Initial angle of bullet(th)in m/s:','Position of gun in metres:'};
name='Simulation on rotating disc';
%SETTING THE DEFAULT VALUES
numlines=1;defaultanswer={'1','1','10.73','0','1'};
options.Resize='on';
options.WindowStyle='normal';
options.Interpreter='tex';
%INPUTTING THE VALUES
answer=inputdlg(prompt,name,numlines,defaultanswer,options);
%CHANGING THE INPUTTED STRING INTO NUMERICAL VALUES
mu=str2num(answer{1});
omg=str2num(answer{2});
vrot=str2num(answer{3});
throt=str2num(answer{4});
r0=str2num(answer{5});
%CONVERTING THE SPEED W.R.T. DISC TO SPEED AS PER INERTIAL FRAME
v0 = sqrt((omg*r0)^2 + vrot^2 + 2*(omg*r0)*vrot*cos(pi/2-throt));
th0 = acos(vrot*cos(throt)/v0);
ifthrot>pi
ifthrot
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x(1) = r0;
y(1) = 0;
mug = mu*g;%CALCULATING POSITION AFTER TIME dT, GETTING THE NEW ACCELERATION
THEN NEW VELOCITY TO CALCULATE THE NEW POSITION RECURSIVELY
for i = 1:N
omgt = omg*i*dt;
xr(i) = x(i)*cos(omgt) + y(i)*sin(omgt);
yr(i) = -1*x(i)*sin(omgt) + y(i)*cos(omgt);
x(i+1) = x(i) + vx*dt;
y(i+1) = y(i) + vy*dt;
vtx = -1*omg*y(i+1);
vty = omg*x(i+1);
Fint = mug/(sqrt((vtx - vx)^2 + (vty - vy)^2));
Fx = Fint*(vtx - vx);
Fy = Fint*(vty - vy);
vx = vx + Fx*dt;
vy = vy + Fy*dt;
end
% PLOTTING THE TRAJECTORY ON A GRAPH
figure(1)
hold off
plot(x,y,'r');lm = max([max(abs(x)) max(abs(y))]);
lm = lm+2;
axis([-lm lm -lm lm]);
axis square
hold on
plot(xr,yr,'b')
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OBSERVATIONSAND
FINDINGS
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CASE 1: Projection Point at R=1m
For the values,
1. Coefficient of friction = 1,2. Gravitational acceleration= 10m/3. = 1 rad/s4. Starting position = 15. Angle ofprojection w.r.t. disc= 0
Graph of Radius of circular loop (Y-axis) versus Speed of projection (X-axis):
0
2
4
6
8
10
12
0 2 4 6 8 10 12
Speed of projection w.r.t.
disc m/s
Radius of Circular loops
Meters
0 1.000
1 1.050
2 1.222
3 1.500
4 1.890
5 2.410
6 3.050
7 3.860
8 4.840
9 6.060
10 7.650
10.4 8.520
10.5 8.790
10.6 9.110
10.68 9.450
10.69 9.500
10.70 9.560
10.71 9.630
10.72 9.710
10.73 9.828
10.735 9.929
10.736 9.985
>10.736 Flies off to infinity
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Tr ajector ies Tr aver sed By the Body at Different Speeds:
*Colour codes in graphs:
Red colour- As observed from inertial frame of reference (outside the disc).
Blue colour- As observed by non-inertial frame of reference (rotating along with the centre of the
disc)
V=0.8 m/s V=2 m/s
V=4 m/s V=6 m/s
V=7 m/s V=9 m/s
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V=10.0 m/sV=10.20 m/s
V=10.60 m/s V=10.736 m/s
V=10.74 m/s V=10.75 m/s
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Var iat ion of Angle of Pr ojection with Escape Velocity
For the values,
1. Coefficient of friction = 1,2. Gravitational acceleration= 10 m/3. = 1 rad/s4. Starting position = 1
Angle of projection w.r.t disc Escape Velocity w.r.t disc
m/s
0 10.736
/8 10.52 /6 10.48
/4 10.46
/3 10.51
/2 10.80
2 /3 11.28
3 /4 11.59
5 /6 11.89
7 /8 12.04
12.45
9 /8 12.73
7 /6 12.79
5 /4 12.83
4 /3 12.79
3 /2 12.44
5 /3 11.85
7 /4 11.53
11 /6 11.22
15 /8 11.08
2 10.73
Graph of Escape velocity w.r.t. disc (Y-axis) versus Angle of Projection (X-axis):
0
2
4
6
8
10
12
14
0 100 200 300 400
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CASE 2: Projection Point at R=5m
For the values,
1. Coefficient of friction = 1,2. Gravitational acceleration= 10 m/3. = 1 rad/s4. Starting position = 55. Angle of projection w.r.t. disc= 0
Graph of Radius of circular loop (Y-axis) versus Speed of projection (X-axis):
0
2
4
6
8
10
12
0 1 2 3 4 5 6 7
Speed of projection w.r.t.
disc m/s
Radius of Circular loops
Meters
0 5.0000.2 5.008
0.4 5.018
0.6 5.014
0.8 5.065
1 5.101
2 5.403
3 5.908
4 6.639
4.5 7.106
5 7.66
5.5 8.3616 9.456
6.07 9.83
6.075 9.903
>6.075 Flies off to infinity
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Var iat ion of Angle of Pr ojection with Escape Velocity
For the values,
1. Coefficient of friction = 1,2. Gravitational acceleration= 10 m/3. = 1 rad/s4. Starting position = 5
Graph of Escape velocity w.r.t. disc (Y-axis) versus Angle of Projection (X-axis):
0
5
10
15
20
0 100 200 300 400
Angle of projection w.r.t disc Escape Velocity w.r.t disc
m/s
0 6.075
/8 5.730
/6 5.690 /4 5.756
/3 5.960
/2 6.870
2 /3 8.570
3 /4 9.790
5 /6 11.21
7 /8 11.97
14.39
9 /8 16.44
7 /6 16.94
5 /4 17.514 /3 17.39
3 /2 14.81
5 /3 10.37
7 /4 8.60
11 /6 6.90
15 /8 6.14
2 6.075
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CASE 3: Projection Point outside at R=10.6mFor the values,
1. Coefficient of friction = 12. Gravitational acceleration= 10 m/3. = 1 rad/s4. Starting position = 10.6 m5. Angle ofprojection w.r.t. disc=
Graph of Radius of circular loop (Y-axis) versus Speed of projection (X-axis):
9.2
9.3
9.4
9.5
9.6
9.7
9.8
9.9
10
10.1
0 2 4 6 8 10 12
Speed of projection w.r.t. disc m/s Radius of Circular loops
Meters11.037 Flies off to infinity
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Tr ajector ies Tr aver sed by the Body at Dif ferent Speeds:
*Colour codes in graphs:
Red colour- As observed from inertial frame of reference (outside the disc).
Blue colour- As observed by non-inertial frame of reference (rotating along with the centre of the
disc)
V=6.0 m/s V=6.08 m/s
V=7 m/s V=8 m/s
V=9 m/s V=10 m/s
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V=10.5 m/s V=10.8 m/s
V=10.9 m/sV=11.0 m/s
V=11.037 m/s V=11.04 m/s
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RESULTSAND
CONCLUSION
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Inferences
1.
The particle behaves in a very unique way on varying the velocity ofprojection. Under a certain limiting value of velocity, the particle
appears to enter a stable circular orbit of radius . At velocitieshigher than the critical value or the escape velocity, the particle leaves
the orbit and flies off to infinity.
This phenomenon can be explained as the friction force (providing
centripetal force) and centrifugal force balance each other. The particle
moves and gets into the circular orbit till the friction force is able to
balance the centrifugal force. = The value of depends on the value of angular velocity of disc,coefficient of friction between the disc and particle and the value of
acceleration due to gravity.
In our simulations, we took the values of=1rad/s, =1 and g=10m/ .Theoretically, the value of should be 10m. And it is verified by thesimulations also. The particle appears to rotate in circular orbit of radiusapproaching 10m just below the escape velocity.
2. When the particle is projected from outside at angle , it entersthe circular orbit for a certain range of velocities. The particle is
observed to go into infinity below and above those critical values.
This behaviour can be explained as the particle will not be able to enter
the region of if velocity of projection is too small, and it will not gointo circular orbit if velocity of projection is too high. In both theviolating cases, particle will go off to infinity, as observed in simulations.
3. The angle of projection of the particle is seen to have an unexpectedeffect on the escape velocity of particle. On increasing the angle , the
escape velocity first decreases, as expected but it increases suddenly at
=/2, instead of decreasing (as expected by common observation). The
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References and Bibliography
Books
Kleppner .D & Kolenkow .R (1999), An Introduction to Mechanics. New Delhi:
Tata McGraw Hill Publications.
Tutorials
MIT open courseware, MATLAB tutorial
http://ocw.mit.edu/index.htm
http://www.mathworks.com/help/techdoc/learn_matlab/bqr_2pl.html
Webpages
Spherical coordinate system, Wikipedia.