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AP50 Fall 2016
Problem Set 3 Solutions
1) Ultralight umbrella estimation
A company thinks that there is a market for ultralight umbrellas. While developing
these umbrellas, the design team was confronted with the following question: Is the
force of the rain falling on an opened umbrella a force that needs to be taken into
consideration in designing an umbrella? You have been hired as a consultant to this
firm. Your task is to estimate the magnitude of this force. You are allowed to use
anything that you feel may assist you. You must be able to justify your solution
before the members of the design team.
I: Getting Started: We want to decide whether or not the force of raindrops on an
open umbrella is large enough to need to be considered when designed an ultralight
umbrella. We draw a sketch.
𝑣⃑ 𝑑𝑟𝑜𝑝
Aumbrella = πr 2 = π ∗ (0.5m)2
Assumptions:
1. The rain falls vertically and collides totally inelastically with the umbrella. It
just sticks there and rolls off (doesn’t bounce back like hail might for
example)
2. Our umbrella has a diameter of about 1m and we will consider it a flat circle
with surface area of A = πr 2 over which the raindrops apply force.
3. In weather forecasts, I have seen severe storm warnings for rainfall
maximum rates of up to about 2-3” in an hour. So let’s assume this for a
worst-case scenario very heavy rainfall rate to start. I can’t imagine even
going out with a small umbrella in this heavy of rain, but let’s try it.
4. The raindrops are water droplets with density 1000 kg/m3
II: Devise Plan: This problem is not well defined. We’ve already made some
assumptions which will help a lot, but let’s try to make our plan more clear.
We know the contact force of a raindrop on the umbrella must have an opposing
force (the contact force of the umbrella up on the raindrop) which we can think of as
equivalent to the force keeping the umbrella open, the design force we want.
We want to know how big the force of raindrops is on the umbrella, to decide if this
is a factor in the design. We will break it up into a few smaller problems.
1) Estimate how fast a single raindrop will be falling when it hits the umbrella;
2) Estimate how many raindrops will hit the umbrella at once (i.e. the rate of
rainfall and size of the umbrella);
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3) Use these values and impulses to estimate the average force that rain will
have while hitting an open umbrella;
III: Execute Plan:
Step 1) First, we want to estimate how fast a single raindrop will be falling when it
hits the umbrella.
We know that the rain is accelerated by gravity so we may be tempted to just use
our free-fall equations. The problem with this approach is that gravity is not the
only force on the falling raindrop. In this case, air resistance is not negligible so we
should not ignore it. Let’s investigate this further…
Let’s pretend first that air resistance is negligible, as we have been, to see what
happens. Assuming a raindrop falls from a height >2000m (an average cloud height http://science-edu.larc.nasa.gov/SCOOL/cldhgt.html), if the only force acting on it is
gravity, using our free-fall projectile equations we find that the force as it reaches
the ground (approximate the distance between where the raindrop starts to
accelerate and where our umbrella is as compared to a length of 2000 m) is:
𝑣⃑ 𝑑𝑟𝑜𝑝 = √2g Δx = √2 ∗
9.8m
s2
∗ 2000m
around 200m/s! which is approx. 450mph. This is CRAZY!
We know we need to consider air resistance since the velocity, as we know from
experience, is MUCH slower than this. The wind speed of a powerful tornado is only
about 130m/s (http://en.wikipedia.org/wiki/Orders_of_magnitude_(speed)) and
we know raindrops fall much slower than that.
We instead find the terminal velocity. This means that we assume the drop will be
slowed down by air resistance and we assume it to be in equilibrium before it hits
the umbrella. Let’s draw a free-body diagram.
⃑F⃑ cad = the contact force by air resistance on the raindrop
⃑F⃑ GEd = the force by Earth’s gravity on the raindrop
⃑⃑ cad = drag force
F
⃑F⃑ GEd = mg
The terminal velocity is constant due to the fact that there is no acceleration and the
raindrop is in equilibrium, therefore:
⃑F⃑ GEd = ⃑F⃑ cad
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But since we haven’t learned about how to calculate the force of drag/air resistance,
let’s see if we can just look up some values for the terminal velocity of raindrops and
see what we find.
This site at http://hyperphysics.phy-astr.gsu.edu/hbase/airfri2.html#c2 explains
not only what terminal velocity is but also has some examples (including a
raindrop). It also contains an interactive form in which we can make some
calculations. For a raindrop of 0.2cm radius, the terminal velocity can be calculated
as 9m/s. Let’s round up to an even 10m/s for our approximation.
Step 2) Next, we want to estimate how many raindrops will hit the umbrella at a
time. To do this we estimate the rate of rainfall and the size of the umbrella. We’ll
start with thinking about a rate we know, for example that 2” of rain in an hour is
what we may expect for a very heavy storm (we want to choose a design constraint
with what we consider a heavy load for an ultralight umbrella to test it). We can
calculate the volume of water hitting the umbrella in that time, and then convert the
units from m3/hr to kg/s that we need to help find force.
2” in an hour
𝐴𝑢𝑚𝑏𝑟𝑒𝑙𝑙𝑎 = 𝜋𝑟 2= 𝜋(0.5m)2
volume of rain hitting umbrella
= Area of umbrella ∗ rainfall rate
time
= πr 2 ∗ 2 inches/hour
2 inches
0.0254m
= π (0.5m)2 ∗ hour ∗ inch = 0.040m3 /hr
mass of rain volume of rain
=
∗ density of water
time
time
=
0.040m3
hour
hour
∗ 3600 sec ∗
1000kg
m3
= 0.011 kg/s
Step 3) Finally, we use the values we have calculated so far and the terminal velocity
of raindrop to estimate the average force of raindrop hitting an open umbrella. If we
consider our system as the raindrops and the umbrella, we know that the rate of
change of the momentum of the raindrops is due to a net force applied to them. The
system is not isolated.
∆p
∑F =
∆t
If we focus on a single drop of rain, we know that it just hits the umbrella and then
slides off with effectively 𝑣⃑ drop−final = 0 in the vertical direction after hitting the
umbrella
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We know, the initial velocity of a drop of rain before hitting the umbrella is about:
𝑣⃑ drop−initial = −10 m/𝑠 (the terminal veloicty from earlier)
Assuming a completely inelastic collision, where md = inertia of a raindrop
∆p = pf − pi = md ∗ (0 – (−10m/s)) = md ∗ 10m/s
If we now look back at the contact force of the raindrop on the umbrella, and use the
c
⃑⃑du
values we have found, we find that the average force F
on the umbrella from all of
the drops of rain is only about 0.1N.
c
⃑F⃑du
=
∆p md ∗ 10m/s
kg
m
=
= 0.011 ∗ 10 = 0.1 N
∆t
∆t
s
s
To get an order of magnitude idea of how large this is, if you take the inertia of 10g
(or 1/10th of an apple) and multiply it by g, this gives 0.1N of force. The force of the
raindrops on the umbrella is tiny and does not need to be taken into account.
IV: Evaluate Result:
First, does this order of magnitude make sense to us? Yes, from personal experience
the amount of impact on our umbrellas from raindrops seems to be less than the
force we feel of gravity pushing down on an apple if we hold it in our hand. Also,
from experience, wind seems much more likely to break an umbrella than rain and
this would likely be a much more important design parameter to consider.
As a better check, let’s write the equation we used to calculate 0.1 N in general terms
and look at limiting cases:
∆p md ∗ 𝑣⃑ drop−terminal
c
⃑⃑du
F
=
=
∆t
∆t
= πr 2 ∗ 𝑣⃑ drop−terminal ∗ rainfall rate ∗ density of water
First, let’s check the units of this to make sure we didn’t forget anything.
m
m
Units = [m2 ] ∗ [ s ] ∗ [ s ] ∗ [kg/m3 ] = [kg*m/s 2 ] = [N]
Looks ok, so now let’s think about limiting cases. For example, if the radius of the
umbrella r → 0, then the force of rain on the umbrella would also go to zero. If
r increases, the force also increases. A larger umbrella catches more raindrops and
feels a larger impulse. What about the terminal velocity of the raindrops,
𝑣⃑ drop−terminal? Well, as we already mentioned, we know that if the raindrops get
bigger, they have larger terminal velocities, and this also increases the force on the
umbrella, as we expect. If rainfall rate → 0, then the force on the umbrella → 0 as we
expect if it is not raining so there are no drops hitting it. Good!
AP50 Fall 2016
2) Spring into action
On a low-friction track, a 0.36-kg cart initially moving to the right at 2.05m/s
collides elastically with a 0.12-kg cart initially moving to the left at 0.13m/s. The
0.12-kg cart bounces off of the other cart and then compresses a spring attached to
the right end of the track.
(𝑎) At the instant of maximum compression of the spring, how much elastic
potential energy is stored in the spring?
(𝑏) If the spring then returns all of this energy to the cart, and the two carts
collide again, what is the final velocity of each cart?
I: Getting Started: We can begin by organizing the information given to us in the
problem into a few sketches.
Initially
After Elastic Collision
Compression of Spring
Second Elastic Collision
After Second Elastic Collision
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II: Devise Plan:
Step 1: Determine the velocity of both carts after the elastic collision using
conservation of momentum and conservation of energy.
Step 2: Determine the kinetic energy of the 0.12kg cart.
Step 3: Determine the Elastic potential energy of the spring using the kinetic energy
obtained in the previous step.
Step 4: Determine the velocities of the two carts after the second elastic collision.
III: Execute Plan:
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IV: Evaluate Result:
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3) Going up?
Penny wants to test out her new scale. Suppose Penny’s inertia is 45 kg. She does
five experiments (A-E). Penny stands on a scale that is…
A:
B:
C:
D:
E:
in her room and notes the scale reading.
at rest with respect to an elevator that is moving upward at a constant speed.
at rest with respect to an elevator that is moving downward at a constant speed.
at rest with respect to an elevator that is accelerating up at 2 m/s2.
at rest with respect to an elevator that is accelerating downward at 2 m/s2.
(𝑎) Draw free-body diagrams for the elevator cabin, Penny, and the scale,
respectively, labeling the forces using the notation introduced in the book. Make
sure the lengths of the force vectors are appropriate to the magnitudes of the forces.
(𝑏) State the magnitudes of as many forces as possible and identify all the forces
that form interaction pairs. Explain briefly how you obtained these magnitudes.
(𝑐) In which, if any, of these experiments (A-E) does the scale reading register a
value that is less than the reading when Penny stands on the scale in her room.
I: Getting Started:
(a) First we draw a sketch of Penny standing on the scale in an elevator cabin.
We draw free body diagrams for the elevator cabin, Penny and the scale. These must
be separate to indicate only the forces on each object.
Elevator Cabin:
Penny:
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Scale:
II: Devise a Plan:
First we have drawn free body diagrams labeling all of the forces (part a). To find
the magnitudes of the forces and identify the force pairs, we will go through
Experiments A-E one by one and find the forces using F = ma. Finally for part c, we
will compare the forces we have found to mg, which is the reading on Penny’s scale
in her room.
III: Execute Plan:
(b) Finding the force pairs and magnitudes
Experiment A:
Experiment B:
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Experiment C:
Experiment D:
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Experiment E:
(c) In Experiment D, the force by the scale on Penny (and thus also the opposite) is
greater than mg (Penny’s inertia times gravity), so she will see a number on the
scale that is larger than she does when she stands on the scale at home! If she
wants to feel like she has lost some weight, she would want to stand on the scale
and take the reading during experiment E. While the elevator accelerates
downwards, the reading will be less than when she takes it at home.
IV: Evaluate Result:
Penny “feels heavier” when the elevator accelerates upwards (experiment D) and
“lighter” when the elevator accelerates downwards. Think of when you are taking
off in a plane. As it accelerates upward, you do feel a force from your seat pushing
you up harder, just like when the elevator accelerates up. You can also check your
answers using: http://dallaswinwin.com/Newtons_Laws/weight_elevator.htm
AP50 Fall 2016
4) Quick shot
A bullet of inertia 𝑚 traveling at speed 𝑣 is fired into a wooden block that has inertia
4𝑚 and rests on a level surface. The bullet passes through the block and emerges
with speed 𝑣/3, taking a negligible amount of the wood with it. The block moves to
the right but comes to rest after traveling a distance 𝑑.
(𝑎) What is the magnitude of the frictional force between the block and the surface
while the block is moving?
(𝑏) What is the ratio of the energy dissipated as the bullet passes through the block
to the energy dissipated by friction between the level surface and the bottom face of
the block?
I: Getting Started:
We assume that the bullet passes through the block so quickly that the block has
moved a negligible amount of distance by the time the bullet is already out the other
side. Drawing a sketch of this first part…
Before the bullet, passes through (figure a)
+x
And just after the bullet passes through, (figure b)
After the bullet passes through, the block then starts to slide and loses kinetic
energy to thermal energy as energy is dissipated by the friction force with the table.
The sliding of the block across the level surface, starting with some initial velocity
and then coming to a stop some distance, d, later is depicted below as well as an
energy diagram for the closed block-Earth system.
(figure c)
(figure d)
v=?
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II: Devise Plan:
1) First, we will use conservation of momentum to find out how fast the block
will be moving after the bullet passes through;
2) Then we will use that value and conservation of momentum to find the
magnitude of the frictional force between the block and the surface;
3) Finally, we will use the fact that some kinetic energy is converted to thermal
energy as the bullet passes through the block to calculate how much energy
is dissipated in the process and compare it to the energy lost due to friction.
III: Execute Plan:
(a) First, we use conservation of momentum to find the velocity of the block just
after the bullet passes through it. From figures a & b, there are no external forces
on the block-bullet system during the short collision time so the momentum is
unchanged.
This gives us the velocity that the block starts with before it slides across the level
surface that it is sitting on. We know that while the block is sliding, in our system of
block + Earth drawn above in figure d, there are no reversible changes in the
configuration (no potential energy change) and no source energy is used (like a
person’s energy, fuel in a car, battery, etc) so this is entirely kinetic energy being
converted to thermal energy (friction generates heat).
We know that the energy is dissipated over a distance d so we can find the friction
force of the level surface on the block in the x- direction as:
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The force is negative because it acts in the opposite direction of the motion of the
block, causing the block to slow down, but the magnitude of the frictional force that
we have been asked to find in part (a) is 𝑚𝑣 2 /18𝑑.
(b) For part b of the problem, we want to compare the energy dissipated by friction
between the level surface and the bottom face of the block to the energy dissipated
as the bullet travels through the block so we need to calculate how much energy is
lost when the bullet travels through the block.
We calculate the initial kinetic energy, Ki (back to figure a above) and compare to it
the final energy, Kf just after the bullet exits the block (figure b).
If we compare this to the energy dissipated in sliding that we found earlier in part
(a), we find that 7 times of energy is dissipated by the bullet going through the block
as compared to by the friction of the block sliding along the surface.
IV: Evaluate Result:
The magnitude of frictional force we found in part (a) is 𝑚𝑣 2 /18𝑑. We can check the
units of this expression to make sure they give us the units of force [N}.
units = [kg][m/s]^2/[m] = [kg*m/s^2] = [N] so this looks ok
This is an easy first check but not good enough to convince us we’re right. Let’s look
at what happens to the frictional force as different variables change and see if it
makes sense. As the inertia m of the block increases, the force of friction increases.
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We know this is true if we try to slide two boxes which are identical but one has
more stuff inside. If we push horizontally and need to counteract the friction force
which is also horizontal, it is harder when the box has greater inertia. If the block is
not moving, (v→0), then there is no friction as we expect. If the distance d over
which the block stops decreases, the frictional force must increase (greater force to
stop it faster). In the end, we have entirely thermal energy which is irreversible.
This makes sense because we do not expect the block to spontaneously gain energy
and start moving again after it has stopped sliding.
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5) Working hard or hardly working?
In the block-and tackle arrangement shown in the figure, three segments
of the single rope pull on the block. m is the inertia of the block.
c
(𝑎) Show that the magnitude Fpr
of the force exerted by a person on the
rope to raise the block at constant speed is 𝑚𝑔/3.
(𝑏) One worker uses this arrangement to lift a heavy block, while another
hauls up an identical block with a straight rope. After both blocks have
been raised to the same second-floor window, the workers argue about
whether they have done the same amount of work.
AP50 Fall 2016
AP50 Fall 2016