Today%20Review:

1
Lecture 25

• Today Review

• Exam covers Chapters 14-17 plus angular momentum,
  rolling motion torque

• Assignment
• HW11, Due Tuesday, May 6th
• For Thursday, read through all of Chapter 18

2
Angular Momentum Exercise

• A mass m0.10 kg is attached to a cord passing
  through a small hole in a frictionless,
  horizontal surface as in the Figure. The mass is
  initially orbiting with speed wi 5 rad / s in a
  circle of radius ri 0.20 m. The cord is then
  slowly pulled from below, and the radius
  decreases to r 0.10 m.
• What is the final angular velocity ?
• Underlying concept Conservation of Momentum

3
Angular Momentum Exercise

• A mass m0.10 kg is attached to a cord passing
  through a small hole in a frictionless,
  horizontal surface as in the Figure. The mass is
  initially orbiting with speed wi 5 rad / s in a
  circle of radius ri 0.20 m. The cord is then
  slowly pulled from below, and the radius
  decreases to r 0.10 m.
• What is the final angular velocity ?
• No external torque implies
• DL 0 or Li Lc
• Ii wi If wf
• I for a point mass is mr2 where r is the
  distance to the axis of rotation
• m ri2wi m rf2 wf
• wf ri2wi / rf2 (0.20/0.10)2 5 rad/s 20
  rad/s

4
Example Throwing ball from stool

• A student sits on a stool, initially at rest, but
  which is free to rotate. The moment of inertia
  of the student plus the stool is I. They throw a
  heavy ball of mass M with speed v such that its
  velocity vector has a perpendicular distance d
  from the axis of rotation.
• What is the angular speed ?F of the
  student-stool system after they throw the ball ?

M
Mv
r
d
?F
I
I
Top view before after
5
Example Throwing ball from stool

• What is the angular speed ?F of the student-stool
  system after they throw the ball ?
• Process (1) Define system (2) Identify
  Conditions
• (1) System student, stool and ball (No Ext.
  torque, L is constant)
• (2) Momentum is conserved (check L r p
  sin q for sign)
• Linit 0 Lfinal - M v d I wf

M
v
d
?F
I
I
Top view before after
6
Ideal Fluid

• Bernoulli Equation ? P1 ½ r v12 r g y1
  constant
• A 5 cm radius horizontal pipe carries water at 10
  m/s into a 10 cm radius. ( rwater 103 kg/m3 )
• What is the pressure difference?
• P1 ½ r v12 P2 ½ r v22
• DP ½ r v22 - ½ r v12
• DP ½ r (v22 - v12 )
• and A1 v1 A2 v2
• DP ½ r v22 (1 (A2/A1 ) 2 )
• 0.5 x 1000 kg/m x 100 m2/s2 (1- (25/100)2 )
• 47000 Pa

7
A water fountain

• A fountain, at sea level, consists of a 10 cm
  radius pipe with a 5 cm radius nozzle. The water
  sprays up to a height of 20 m.
• What is the velocity of the water as it leaves
  the nozzle?
• What volume of the water per second as it leaves
  the nozzle?
• What is the velocity of the water in the pipe?
• What is the pressure in the pipe?
• How many watts must the water pump supply?

8
A water fountain

• A fountain, at sea level, consists of a 10 cm
  radius pipe with a
• 5 cm radius nozzle. The water sprays up to a
  height of 20 m.
• What is the velocity of the water as it leaves
  the nozzle?
• Simple Picture ½mv2mgh ? v(2gh)½
  (2x10x20)½ 20 m/s
• What volume of the water per second as it leaves
  the nozzle?
• Q A vn 0.0025 x 20 x 3.14 0.155 m3/s
• What is the velocity of the water in the pipe?
• An vn Ap vp ? vp Q /4 5 m/s
• What is the pressure in the pipe?
• 1atm ½ r vn2 1 atm DP ½ r vp2 ? 1.9 x
  105 N/m2
• How many watts must the water pump supply?
• Power Q r g h 0.0155 m3/s x 103 kg/m3 x 9.8
  m/s2 x 20 m
• 3x104 W (Comment on syringe
  injection)

9
Fluids Buoyancy

• A metal cylinder, 0.5 m in radius and 4.0 m high
  is lowered, as shown, from a massles rope into a
  vat of oil and water. The tension, T, in the
  rope goes to zero when the cylinder is half in
  the oil and half in the water. The densities of
  the oil is 0.9 gm/cm3 and the water is 1.0 gm/cm3
• What is the average density of the cylinder?
• What was the tension in the rope when the
  cylinder was submerged in the oil?

10
Fluids Buoyancy

• r 0.5 m, h 4.0 m
• roil 0.9 gm/cm3 rwater 1.0 gm/cm3
• What is the average density of the cylinder?
• When T 0 Fbuoyancy Wcylinder
• Fbuoyancy roil g ½ Vcyl. rwater g ½ Vcyl.
• Wcylinder rcyl g Vcyl.
• rcyl g Vcyl. roil g ½ Vcyl. rwater g ½
  Vcyl.
• rcyl ½ roill. ½ rwater
• What was the tension in the rope when the
  cylinder was submerged in the oil?
• Use a Free Body Diagram !

11
Fluids Buoyancy

• r 0.5 m, h 4.0 m Vcyl. p r2 h
• roil 0.9 gm/cm3 rwater 1.0 gm/cm3
• What is the average density of the cylinder?
• When T 0 Fbuoyancy Wcylinder
• Fbuoyancy roil g ½ Vcyl. rwater g ½ Vcyl.
• Wcylinder rcyl g Vcyl.
• rcyl g Vcyl. roil g ½ Vcyl. rwater g ½
  Vcyl.
• rcyl ½ roill. ½ rwater 0.95 gm/cm3
  
• What was the tension in the rope when the
  cylinder was submerged in the oil?
• Use a Free Body Diagram!
• S Fz 0 T - Wcylinder Fbuoyancy
• T Wcyl - Fbuoy g ( rcyl .- roil )
  Vcyl
• T 9.8 x 0.05 x 103 x p x 0.52 x 4 .0 1500 N

12
A new trick

• Two trapeze artists, of mass 100 kg and 50 kg
  respectively are testing a new trick and want to
  get the timing right. They both start at the
  same time using ropes of 10 meter in length and,
  at the turnaround point the smaller grabs hold of
  the larger artist and together they swing back to
  the starting platform. A model of the stunt is
  shown at right.
• How long will this stunt require if the angle is
  small ?

13
A new trick

• How long will this stunt require?
• Period of a pendulum is just
• w (g/L)½
• T 2p (L/g)½
• Time before ½ period
• Time after ½ period
• So, t T 2p(L/g)½ 2p sec
• Key points Period is one full swing
• and independent of mass
• (this is SHM but very different than a spring.
  SHM requires only a linear restoring force.)

14
Example

• A Hookes Law spring, k200 N/m, is on a
  horizontal frictionless surface is stretched 2.0
  m from its equilibrium position. A 1.0 kg mass
  is initially attached to the spring however, at a
  displacement of 1.0 m a 2.0 kg lump of clay is
  dropped onto the mass. The clay sticks.
• What is the new amplitude?

15
Example

• A Hookes Law spring, k200 N/m, is on a
  horizontal frictionless surface is stretched 2.0
  m from its equilibrium position. A 1.0 kg mass
  is initially attached to the spring however, at a
  displacement of 1.0 m a 2.0 kg lump of clay is
  dropped onto the mass.
• What is the new amplitude?
• Sequence SHM, collision, SHM
• ½ k A02 const.
• ½ k A02 ½ mv2 ½ k (A0/2)2
• ¾ k A02 m v2 ? v ( ¾ k A02 / m )½
• v (0.752004 / 1 )½ 24.5 m/s
• Conservation of x-momentum
• mv (mM) V ? V mv/(mM)
• V 24.5/3 m/s 8.2 m/s

16
Example

• A Hookes Law spring, k200 N/m, is on a
  horizontal frictionless surface is stretched 2.0
  m from its equilibrium position. A 1.0 kg mass
  is initially attached to the spring however, at a
  displacement of 1.0 m a 2.0 kg lump of clay is
  dropped onto the mass. The clay sticks.
• What is the new amplitude?
• Sequence SHM, collision, SHM
• V 24.5/3 m/s 8.2 m/s
• ½ k Af2 const.
• ½ k Af2 ½ (mM)V2 ½ k (Ai)2
• Af2 (mM)V2 /k (Ai)2 ½
• Af2 3 x 8.22 /200 (1)2 ½
• Af2 1 1½ ? Af2 1.4 m
• Key point KU is constant in SHM

17
Fluids Buoyancy SHM

• A metal cylinder, 0.5 m in radius and 4.0 m high
  is lowered, as shown, from a rope into a vat of
  oil and water. The tension, T, in the rope goes
  to zero when the cylinder is half in the oil and
  half in the water. The densities of the oil is
  0.9 gm/cm3 and the water is 1.0 gm/cm3
• Refer to earlier example
• Now the metal cylinder is lifted slightly from
  its equilibrium position. What is the
  relationship between the displacement and the
  ropes tension?
• If the rope is cut and the drum undergoes SHM,
  what is the period of the oscillation if
  undamped?

18
Fluids Buoyancy SHM

• Refer to earlier example
• Now the metal cylinder is lifted Dy from its
  equilibrium position. What is the relationship
  between the displacement and the ropes tension?
• 0 T Fbuoyancy Wcylinder
• T - Fbuoyancy Wcylinder
• T -rog(h/2Dy) AcrwgAc(h/2-Dy) Wcyl
• T -ghAc(ro rw)/2 Dy gAc(ro -rw) Wcyl
• T -Wcyl Dy gAc(ro -rw) Wcyl
• T g Ac (rw -ro) Dy
• If the is rope cut, net force is towards
  equilibrium position with a proportionality
  constant
• g Ac (rw -ro) with g10 m/s2
• If F - k Dy then k g Ac (ro -rw) p/4
  x103 N/m

19
Fluids Buoyancy SHM

• A metal cylinder, 0.5 m in radius and 4.0 m high
  is lowered, as shown, from a rope into a vat of
  oil and water. The tension, T, in the rope goes
  to zero when the cylinder is half in the oil and
  half in the water. The densities of the oil is
  0.9 gm/cm3 and the water is 1.0 gm/cm3
• If the rope is cut and the drum undergoes SHM,
  what is the period of the oscillation if
  undamped?
• F ma - k Dy and with SHM . w (k/m)½
• where k is a spring constant and m is the
  inertial mass (resistance to motion), the
  cylinder
• So w (1000p /4 mcyl)½
• (1000p / 4rcylVcyl)½ ( 0.25/0.95 )½
  
• 0.51 rad/sec
• T 3.2 sec

20
Underdamped SHM
if
If the period is 2.0 sec and, after four cycles,
the amplitude drops by 75, what is the time
constant?
Four cycles implies 8 sec So 0.25 A0 A0
exp(-4 b/m) ln(1/4) -4 1/t t -4 / ln(1/4)
2.9 sec
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Ch. 12
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Hookes Law Springs and a Restoring Force

• Key fact w (k / m)½ is general result where
  k reflects a
• constant of the linear restoring force and m is
  the inertial response
• (e.g., the physical pendulum where w (k / I)½

23
Simple Harmonic Motion
Maximum potential energy
Maximum kinetic energy
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Resonance and damping

• Energy transfer is optimal when the driving force
  varies at the resonant frequency.

• Types of motion
• Undamped
• Underdamped
• Critically damped
• Overdamped

25
Fluid Flow
26
Density and pressure
27
Response to forces
28
States of Matter and Phase Diagrams
29
Ideal gas equation of state
30
pV diagrams
31
Thermodynamics
32
Work, Pressure, Volume, Heat
T can change!
In steady-state Tconstant and so heat in equals
heat out
33
Gas Processes
34
Lecture 25

• Exam covers Chapters 14-17 plus angular momentum,
  rolling motion torque

• Assignment
• HW11, Due Tuesday, May 6th
• For Thursday, read through all of Chapter 18