Physics 2211: Lecture 12 Todays Agenda

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Physics 2211 Lecture 12Todays Agenda

• Dynamics of many-body systems
• Atwoods machine
• General case of two attached blocks on inclined
  planes
• Some interesting problems

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Lecture 12, Act 1Two-body dynamics

• A block of mass m, when placed on a rough
  inclined plane (m gt 0) and given a brief push,
  keeps moving down the plane with constant speed.
• If a similar block (same m) of mass 2m were
  placed on the same incline and given a brief
  push, it would

(a) stop (b) accelerate (c)
move with constant speed
m
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Many-body Dynamics

• Systems made up of more than one object
• Objects are typically connected
• By ropes pulleys today
• By rods, springs, etc. later on

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Atwoods Machine
Masses m1 and m2 are attached to an ideal
massless string and hung as shown around an ideal
massless pulley.
Fixed Pulley

• Find the accelerations, a1 and a2, of the masses.
• What is the tension in the string T ?

y
T1
T2
m1
a1
m2
a2
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Atwoods Machine...

• Draw free body diagrams for each object
• Applying Newtons Second Law ( y -components)
• T1 - m1g m1a1
• T2 - m2g m2a2
• But T1 T2 T since pulley is ideal
• and a1 -a2 -a.since the masses are
  connected by the string

Free Body Diagrams
T1
T2
y
a1
a2
m2g
m1g
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Atwoods Machine...

• T - m1g -m1 a (a)
• T - m2g m2 a (b)
• Two equations two unknowns
• we can solve for both unknowns (T and a).
• subtract (b) - (a)
• g(m1 - m2 ) a(m1 m2 )
• a
• add (b) (a)
• 2T - g(m1 m2 ) -a(m1 - m2 )
• T 2gm1m2 / (m1 m2 )

-
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Atwoods Machine...

• So we find

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Is the result reasonable? Check limiting
cases!

• Special cases
• i.) m1 m2 m a 0 and T mg. OK!
• ii.) m2 or m1 0 a g and T 0.
  OK!
• Atwoods machine can be used to determine g (by
  measuring the acceleration a for given masses).

-
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Attached bodies on two inclined planes
smooth peg
m2
m1
?1
?2
All surfaces frictionless
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How will the bodies move?
From the free body diagrams for each body, and
the chosen coordinate system for each block, we
can apply Newtons Second Law
x
y
Taking x components 1) T1 - m1g sin ?1 m1
a1X 2) T2 - m2g sin ?2 m2 a2X? But T1 T2
T and a1X -a2X a (constraints)
x
y
T1
T2
N
N
m2
m1
?2
?1
m2g
m1g
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Solving the equations
Using the constraints, solve the equations. T -
m1gsin ?1 -m1 a (a) T - m2gsin ?2 m2
a (b) Subtracting (a) from (b) gives
m1gsin ?1 - m2gsin ?2 (m1m2 )a So
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Special Case 1
Boring
m2
m1
If ?1 0 and ?2 0, a 0.
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Special Case 2
T
Atwoods Machine
T
m1
m2
If ?1 900 and ?2 900,
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Special Case 3
m1
m2
-
If ?1 0 and ?2 90,
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Lecture 12, Act 2Two-body dynamics

• In which case does block m experience a larger
  acceleration? In (1) there is a 10 kg mass
  hanging from a rope. In (2) a hand is providing
  a constant downward force of 98.1 N. In both
  cases the ropes and pulleys are massless.

m
a
F 98.1 N
Case (1)
Case (2)
(a) Case (1) (b) Case (2) (c)
same
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Problem Two strings Two Masses onhorizontal
frictionless floor

• Given T1, m1 and m2, what are a and T2?
• T1 - T2 m1a (a)
• T2 m2a (b)
• Add (a) (b) T1 (m1 m2)a
  a

• Plugging solution into (b)

a
x
m1
m2
T2
T1
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Lecture 12, Act 3Two-body dynamics

• Three blocks of mass 3m, 2m, and m are connected
  by strings and pulled with constant acceleration
  a. What is the relationship between the tension
  in each of the strings?

(a) T1 gt T2 gt T3 (b) T3 gt T2 gt T1
(c) T1 T2 T3
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Recap of todays lecture

• Dynamics of many-body systems.
• Atwoods machine.
• General case of two attached blocks on inclined
  planes.
• Some interesting special cases.
• Read Section 5.2 in Tipler.