Download Chapter 6 Kinetic-Energy Transformers (Teeter-Totters)

Chapter 6
Kinetic-Energy Transformers
(Teeter-Totters)
Problem
400 lb
100 lb
3 ft
fulcrum
Two circus performers, each weighing 200 lb jump off a platform 3 ft above a teetertotter and propel a 100-lb performer into the air, where he lands in a chair supported by a
fourth performer. How high can the 100-lb performer go? Where should the fulcrum of
the teeter-totter be placed so all the energy of the falling 400 lb is transferred to the
100-lb guy?
How High?
100
lb
d2 = ?
400 lb m1g
d1 = 3 ft
100
lb
m2g
We’ll represent the performers by idealized weights. The falling weight is mass m1 with
weight
m1g = 400 lb,
where g is the acceleration due to gravity g = 32 ft/s2. Then m1 = (400 lb)/( 32 ft/s2) =
12.5 slugs. Similarly, the weight
m2g = 100 lb
to be thrown in the air (on the left) has mass m2 = (100 lb)/( 32 ft/s2) = 3.125 slugs.
Mass m1 will fall a distance d1 = 3 ft. Therefore its potential energy (See Lesson 3) is
EP1 = m1g·d1 = (400 lb)(3 ft) = 1200 ft-lb.
If all this energy is transferred to mass m2, then it ends up (at the top of its rise) with the
same potential energy: EP2 = 1200 ft-lb. But
EP2 = m2g·d2 = (100 lb)d2 = 1200 ft-lb.
Therefore
d2 = (1200 ft-lb)/(100 lb) = 12 ft.
Total Energy Transfer
Where should the fulcrum of the teeter-totter be placed so mass m1 loses all its energy
(comes to a stop) and transfers it all to m2? We’ve seen total transfer of energy before in
Lesson 1, where one moving ball hits another at rest, and after the collision the first ball
is at rest, and the second ball is moving. This happened only when the two balls had the
same mass.
Imagine an experiment in outer space so there’s no gravity. Mass m1 with velocity –1 is
heading for an equal mass m2, which is standing still. (Since the masses have no weight
in outer space, we refer to them being 12.5 slugs rather than 400 lb.) After the collision,
m1 is standing still, and m2 has a velocity of –1. All the energy of m1 has been
transferred to m2!
After
Before
m1
m2
–1
12.5
m1
12.5
0
0
12.5
m2
–1
12.5
A teeter-totter can be used to reverse the direction of the energy. (The fulcrum is held
fixed somehow.) If the two arms of the teeter-totter are equal length, then m1 will still
“feel” that it is colliding with an equal mass. Again, all the energy of m1 is transferred to
m2.
Before
After
m1
m2
0
12.5
12.5
–1
m2
12.5
1
m1
12.5
0
First Try at Fulcrum Position
If the two weights are not equal (m1g = 400 lb, and m2g = 100 lb), what should the ratio of
the teeter-totter arms be so m1 still “feels” an equal mass? We know that the teeter-totter
will balance if the left arm has length L2 = 16 ft, and the right arm has length L1 = 4.
m2g
m1g
100
lb
L2
= 16 ft
400 lb
L1
= 4 ft
We can show this by removing m1 and finding the force F1 that will support m2.
d2 = ?
F1 = ?
100
lb
F2= 100 lb
d1= 0.4 ft
L2
= 16 ft
L1
= 4 ft
Let F1 push the right side down by a distance d1 = 0.4 ft, causing the 100-lb weight to go
up by a distance d2. Since the distances d2 and d1 must have the same ratio as L2 and L1
have, then
d2 = 4·d1 = 4·0.4 ft = 1.6 ft.
The lifting force F2 on the left is 100 lb, so the work done on the left is
W2 = F2d2 = (100 lb)(1.6 ft) = 160 ft-lb.
Then the work done on the right must also be 160 ft-lb.
W1 = F1d1 = F1·(0.4 ft) = 160 ft-lb.
Therefore
F1 = (160 ft-lb)/(0.4 ft) = 400 lb,
and 400 lb on the right will balance with 100 lb on the left if the teeter-totter arms have
the ratio L2/ L1 = 4.
But we aren’t interested in having F1 balance the 100-lb weight. We want F1 to
accelerate the 100-lb weight. Now, the acceleration of gravity is the same on the left and
on the right: g = 32 ft/s2. But when F1 accelerates the right end of the teeter-totter, the
left end will have a different acceleration.
Let’s go back to outer space so gravity isn’t an issue. (We use the mass m2 = 3.125 slugs
for the 100-lb weight.) As F1 pushes through a distance d1, we’ll pay attention to the
F1 = ?
d1= 0.1 ft
a1 = 0.2 ft/s2
m
d2 =0.4
a2 =0.8 ft/s2
2
3.125
F2
L2
= 16 ft
L1
= 4 ft
acceleration a1 there. Say a1 = 0.2 ft/s2. If we accelerate for T = 1 sec, then the final
velocity will be vfinal = a1T = (0.2 ft/s2)(1 sec) = 0.2 ft/s. The average velocity will be
half of this: vave = ½vfinal = 0.1 ft/s, and the distance gone in that time T is d1 = vaveT =
0.1 ft.
Both the distance and the acceleration on the left will be greater by a factor L2/ L1 = 4.
That is,
d2 = 4· d1 = 4·0.1 ft = 0.4 ft,
a2 = 4· a1 = 4·0.2 ft/s2 = 0.8 ft/s2.
and
What force F2 will it take to accelerate m2 at 0.8 ft/s2?
F2 = m2a2 = (3.125 slugs)( 0.8 ft/s2) = 2.5 lb.
What force F1 on the right will it take to produce F2 = 2.5 lb on the left? The ratio F1/ F2
is the same as the ratio L2/ L1 = 4. Therefore
F1 = 4·F2 = 4·(2.5 lb) = 10 lb.
Now we can calculate how much mass the force F1 “feels.” Since F = ma, the mass felt
at the right is
mfelt = F1/a1 = (10 lb)/(0.2 ft/s2) = 50 slugs.
[This mass, if it were a physical thing, would have a weight mfeltg = (50 slugs)(32 ft/s2) =
1600 lb!]
But for all the energy to be transferred, we wanted mfelt to equal the m1 = 12.5 slugs that
falls. Instead we got mfelt = 4· m1. What happened? Well, the ratio L2/L1 = 4 became a
factor twice in determining mfelt, so we got a factor of 42 = 16 rather that the desired
factor of 4. If we go through the calculations above, we find that
mfelt = (L2/L1)2·m2 = 42·(3.125 slugs) = 50 slugs.
Second Try at Fulcrum Position
In order for mfelt to equal m1, we must choose
(L2/L1)2 = mfelt/m2 = m1/m2 = 4,
and
L2/L1 = 2.
An example is shown below with L2 = 14 ft and L1 = 7 ft. Now the 400-lb weight falling
d1 = 3 ft can transfer all its energy to the 100-lb weight, which then rises d2 = 12 ft.
100
lb
d2 = 12
400 lb m1g
d1 = 3 ft
100
lb
m2g
L2
= 14 ft
L1
= 7 ft
Problems
Problem 1
A mass m1 = 12.5 slugs with velocity –5 hits a stationary mass m2 = 50 slugs. What will
the velocity of m2 be after the collision? Calculate the kinetic energy of m1 before the
collision and the kinetic energy of m2 before the collision. What fraction of the kinetic
energy of m1 has been transferred to m2? (See Lesson 1 for solving for the “After”
velocities.)
After
Before
m1
m2
12.5
50
–5
m1
12.5
?
0
m2
50
?
Problem 2
We saw in “First Try at Fulcrum Position” that if m2 = 3.125 slugs, L2 = 16 ft and L1 =
4 ft, then mfelt is 50 slugs. Use the results of Problem 1 to find the fraction of the
potential energy of the falling m1 (12.5 slugs) that will be transferred to m2. How high
will m2 go when m1 falls 3 ft?
3.125
d2 = ?
12.5
m1
d1 = 3 ft
3.125
m2
L2
= 16 ft
L1
= 4 ft
Problem 3
We reduce the weight of m1 to 300 lb (see figure below). Calculate the proper value for
L2 below so that all the energy is transferred to m2 with weight 100 lb. (Use the method
in “Second Try at Fulcrum Position.”) Find the height d2.
100 lb
d2 = ?
300 lb
m1g
d1 = 3 ft
100 lb
m2g
L2
=?
L1
= 4 ft
Experiments
Experiment 1
2 oz
d2
8 oz
m1g
d1
2 oz
m2g
L2
L1
Using the m1g = 8 oz and m2g = 2 oz weights provided, set L2 = 2·L1, and measure the
ratio d2/d1. (Have a partner hold rulers at d1 and d2.) The ratio will be somewhat less
than 4 because some of the energy from m1 is going into the teeter-totter’s motion.
Calculate what fraction of the energy from m1 is going to m2.
Experiment 2
Repeat Experiment 1, but setting L2 = 4·L1. Compare the measured ratio d2/d1 with the
results of Problem 2. Calculate what fraction of the energy is going to m2.