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Physics 2211: Lecture 20 Today's Agenda

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Planet 1 is stationary, while Planet 2 is rotating with an ... A low budget space program decides to launch a 10,000 kg spaceship into space using a big spring. ... – PowerPoint PPT presentation

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Title: Physics 2211: Lecture 20 Today's Agenda


1
Physics 2211 Lecture 20Today's Agenda
  • Problems using work / energy theorem
  • Escape velocity
  • Space spring
  • Loop the loop
  • Vertical springs

2
Lecture 20, 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)
3
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. (Ignore air
    resistance.)

4
Problem Space Spring...
  • WEXT WNC 0, therefore mechanical energy is
    conserved. Since K 0 both initially and at the
    maximum height (v 0) we know
  • Ubefore Uafter
  • (US UG )before (US UG )after

5
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
6
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
7
Problem Loop the loop
  • Draw a FBD of the mass at the top of the loop
  • FTOT -mg - N
  • ma -mv2/R
  • mg N mv2/R
  • If it just makes it, N 0.
  • mg mv2/R

8
Problem Loop the loop
  • Now notice that KU energy is conserved gt ?K
    -?U.
  • ?U -mgy -mg(H-2R)
  • ?K 1/2 mv2 1/2 mRg

9
Lecture 20, 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) 2R (b) 2.5R (c)
3R
H
R
10
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).

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

y 0
y ye
11
Vertical Springs
(a)
(b)
  • The potential energy of the spring-mass-earth
    system is

y
k
y 0
y ye
choose C to make U 0 at y ye
m
12
Vertical Springs
(a)
(b)
  • So

y
k
y 0
which can be written
y ye
m
13
Vertical Springs
(a)
(b)
y
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
?
14
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.

y
k
y 0
m
15
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
16
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
17
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
18
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
19
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.

y
k
y d
y 0
v
y -d
20
Vertical SpringsExample Problem
y
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
21
Recap of todays lecture
  • Problems using work / energy theorem
  • Escape velocity
  • Space spring
  • Loop the loop
  • Vertical springs
  • Begin reading Chapter 8 in Tipler
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