Download solutions to Midterm 2 - University of California, Berkeley

1
University of California, Berkeley
Physics 7A, Section 1, Spring 2009 (Yury Kolomensky)
SOLUTIONS TO THE SECOND MIDTERM
Maximum score: 100 points
1. (10 points) Blitz
This is a set of simple questions to warm you up. The
problem consists of five questions, 2 points each.
1. Circle correct answer. Imagine a basketball
bouncing on a court. After a dozen or so
bounces, the ball stops. From this, you can conclude that
(a) Energy is not conserved.
(b) Mechanical energy is conserved, but only
for a single bounce.
(c) Total energy in conserved, but mechanical
energy is transformed into other types of
energy (e.g. heat).
(d) Potential energy is transformed into kinetic
energy
(e) None of the above.
Answer: (c)
2. Circle correct answer. In an elastic collision
(a) hoop, disk, sphere
(b) disk, hoop, sphere
(c) sphere, hoop, disk
(d) sphere, disk, hoop
(e) hoop, sphere, disk
Answer: (d). The object with highest moment of
inertia has the largest kinetic energy at the bottom, and would reach highest elevation. The object with the smallest moment of inertia (sphere)
will therefore reach lowest altitude, and the object with the largest moment of inertia (hoop)
will be highest.
4. A particle attached to the end of a massless rod
of length R is rotating counter-clockwise in the
horizontal plane with a constant angular velocity
ω as shown in the picture below.
ω
(a) momentum is not conserved but kinetic energy is conserved;
(b) momentum is conserved but kinetic energy
is not conserved;
R
(c) both momentum and kinetic energy are
conserved;
(d) neither kinetic energy nor momentum is
conserved;
(e) none of the above.
Answer: (c)
3. Circle correct answer. A hoop, a uniform disk,
and a uniform sphere, all with the same mass
and outer radius, start with the same speed and
roll without sliding up identical inclines. Rank
the objects according to how high they go, least
to greatest.
Which direction does the vector ~ω point ?
(a) Tangential to the circle, in the counterclockwise direction
(b) Along the radius, towards the center of the
circle
(c) Along the radius, away from the center of
the circle
(d) Normal to the page, towards you
(e) Normal to the page, away from you
2
Answer: (d), according to the right-hand rule
5. Clean-and-jerk is an Olympic weightlifting
event. In the first phase (clean), a competitor
stands up with a heavy barbell from the squat
position, lifting the bar by about 1 m in less than
1 second. The Olympic record for this event is
263.5 kg (580.3 pounds), set by Hossein Reza
Zadeh of Iran in 2004 Olympics in Athens. Estimate (to 1 digit) the power delivered to the
barbell by Zadeh during this lift in horsepower
(1 hp = 750 W).
Answer: 3 hp. 4 hp would also be an allowed
answer (e.g., if you used g = 10 m/s2 , or accounted for the elevation of the weightlifter’s
body).
2. (15 points) Grapevine Grade
Grapevine grade is a famous place on interstate I-5 between Bakersfield and Los Angeles, where a 5 mile (8
km) stretch of the road has an average incline of 6%
(3.4◦ ). Imagine a heavy duty pickup truck towing a
trailer up Grapevine Grade at the constant speed of 65
mph (29 m/s). At this speed, the air drag force acting on
the truck is about Fdrag = 1.5 kN. The total mass of the
truck and the trailer is m = 10 tons (104 kg).
(a) (10 points) Estimate the power output of the
truck’s engine (assume that all power goes into
locomotive motion of the truck). Express you answer in Horsepower (1 hp = 750 W)
(b) (5 points) The modern diesel engines are about
50% efficient, i.e. half of the energy from burning diesel fuel goes to locomotive motion, and
half goes to waste. Burning one gallon of diesel
fuel produces about 150 MJ of energy. How
many gallons of diesel is spent on the 8 km trip
up Grapevine Grade ?
2. Solution
(a) There are four forces acting on the truck and the
trailer: weight mg, directed downward, normal force
from the surface, air drag (pointing along the surface
in the direction opposite to the velocity of the truck),
and the propulsion force Fpr . The propulsion force is
the force of static friction between the truck’s tires and
ground, and it prevents the wheels from slipping. Since
the velocity of the truck is constant, the forces are balanced. The balance of forces along the slope leads to
the equation:
Fdrag + mg sin θ = Fpr = 7.3 kN
The power provided by the force Fpr (which ultimately
comes from the truck’s drivetrain) is
P = Fpr v = 2.1 · 105 J = 280 hp
(b) The useful work done by the propulsion force is
W = Fpr L = 58 MJ
One gallon of diesel fuel produces 0.5∗150 MJ = 75 MJ
of useful work. So the entire trip would use up 60/75 =
0.8 gal of diesel.
3. (20 points) Dangerous Demo
You may remember this demonstration from earlier in the semester. A 200 g wooden block is placed on top of a
muzzle of a 0.22” rifle, mounted vertically next to a meter stick (see picture below). The rifle fires a 3 g bullet into
the block. The bullet stops inside the block, and we observe that the block (with the bullet inside) lifts up to the
elevation h = 240 cm above the muzzle (this is not what happened in class as we used a low-velocity lighter bullet).
This is a convenient way to measure the muzzle velocity (velocity with which it exits the rifle) of the bullet.
3
(a) (10 points) What velocity does the bullet have just before it
hits the block ? That is the muzzle velocity.
(b) (10 points) Obviously, the rifles are not normally used this
way. Usually, you would hold the rifle in your hands, press
the butt end against your shoulder, and fire. When the bullet is fired, you feel the rifle recoil, i.e. the butt of the rifle
strike you in the shoulder. Why does the rifle recoil ? What
average force does the rifle exert on the shoulder while the
bullet accelerates ? You will need to know the length of the
rifle barrel (68.5 cm).
3. Solution
(a) This part should be familiar, since it has been
worked out in class when we discussed inelastic collisions.
First, momentum conservation relates the velocity v
of the bullet just before the collision to the velocity V
of the block-bullet combination just after the collision:
pbefore = mv = pafter = (m + M )V
where m = 3 g is the mass of the bullet, and M = 200 g
is the mass of the block. The collision is fully inelastic:
the bullet is embedded in the block. So if we know the
velocity V of the block after the collision, we can find
out the velocity of the bullet:
m+M
V
(1)
v=
m
Since we know the elevation of the block, we can relate it to the velocity V using the conservation of energy.
At the muzzle, the potential energy of the block-bullet
combo is zero, and the kinetic energy is
m+M 2
V
Kinitial =
2
The block rises to the elevation h = 2.4 m, where its
kinetic energy is zero, and its potential energy is
Ufinal = (m + M )gh
Energy conservation requires
Uinitial + Kinitial = Ufinal + Kfinal
Plugging in the values, we can find V :
V =
p
2gh
(which is the standard answer from kinematics as well),
and plug it into Eq. (1):
v=
m+Mp
2gh = 464 m/s .
m
(b) The recoil of the rifle is another demonstration of
momentum conservation. Initially, the rifle and the bullet are at rest, and their total momentum is zero. After the bullet is fired and before it and the rifle have
a chance to interact with the surrounding bodies (e.g.
your shoulder), the momentum of the system has to remain the same. So if the bullet is traveling fast away
from you, the rifle has to recoil towards you, to keep the
total momentum zero. Ignoring the powder gas (which
escapes through the holes on the side of the muzzle and
carries no forward momentum – I did not tell you about
it in the problem, and it’s safe to ignore this), the momentum of the bullet is equal in magnitude and opposite
in direction to the momentum the rifle picks up.
4
You can also think about it in terms of the 3rd Newton’s law. Since the bullet is accelerated, there must be a
force applied to it. The source of this force, ultimately,
is the rifle, or more precisely, the gun powder gas that
pushes against the rifle and accelerates the bullet. By
3rd Newton’s law, the force that accelerates the bullet
has to have a pair – the force acting on the rifle in the
opposite direction.
To compute the force acting on the shoulder, we will
put a 3rd Newton’s law to good use. The force that accelerates the bullet is equal and opposite to the force
that is applied to the rifle and, through the buttstock,
transfered to the shoulder. So if we find the force that
accelerates the bullet, we get the force acting on the rifle and the shoulder.
The easiest way to find the average force is to invoke
energy conservation (recall problem #6 on the Problem
of the Week page). The kinetic energy the bullet picks
up is
K=
mv 2
2
and it must be equal to the work done by external forces
– which provide acceleration:
K = Wext = hF il
where l = 68.5 cm is the distance over which the bullet
is accelerated, i.e. the length of the barrel. The average
force is hence
hF i =
mv 2
= 470 N
2l
∆p = mv
To find the time interval ∆t, we can use kinematics
equations for the bullet:
∆t2
2
v = hai∆t
l = hai
where hai is the average acceleration. These can be
solved to express
∆t =
hF i =
mv
mv 2
=
= 470 N
∆t
2l
which is the same result.
4. (30 points) Daredevil Gymnast
A gymnast of mass m = 60 kg is rotating around a
horizontal bar, holding on to it with his arms (this skill
is called the giants). When his body is vertical and is
below the bar, his angular velocity is ω0 = 7 sec−1 .
In the following, treat the gymnast’s body as a uniform
solid rod of length L = 2 m (arms to toes), rotating
around its end. The moment of inertia of such a rod is
I = mL2 /3. Ignore friction and air resistance.
∆p = hF i∆t
where ∆p is the change in momentum picked up by the
rifle (it is equal and opposite to the momentum picked
2l
v
Then, the average force is
)
This is a significant force (a bit over 100 lbs !) The rifle recoil can be quite tiring after a few rounds, especially for higher power weapons. Have you noticed that
in (realistic) movies, a sharpshooter never puts his eye
right up against the scope of a rifle (like you would be
inclined to do looking through binoculars) ? He would
not want to get a black eye, or, worse yet, end up with a
broken eye socket !
The other way to find the force is to use the impulse
theorem:
up by the bullet, if we ignore the powder gases), and
∆t is the time it takes for the bullet to fully accelerate
(i.e. leave the barrel). The change in momentum is (by
magnitude)
(a)
5
which can be solved for N :
)
Nbot = m(g + acp,bot ) = m g +
)
)
90o
(c)
(a) (5 points) What force does the gymnast apply to
the bar in the bottom position ? Is he pushing
himself away from the bar, or pulling toward it ?
(b) (10 points) What force does the gymnast apply to
the bar in the top position ? Is he pushing himself
away from the bar, or pulling toward it ?
(c) (15 points) Suppose that after a few giants
around the bar the gymnast releases the bar when
he is at 90◦ with the vertical. His center of mass
continues moving vertically while he performs a
layout somersault by rotating around his center
of mass in a prone position (see picture). How
high over the bar would his center of mass be at
the highest point ? How many revolutions would
the gymnast complete before falling back to the
bar elevation ?
4. Solution
(a) The acceleration of the center of mass of the gymnast rotating with angular velocity ωbot = 7 rad/sec is
2
acp,bot = ωbot
Rcm
where Rcm = L/2 is the distance between the center of
mass of the gymnast (approximated to be a uniformly
dense solid rod) and the axis of rotation. There are two
forces acting on the gymnast: his weight mg, directed
downward, and the normal force of the bar Nbot . We
will assume that it is directed toward the bar, i.e. upward. The 2nd Newton’s law relates these forces to the
centripetal acceleration:
Nbot − mg = macp,bot
2
= 3.5 kN
(6 times his weight !) By 3rd Newton’s law, this is
the force the gymnast is applying to the bar, and he is
clearly pulling himself toward the bar to stay on a circular orbit.
(b) This is a two-part question. First, assuming that the
gymnast has angular velocity ωtop 6= ωbot at the top of
the swing, the normal force of the bar Ntop and, consequently, the force which which the gymnast pulls the
bar, can be found from the 2nd Newton’s law:
)
(b)
2 L
ωbot
(2)
Ntop + mg = macp,top
(3)
2 L/2 is the centripetal acceleration
where acp,top = ωtop
of the center of mass in the top position (directed downward). Notice the difference in sign of the normal force
with respect to Eq. (2): we assume that both the weight
of the gymnast and the normal force are directed downward. We will be able to test this assumption: if the normal force happens to be directed upward, it will come
out negative when we solve Eq. (3).
Now we need to find ωtop . Since there is no friction
involved, the total mechanical energy of the gymnast is
conserved. On the bottom of the swing (when the gymnast is vertical and is below the bar), his kinetic energy
is
Kbot =
2
Iωbot
2
and his potential energy (taking the bar elevation to be
zero) is
Ubot = −mg
L
2
When the gymnast is at his highest point over the bar,
his potential energy is
Utop = +mg
L
2
and his kinetic energy is
Ktop =
2
Iωtop
2
Energy conservation then implies the equation
2
2
Iωtop
Iωbot
L
L
− mg =
+ mg
2
2
2
2
(4)
6
Now we can plug ωside , I = mL2 /3, and I0 = mL2 /12
into Eq. (5) and solve for H:
which can be solved for ωtop :
ωtop =
s
2
ωbot
2mgL
=
−
I
r
2 −
ωbot
6g
= 4.4 sec−1
L
We can now plug that into Eq. (3) and solve for Ntop :
Ntop
2 L
− 4g = 540 N
= m(acp,top − g) = m ωbot
2
So indeed, the gymnast is pulling himself toward the
bar, with the force that is slightly less than his own
weight.
(c) This is another application of the energy conservation. As the gymnast lets go of the bar, the velocity of
his center of mass is directed upward. Since the only
force acting on him now is gravity, the center of mass
will continue moving upward to some elevation H. At
the same time, his body will be rotating around its center of mass. Since gravity exerts no torque relative to
the center of mass, the angular velocity relative to the
center of mass remains constant and equal to what it
2 . At the top of
was when the gymnast let go, i.e., ωside
the upward trajectory (elevation H above the bar), the
gymnast has potential energy
U (H) = mgH
and kinetic energy of
K(H) =
2
I0 ωside
2
where I0 = I − mL2 /4 = mL2 /12 (according to the
parallel axis theorem). Equaling it to the total mechan2 /2, we get an equaical energy Eside = Kside = Iωside
tion for H:
2
2
Iωside
I0 ωside
= mgH +
2
2
2
ωside
H =
(I − I0 )
2mg
2
− mg
L
=
2
ωside =
2
ωbot
mgL
=
−
I
That’s quite impressive !
Finally, to find the number of revolutions, we need
to figure out the amount of time it takes the gymnast to
fly up to elevation H and come back down. The motion
of center of mass is a free fall. Therefore, it takes
tfall =
s
2H
= 0.6 s
g
to fall from elevation H above the bar to the bar level,
and therefore the total flight time.is
ttotal = 2tfall = 1.2 s
The number of revolutions is
N=
ωside ttotal
=1
2π
So the gymnast makes (almost) exactly one revolution.
Indeed, if he adds a half-twist, he would be facing the
bar again and would be able to catch it. That is one of a
gymnastics release moves.
5. (25 points) Another Lecture Demo
This was one of the most recent lecture demonstrations.
A spool consists of an inner hollow cylinder of radius
r = 5 cm and mass mc = 4 kg and two solid disks of
radius R = 10 cm and mass md = 1 kg each, attached
coaxially to the ends of the cylinder. A rope is wound
tightly around the inner cylinder, and a force F = 16 N
is applied to the end of the rope in the horizontal direction, as shown on the picture below. The spool rolls
without slipping on a horizontal table under the influence of this force.
F
r
R
2
Iωside
2
(a) (5 points) Which direction does the spool roll ?
which is solved for
s
2 L2
ωside
= 1.7 m
8mg
(5)
We still need to find ωside , and that is straightforwardly
done using energy conservation, similar to Eq. (4):
2
Iωbot
H=
r
2 −
ωbot
3g
= 5.9 sec−1
L
(b) (10 points) Find acceleration of the center of
mass of the spool, and its angular acceleration.
7
(c) (10 points) Find the smallest coefficient of static
friction between the spool and the table for the
roll to proceed without slipping.
5. Solution
(a) We worked out this problem in some detail in lecture, though there is one important difference in the
setup above: the rope goes over the top of the spool.
There are two forces acting on the spool: F, which
points to the right, and the friction force Ff r , which
points to the left. Both produce torque that would rotate the spool in the clockwise direction. Therefore, the
spool would rotate in the clockwise direction, and roll
to the right, in the direction of F.
(b) Now we can be numeric. We start with the 2nd Newton’s law:
F + Ff r + M g + N = M a
where a is the linear acceleration of the center of mass
(points to the right), N is the normal force, and M =
mc + 2md = 6 kg is the total mass of the spool. We can
project it onto the horizontal X and vertical Y axes:
M a = F − Ff r (X)
a = =
α =
(6)
0 = N − M g (Y )
(7)
Third equation comes from the balance of torques:
Iα = F r + Ff r R
(8)
where α = a/R is the angular acceleration, and I is the
moment of inertia of the spool. Eq. (6) and Eq. (8) can
be solved for a:
a=
F 1 + r/R
M 1 + I/M R2
(9)
The final step is to find the moment of inertia I. The
spool consists of a hollow cylinder with moment of inertia Ic = mc r 2 and two solid disks, each with moments
of inertia Id = md R2 /2. Therefore, the total moment
of inertia is
I = Ic + 2Id = mc r 2 + md R2
Finally, the linear and angular accelerations are
F
1 + r/R
= 3 m/s2
2
[mc + 2md ] [1 + mc r /((mc + 2md )R2 ) + md /(mc + 2md )]
a
= 30 rad/s2
R
(10)
(11)
µs ≥
The sign of the friction force is very curious. What it
means is that the friction force actually points in the
direction opposite to what we assumed, i.e., it points
to the right ! That means that if the spool slips, it will
spin rather than slide. You can explicitly check that our
expression for acceleration is still valid. The condition
of no slipping is that the coefficient of friction should
indeed be positive and larger than
N = M g = 59 N
F − Ma
= M a − F = 0.03
µs ≥ Mg Mg
(c)
The spool does not slip as long as the friction force
Ff r has not reached its maximal value Ff r max = µs N ,
where N is the magnitude of the normal force. Hence,
we have the inequality for the coefficient of static friction:
Ff r
N
Given acceleration a from Eq. (11), Ff r and N can be
computed from Eq. (6) and Eq. (7):
Ff r = F − M a = −2 N