Example 2: Particle on String Pulled Through Hole
1
2.003J/1.053J Dynamics and Control I, Spring 2007
Professor Thomas Peacock
2/14/2007
Lecture 3
Dynamics of a Single Particle: Angular
Momentum
Example 2: Particle on String Pulled Through
Hole
Figure 1: Particle on string pulled through hole. Tabletop with hole B. A string
comes out with an attached mass. The particle is traveling around with an
angular velocity θ̇. Figure by MIT OCW.
Assume: Frictionless surface. Inextensible String.
Pull string through hole at B such that:
r(t0 ) = L
r(t1 ) = L/2
dr
dt (t0 )
dr
dt (t1 )
= 0
=0
If θ̇(t0 ) = θ˙0 , what is θ̇(t1 ) = θ˙1 ?
Cite as: Thomas Peacock and Nicolas Hadjiconstnatinou, course materials for 2.003J/1.053J Dynamics and
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Dynamics of systems of particles
2
Discussion
If we use linear momentum, will need to describe forces between m and string.
Thinking about angular momentum about the point B:
τ B = ḣB + v B × mv ← Angular momentum principle
hB = r × mv = Angular Momentum
Now:
τ B = r × F ← Forces acting on particleτB = 0 because r � F
τB = ḣB + v B × mv ⇒ Angular momentum about B is constant ḣB = 0.
τ B = 0 (from above)
v B = 0 because B is not moving
∴ hB = Constant
In Cartesian Coordinates
r = r cos θˆı + r sin θĵ
p = mv = mr˙ = −mrθ̇ sin θı̂ + mrθ̇ cos θĵ
a. hB (t0 ) = r × p = LmLθ˙0 k̂(k̂ is unit vector in z-direction: out of page).
b. hB (t1 ) = L2 m L2 θ˙1 k̂
Setting (a) = (b): θ˙1 = 4θ˙0 , and velocity of particle v1 = 2v0 =
L ˙
2 4θ0
= 2Lθ˙0 .
Energy is not conserved: why? The pulling force (tension) does work.
Dynamics of systems of particles
Forces on each particle may be composed as follows
F i = F iext + F int
i
Cite as: Thomas Peacock and Nicolas Hadjiconstnatinou, course materials for 2.003J/1.053J Dynamics and
Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology.
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Dynamics of systems of particles
3
Figure 2: Dynamics of systems of particles. Figure by MIT OCW.
Fi : Resultant force acting on mi
Fiext : External forces (e.g. gravity)
Fi
int : Internal forces between particles (e.g. charge attraction)
F iint =
n
�
f ij Force on particle i due to particle j
j=1
Newton’s Third Law
f ij = −f ji
Thus:
n
�
F iint =
n �
n
�
f ij = 0
i=1 j=1
j�=i
i=1
Sum of all internal forces is zero, therefore:
n
�
Fiint = 0
i=1
Total internal torques is also zero: demonstrate by considering an arbitrary pair
of particles:
Cite as: Thomas Peacock and Nicolas Hadjiconstnatinou, course materials for 2.003J/1.053J Dynamics and
Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology.
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Dynamics of systems of particles
4
Figure 3: Arbitrary pair of particles subject to individual forces. Figure by MIT
OCW.
τ B = r i/B × f ij + r j/B × f ji = (r i/B − rj/B ) × f ij
but (r i/B − r j/B � f ij )
∴ τ int
B = 0 No net internal torque
Center of mass
�n
�n
mi r i
mi r i
i=1
= i=1
r c = �n
M
m
i
i=1
M : Total Mass of System
�n
Note that this relation can also be written as i=1 mi (r i − rc ) = 0 i.e. center
of mass is the point about which the total mass moment is zero.
Newton’s Laws for Systems of Particles
(Williams: C-1 to C-3.6)
Derivation needed to prevent mistakes in applying the laws later. Will be able
to use results for rigid bodies.
Cite as: Thomas Peacock and Nicolas Hadjiconstnatinou, course materials for 2.003J/1.053J Dynamics and
Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology.
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Dynamics of systems of particles
5
Linear Momentum Principle (for a single particle)
Fi =
d
p
dt i
F i : Total Force on particle i
pi : Linear momentum of particle i
n
�
+
F ext
i
d �
d
pi = p
dt i=1
dt
n
F iint =
i=1
i=1
F int
i
n
�
=0
d
p
dt
F ext =
F ext : Sum of F external for whole system.
Note that total linear momentum:
p=
If
�n
ext
i=1 F i
n
�
pi =
ext
d � mi ri
dt i=1 M
n
mi vi = M v c where v c = ṙc =
i=1
i=1
=F
n
�
= 0 ⇒ p =constant; therefore, v c =constant.
Example: You have a ball as a ice skater. Throw object, both ball and skater
move, but center of mass stays the same, does not move.
Angular Momentum Principle
From Newton II F i =
d
dt pi
Torque:
�
d
p
dt i
�
ri × F i = ri ×
Sum over all particles.
n
�
i=1
�
=
r × F ext
i
�
�
ri ×
i=1
d
p
dt i
Later will need vectors to center of mass.
n
�
τ ext
iB = Sum of all external torques about B
i=1
Cite as: Thomas Peacock and Nicolas Hadjiconstnatinou, course materials for 2.003J/1.053J Dynamics and
Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology.
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Dynamics of systems of particles
6
Figure 4: A system of particles subject to a force. Figure by MIT OCW.
n
�
τ ext
iB =
i=1
n
�
�
r ×
i=1
n
�
n
n
�
�
d �
d
d �
( r i )pi
pi =
(r i × pi ) −
dt
dt
dt
i=1
i=1
� d
d �
hiB −
(r − rB ) × pi
dt i=1
dt i
i=1
n
τ ext
iB =
i=1
τBext =
n
n
n
�
�
d
HB −
v i × pi +
v B × pi
dt
i=1
i=1
v B is the same for each pi .
n
�
v B × pi = v B ×
i=1
n
�
pi = v B × p
i=1
So, finally we have:
τ ext
B =
d
H + vB × P
dt B
τ ext
B : Total External Torque
d
dt H B : Total Angular Momentum
v B × p: Total Linear Momentum
Next time: Consequences of this expression and work-energy principle.
Cite as: Thomas Peacock and Nicolas Hadjiconstnatinou, course materials for 2.003J/1.053J Dynamics and
Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology.
Downloaded on [DD Month YYYY].