Physics 111: Lecture 12 Today's Agenda

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Physics 111 Lecture 12Today's Agenda

• Problems using work/energy theorem
• Spring shot
• Escape velocity
• Loop the loop
• Vertical springs
• Definition of Power, with example

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Problem Spring Shot

• A sling shot is made from a pair of springs each
  having spring constant k. The initial length of
  each spring is x0. A puck of mass m is placed at
  the point connecting the two springs and pulled
  back so that the length of each spring is x1. The
  puck is released. What is its speed v after
  leaving the springs? (The relaxed length of each
  spring is xr).

xr
x1
x0
m
m
m
v
3
Problem Spring Shot

• Only conservative forces are at work, so KU
  energy is conserved. EI EF ?K
  -?Us

x1
x0
m
m
4
Problem Spring Shot

• Only conservative forces are at work, so KU
  energy is conserved. EI EF ?K
  -?Us

m
m
at rest
v
5
Problem Spring Shot
Spring Shot

• Only conservative forces are at work, so KU
  energy is conserved. EI EF ?K
  -?Us

m
v
6
Problem How High?

• A projectile of mass m is launched straight up
  from the surface of the earth with initial speed
  v0. What is the maximum distance from the center
  of the earth RMAX it reaches before falling back
  down.

RMAX
m
RE
v0
M
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Problem How High...

• All forces are conservative
• WNC 0
• ?K -?U

• And we know

RMAX
m
RE
v0
hMAX
M
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Problem How High...
RMAX
m
RE
v0
hMAX
M
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Escape Velocity

• If we want the projectile to escape to infinity
  we need to make the denominator in the above
  equation zero

We call this value of v0 the escape velocity, vesc
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Escape Velocity

• Remembering that we find the
  escape velocity froma planet of mass Mp and
  radius Rp to be(where G 6.67 x 10-11 m3
  kg-1 s-2).

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Lecture 12, Act 1Escape Velocity

• Two identical spaceships are awaiting launch on
  two planets with the same mass. Planet 1 is
  stationary, while Planet 2 is rotating with an
  angular velocity ?.
• Which spaceship needs more fuel to escape to
  infinity?

(a) 1 (b) 2 (c) same

• ?

(1)
(2)
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Lecture 12, Act 1Solution

• Both spaceships require the same escape velocity
  to reach infinity.
• Thus, they require the same kinetic energy.
• Both initially have the same potential energy.
• Spaceship 2 already has some kinetic energy due
  to its rotational motion, so it requires less
  work (i.e. less fuel).

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Lecture 12, Act 1Aside
v ?r

• This is one of the reasons why all of the worlds
  spaceports are located as close to the equator as
  possible.

r2 gt r1
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Lecture 12, Act 1Algebraic Solution
WNC ?K ?U ?E
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Problem Space Spring

• A low budget space program decides to launch a
  10,000 kg spaceship into space using a big
  spring. If the spaceship is to reach a height RE
  above the surface of the Earth, what distance d
  must the launching spring be compressed if it has
  a spring constant of 108 N/m.

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Problem Space Spring...

• Since gravity is a conservative force, energy is
  conserved. Since K 0 both initially and at the
  maximum height (v 0) we know
• Ubefore Uafter
• (US UG )before (UG )after

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Problem Space Spring
So we find

• For the numbers given, d 79.1 m
• But dont get too happy...

F kd ma a kd/m a 79.1 x 106 m/s2 a
791000 g unhappy astronaut!
a
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Problem Loop the loop

• A mass m starts at rest on a frictionless track a
  distance H above the floor. It slides down to
  the level of the floor where it encounters a loop
  of radius R. What is H if the mass just barely
  makes it around the loop without losing contact
  with the track.

H
R
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Problem Loop the loop

• Draw a FBD of the mass at the top of the loop
• FTOT -(mgN) j
• ma -mv2/R j
• If it just makes it, N 0.
• mg mv2/R

v
j
H
R
i
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Problem Loop the loop
Loop the Loop

• Now notice that KU energy is conserved. ?K
  -?U.
• ?U -mg(h) -mg(H-2R)
• ?K 1/2 mv2 1/2 mRg

h H - 2R
v
H
R
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Lecture 12, Act 2Energy Conservation

• A mass starts at rest on a frictionless track a
  distance H above the floor. It slides down to
  the level of the floor where it encounters a loop
  of radius R. What is H if the normal force on
  the block by the track at the top of the loop is
  equal to the weight of the block ?

(a) 3R (b) 3.5R (c)
4R
H
R
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Lecture 12, Act 2Solution

• Draw a FBD of the mass at the top of the loop
• FNET -(mgN) j
• ma -mv2/R j
• In this case, N mg.
• 2mg mv2/R

v
j
H
R
i
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Lecture 12, Act 2Solution

• Use the fact that KU energy is conserved DK
  -DU.
• DU -mg(h) -mg(H - 2R), DK 1/2 mv2 mRg
• mg(H - 2R) mRg

h H - 2R
v
H
R
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Vertical Springs
(a)
(b)

• A spring is hung vertically. Its relaxed
  position is at y 0 (a). When a mass m is hung
  from its end, the new equilibrium position is ye
  (b).

j
k

• Recall that the force of a spring is Fs -kx.
  In case (b) Fs mg and x ye-kye - mg
  0 (ye lt 0)

y 0
y ye
-kye
mg
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Vertical Springs
(a)
(b)

• The potential energy of the spring-mass system is

j
k
y 0
y ye
choose C to make U0 at y ye
m
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Vertical Springs
(a)
(b)

• So

j
k
y 0
which can be written
y ye
m
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Vertical Springs
(a)
(b)
j
k

• So if we define a new y? coordinate system such
  that y? 0 is at the equilibrium position, ( y?
  y - ye ) then we get the simple result

y? 0
m
?
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Vertical Springs
(a)
(b)

• If we choose y 0 to be at the equilibrium
  position of the mass hanging on the spring, we
  can define the potential in the simple form.
• Notice that g does not appear in this
  expression!!
• By choosing our coordinates and constants
  cleverly, we can hide the effects of gravity.

j
k
y 0
m
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160
140
U of Spring
120
U
100
US 1/2ky2
80
60
40
20
0
-10
-8
-6
-4
-2
0
2
4
6
8
10
-20
y
-40
-60
30
160
140
U of Gravity
120
U
100
80
60
UG mgy
40
20
0
-10
-8
-6
-4
-2
0
2
4
6
8
10
-20
y
-40
-60
31
160
UNET UG US
140
U of Spring Gravity
120
U
100
US 1/2ky2
80
60
UG mgy
40
20
0
-10
-8
-6
-4
-2
0
2
4
6
8
10
-20
y
-40
-60
ye
0
shift due to mgy term
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160
Choose C such as to show that the new equilibrium
position has zero potential energy
140
U of Spring Gravity
120
U
100
80
60
40
UNET UG US C
20
0
-10
-8
-6
-4
-2
0
2
4
6
8
10
-20
US 1/2ky2
y
-40
-60
ye
0
shift due to mgy term
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Lecture 12, Act 3Energy Conservation

• In (1) a mass is hanging from a spring.
  In (2) an identical mass is held at the height of
  the end of the same spring in its relaxed
  position.
• Which correctly describes the relation of the
  potential energies of the two cases?

(a) U1 gt U2 (b) U1 lt U2 (c) U1 U2
case 2
case 1
d
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Lecture 12, Act 3Solution

• In case 1, it is simplest to choose the mass to
  have zero total potential energy (sum of spring
  and gravitational potential energies) at its
  equilibrium position.

The answer is (b) U1 lt U2.
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Vertical SpringsExample Problem

• If we displace the mass a distance d from
  equilibrium and let it go, it will oscillate up
  down. Relate the maximum speed of the mass v to
  d and the spring constant k.
• Since all forces are conservative,E K U is
  constant.

j
k
y d
y 0
v
y -d
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Vertical SpringsExample Problem
Spring

• At the initial stretched positionand K 0
  (since v0).

j

• Since EKU is conserved,will always be true !

k
y d

• Energy is shared between the K and U terms.
• At y d or -d the energy is all potential
• At y 0, the energy is all kinetic.

y 0
v
y -d
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Power
Ladder

• We have seen that W F.?r
• This does not depend on time!

F
?r

• Units of power J/sec N-m/sec Watts

v
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Power

• A 2000 kg trolley is pulled up a 30 degree hill
  at 20 mi/hrby a winch at the top of thehill.
  How much power is thewinch providing?
• The power is P F.v T.v
• Since the trolley is not accelerating, the net
  force on it must be zero. In the x direction
• T - mg sin ? 0
• T mg sin ?

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Power

• P T.v Tv since T is parallel to v
• So P mgv sin ?
• v 20 mi/hr 8.94 m/sg 9.81 m/s2m 2000
  kgsin ? sin(30o) 0.5
• and P (2000 kg)(9.81 m/s2)(8.94 m/s)(0.5)
  87,700 W

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Recap of todays lecture

• Problems using work/energy theorem
• Spring shot
• Escape velocity
• Loop the loop
• Vertical springs
• Definition of Power, with example (Text 6-3)
• Look at textbook problems Chapter 8 81, 85,
  95 Chapter 11 49