Download Physics 18 Spring 2011 Homework 4

Physics 18 Spring 2011
Homework 4 - Solutions
Wednesday February 9, 2011
Make sure your name is on your homework, and please box your final answer. Because
we will be giving partial credit, be sure to attempt all the problems, even if you don’t
finish them. The homework is due at the beginning of class on Wednesday, February
16th. Because the solutions will be posted immediately after class, no late homeworks can
be accepted! You are welcome to ask questions during the discussion session or during office
hours.
1. A particle of mass m moving along the x axis experiences the net force Fx = ct, where
c is a constant. The particle has a velocity v0x at time t = 0. Find an algebraic
expression for the particles velocity vx at a later time t.
————————————————————————————————————
Solution
Newton tells us that the force is mass times acceleration, Fx = max . So, the acceleract
. To get the velocity we just need to integrate the acceleration,
tion is ax = Fx /m = m
since ax = v̇x . Doing so gives
ct2
vx =
,
2m
where we have set v0x = 0 as given in the initial conditions.
1
2. Estimate the widest stance you can take when standing on a dry, icy surface. That is,
how wide can you safely place your feet on not slip into an undesired “split?” Let the
coefficient of static friction of rubber on ice be roughly 0.25.
————————————————————————————————————
Solution
Consider the forces acting on your foot, when your feet are
spread apart at an angle θ, as shown in the figure to the right.
The frictional force, F~f , preventing you from sliding points to
the right, the normal force F~N points vertically, while the force of
your weight, F~W , points at an angle, since your leg isn’t straight
up and down. Note that FW 6= mg, in general, since your
weight is spread over two feet. In particular, it would likely
be FW = mg/2, but we’ll just leave it as FW for now. Let’s
work out the components,
P
P Fx = −FW sin θ + Ff = 0 ⇒ Ff = FW sin θ
Fy = −FW cos θ + FN = 0 ⇒ FN = FW cos θ.
FN
Ff
Dividing Ff /FW gives
Ff
= tan θ.
FN
θ
FW
Notice that the weight cancels out - the angle doesn’t depend
on how much you weigh!
Now, the frictional force is given by Ff = µs FN , in the static case, where µs is the
coefficient of static friction. So, Ff /FN = µs . Thus,
µs = tan θ ⇒ θ = tan−1 (µs ).
For the ice and shoe combination, µs = 0.25, and so
θ = tan−1 (µs ) = tan−1 (0.25) = 14◦ ,
which corresponds to an angle of 28◦ between your legs.
2
3. A block of mass m is pulled at a constant
velocity across a horizontal surface by
a string as shown in the figure. The
magnitude of the frictional force is (a)
µk mg, (b) T cos θ, (c) µk (T − mg), (d)
µk T sin θ, or (e) µk (mg − T sin θ).
————————————————————————————————————
Solution
The block is being pulled at a constant velocity, so the acceleration is zero. This
means that all the forces balance. Since the block isn’t being pulled off the ground, the
gravitational force is balanced by the vertical component of the tension in the string.
The friction is fighting the motion to the left, providing a force acting to the right.
The only force acting to the left is the horizontal component of the tension, −T cos θ.
The frictional force has to cancel this, so Ff = T cos θ. So, answer (b) is correct.
3
4. A 12 kg turtle rests on the bed of a zookeeper’s truck, which is traveling down a country
road at 55 mi/h. The zookeeper spots a deer in the road, and slows to a stop in 12
s. Assuming constant acceleration, what is the minimum coefficient of static friction
between the turtle and the truck bed surface that is needed to prevent the turtle from
sliding?
————————————————————————————————————
Solution
If the deceleration of the truck is too big,
the turtle will start sliding. Let’s look at
the different forces on the turtle.
P
−Ff
= −max
P Fx =
Fy = −Fg + FN =
0,
y
FN
where the minus sign on the acceleration
comes from the fact that the truck is decelerating. So, these say that Fg = FN = mg,
and that the frictional force is Ff = max .
Now, the frictional force is Ff = µs FN =
µs mg, where µs is the coefficient of sliding
friction. Thus,
µs =
ax
.
g
x
Ff
So, we just need to determine the acceleration of the truck. The truck starts at a
velocity of 55 mph, which is the same as
vi = 55 miles/hr × 1609 meters/mile × 1
hr/3600 s = 25 m/s. The final velocity is
zero, since the truck stops. If it takes a
time t (which is 12 seconds) to stop, then
the acceleration is ax = v/t. Thus,
µs =
Fg
ax
v
25
=
=
= 0.21
g
gt
9.8 × 12
So, we need a coefficient of at least 0.21 to keep the turtle from sliding.
4
release point of 45 cm. What is the a
maximum height the pig turns aroun
fast is it moving when it arrives at th
5. You and your friends push a 75.0 kg greased pig up an aluminum slide at the county
fair, starting from the low end of the slide. The coefficient of kinetic friction between
the pig and slide is 0.070.
Picture the Problem The free-body
sometime after you and your friends
momentarily stopped moving up th
equations and Newton’s 2 law to
(b) At the maximum height the pig turns around and begins to slip down the slide,
how fast is it moving when itthe
arrivespig’s
at the low
end of thewhen
slide?
speed
it returns to bo
————————————————————————————————————
assigns a coordinate system and var
Solution in solving this problem.
consider
(a) All of you pushing together (parallel to the incline) manage to accelerate the pig
from rest at the constant rate of 5.0 m/s2 over a distance of 1.5 m, at which point
you release the pig. The pig continues up the slide, reaching a maximum vertical
height above its release point of 45 cm. What is the angle of inclination of the
nd
slide?
(a) Let’s look at the forces acting on the pig.
P
P Fx = −Fk − Fg sin θ = max
Fy = −Fg cos θ + FN = 0.
y
Solving the second gives FN = Fg cos θ =
mg cos θ. Now, the frictional force is Fk =
µk FN = µk mg cos θ. Plugging this in to
the first gives, after canceling off the mass,
r
fk
−g (µk cos θ + sin θ) = ax .
This is the acceleration the pig feels as it
moves up the ramp. How far along the
ramp does he move? From the constant acceleration equation, vf2 = vi2 + 2a∆x, then
if he stops moving at the end of the ramp,
v2
vf = 0, and so d = − 2ai .
Plugging in for a gives
vi2
d=
.
2g (µk cos θ + sin θ)
r
Fn
x
r
Fg θ
(a) Apply ∑ F = ma to the pig:
Finally, if the distance along the ramp is d, then the height is h = d sin θ. Thus,
the height the pig slides up is
h=
vi2 sin θ
vi2
=
,
2g (µk cos θ + sin θ)
2g (1 + µk cot θ)
after dividing through by sin θ. Solving for the angle gives
µ
f = μ k Fn and Fg = mg in
Substitute
. k
tan θ = −1
equation (1) to obtain:
k
vi2
2gh
5
Solving equation (2) for Fn yields:
Now, we just need the initial velocity. Since the pig is pushed with an acceleration
√
of 5 m/s2 , then after a distance of 1.5 m, he’s traveling at a speed of vf = 2ad =
√
2 × 5 × 1.5 = 3.9 m/s. So, we have everything that we need,


#
"
µ
0.070
k
= tan−1 (0.01) = 5.7◦ .
 = tan−1
θ = tan−1  2
15
vi
−
1
−1
2×9.8×.45
2gh
(b) Now the pig is sliding back down the ramp. In this case the frictional force opposes
the slide down and changes his acceleration to
a = −g (sin θ − µk cos θ) ,
and is constant. So, since the pig starts from rest at the top, and is moving at a
speed vf at the bottom, if he moves a distance −∆x, then,
p
p
vf = 2a × (−∆x) = 2g (sin θ − µk cos θ) ∆x.
What’s ∆x? This is the total distance than the pig slid up, d = h/ sin θ, plus the
extra 1.5 meters that he started off sliding down (call that distance d0 ). The total
distance is ∆x = sinh θ + d0 . Putting everything together gives
s
p
vf = 2g (sin θ − µk cos θ) ∆x =
h
2g (sin θ − µk cos θ) d0 +
.
sin θ
Plugging in all the numbers gives
q
2g (sin θ − µk cos θ) d0 + sinh θ
vf =
q
=
2 × 9.8 (sin 5.7◦ − 0.07 × cos 5.7◦ ) 1.5 +
=
1.87 m/s.
6
.45
sin 5.7◦
6. A pilot with a mass of 50 kg comes out of a vertical dive in a circular arc such that at
the bottom of the arc her upward acceleration is 3.5g.
454 Chapter 5
(a) How does the magnitude of the force exerted by the airplane seat on the pilot at
the bottom of the arc compare to her weight?
78 •• (b)AUse
pilot
with a mass of 50 kg comes out of a vertical dive in a circular
Newton’s laws of motion to explain why the pilot might be subject to a
arc such thatblackout.
at the bottom
of that
the an
arcabove
hernormal
upward
acceleration
is 3.5
g. (lower
a) How
This means
volume
of blood “pools”
in her
limbs. How
an inertial
reference
observer describe
cause
of the
does the magnitude
ofwould
the force
exerted
by frame
the airplane
seat onthethe
pilot
at the
bottom of theblood
arc pooling?
compare to her weight? (b) Use Newton’s laws of motion to
————————————————————————————————————
explain why
the pilot might be subject to a blackout. This means that an above
normal volume of blood ″pools″ in herSolution
lower limbs. How would an inertial
reference frame observer describe the cause of the blood pooling?
(a) We can draw a free-body diagram for the
pilot, which is seen to the right. Applying
Picture theNewton’s
Problem
The
diagram
laws to
the system
gives
shows the forces X
acting on the pilot
Fy = FN − Fg = macent ,
when her plane is at the lowest point
the centripetal acceleration.
Fnacent
is isthat
the
force the
of its dive.where
The weight
the pilot feels comes from
airplane seatthe exerts
ontheher.
We’ll
strength of
normal
force from the
nd
seat pushing
backfor
up on
her. Solving for
apply Newton’s
2 law
circular
the normal force gives
motion to determine Fn and the
= m (gfollowed
+ acent ) .
radius of the circularFN path
by the airplane.
The ratio of the normal force to her normal
r
r
Fn
r
r
Fg = m g
weight mg is
(a) Apply
pilot:
Because
acent
g + acent
FN
=1+
.
=
Fg
g
g
Since
the=centripetal
acceleration
at theFbottom
of her arc is a n ==m
3.5g,
g +then
ac
Fradial
maradial to
the
n − mg = mac ⇒ Fcent
the ratio is 1 + 3.5g/g = 4.5. So, she feels 4.5 times heavier than usual.
(
∑
)
(b) The pilot might black out because the blood can’t flow up to her brain. An inertial
observer watching from outside the plane doesn’t see any additional “centrifugal”
force pushing the blood down to her feet. Instead, they see the blood trying to
Fn = marc.g Because
+ 3.5 g the= pilot
4.5mg
ac =continue
3.5 g :along it’s path tangent to the circular
pulls up in
the arc, her feet travel up, while the blood continues along horizontally; so, the
blood seems to be “pulled” down towards her feet.
The ratio of Fn to her weight is:
(
)
Fn 4.5mg
=
= 4.5
mg
mg
(b) An observer in an inertial reference frame would see the pilot’s blood continue
to flow in a straight line tangent to the 7circle at the lowest point of the arc. The
pilot accelerates upward away from this lowest point and therefore it appears,
7. A small bead with a mass of 100 g slides
without friction along a semicircular wire
with a radius of 10 cm that rotates about
a vertical axis at a rate of 2.0 revolutions
per second. Find the value of θ for which
the bead will remain stationary relative to
the rotating wire.
————————————————————————————————————
464 Chapter 5
Solution
Picture the Problem The semicircular
Let’s wire
start of
by radius
looking10atcm
thelimits
forces the
on the
bead,
motion
as we’ve done before. The normal force points
of the bead in the same manner as
perpendicularly to the wire, as seen in the figure.
would
a 10-cm
string around
attached
to the
Because
the bead
is spinning
in the
horifixed
at the center
the x
zontalbead
plane,and
there
is a centripetal
forceof
in the
direction.
So,
semicircle.
The horizontal component
ofP
the normal force the wire exerts
on
mv 2
F
=
−F
sin
θ
=
x
N
r
P
the bead
centripetal force.
Fy =is Fthe
0. The
N cos θ − mg =
application of Newton’s 2nd law, the
mg
.
Solving
the bottom
FN =
cos θ
definition
of theequation
speed ofgives
the bead
in its
Plugging this into the first gives for the angle
orbit, and the relationship of the
frequency of a circular
v 2 motion to its
tan θ = .
period will yield the rg
angle at which the
will know
remain
relativewe
to do know the period, T = 2πr/v. So,
Now, bead
we don’t
thestationary
speed. However,
v = 2πr/T
. We are
told the frequency of rotations, f = T −1 . And so, plugging this
the rotating
wire.
back into the expression for the angle gives
Apply ∑ F = ma to the bead:
2
v
4π 2 r
4π 2 f 2 r
F.x = Fn sin θ = m
tan θ =
=
gT 2
g
r
∑
Finally, the radius of the ring is L, but the beadand
rotates at a distance r. From the
Fy = Fgives
mgθ ==04π 2 f 2 L/g,
geometry, we see that r = L sin θ. Making this substitution,
1/−cos
n cosθ
after canceling off the mutual sine term. So, finally solving for the angle gives
2 Eliminate Fn from
the
force
g
9.8 v
−1
−1
tan
θ
=
θ = cos
= cos
= 52◦ .
equations to obtain: 4π 2 f 2 L
4π 2 × 22 ×rg
0.1
∑
The frequency of the motion is the
8
reciprocal of its period T. Express
the speed of the bead as a function of
v=
2πr
T
8. In 1976, Gerard O’Neill proposed that large space stations be built for human habitation in orbit around Earth and the moon. Because prolonged free-fall has adverse
medical effects, he proposed making the stations in the form of long cylinders and
spinning them around the cylinder axis to provide the inhabitants with the sensation
of gravity. One such O’Neill colony is to be built 5.0 miles long, with a diameter of
0.60 mi. A worker on the inside of the colony would experience a sense of “gravity”
because he would be in an accelerated reference frame due to the rotation.
(a) Show that the “acceleration of gravity” experience by the worker in the O’Neill
colony is equal to his centripetal acceleration.
(b) If we assume that the space station is composed of several decks that are at
varying distances (radii) from the axis of rotation, show that the “acceleration of
gravity” becomes weaker the closer the worker gets to the axis.
(c) How many revolutions per minute would this space station have to make to give
an “acceleration of gravity” of 9.8 m/s2 at the outermost edge of the station?
————————————————————————————————————
Solution
(a) An astronaut standing on the “ground” in the spinning station will experience
two forces - one is the centripetal force which comes from spinning around in a
circle, and the other is the normal force from the ground pushing back up on his
feet. As long as he’s not accelerating (which is the case if he’s standing on the
ground), then these two forces will cancel. Since the normal force is what we feel
as our weight, then the centripetal force is what he’ll feel as gravity.
(b) The acceleration of “gravity” is given by the centripetal acceleration, as we’ve
2
discussed. So, we can write g = vr . The different decks are traveling at different
speeds (decks at bigger radii have to go faster to go around in the same amount of
time), but they all have the same orbital period, T . The period is just T = 2πr/v,
and so T = 2πr/v. Thus, the acceleration is
2
4π
v2
=
r.
g=
r
T2
So, the acceleration gets bigger as we go further out.
(c) The number of revolutions p
per second, f = 1/T , and so g = (2πf )2 r. Solving for
1
the frequency gives f = 2π gr . The diameter is given in miles. Recalling that 1
mile = 1609 meters, we can solve for the frequency.
r
r
1 g
1
9.8
f=
=
= 0.023 rev/sec.
2π r
2π .3 × 1609
Multiplying this by 60 gives us f = 1.36 revolutions per minute, which seems
fairly slow (but it is a big station).
9
9. The position of a particle of mass m = 0.80 kg as a function of time is given by
~r = xî + y ĵ = (R sin ωt) î + (R cos ωt) ĵ, where R = 4.0 m and ω = 2πs−1 .
(a) Show that the path of this particle is a circle of radius R, with its center at the
origin of the xy plane.
(b) Compute the velocity vector. Show that vx /vy = −y/x.
(c) Compute the acceleration vector and show that it is directed toward the origin
and has the magnitude v 2 /R.
(d) Find the magnitude and direction of the net force acting on the particle.
————————————————————————————————————
Solution
p
(a) The magnitude of the displacement is r = x2 + y 2 . Plugging in the components
gives
p
p
r = x2 + y 2 = R2 cos2 ωt + R2 sin2 ωt = R.
So, the magnitude of the displacement doesn’t change with time - the particle is
always at a distance r. In other words, it’s moving in a circle of radius R. Because
x and y both have a range from ±R, the circle is centered at the origin - if it was
centered at some other point, say at x = 4, then the range of x would be from
4 + R to 4 − R. Since this isn’t the case, the circle is centered at the origin.
(b) The velocity vector is ~v = ~r˙ , so we have to compute the time-derivative of the
position vector. Doing so gives
~r˙ = (Rω cos ωt) î − (Rω sin ωt) ĵ.
Computing vy /vx = −(Rω sin ωt)/(Rω cos ωt) = −R sin ωt/R cos ωt = −y/x.
(c) The acceleration vector is ~a = ~r¨ = ~v˙ , and so we just need to take one more
derivative of the velocity. This gives
2
2
~a = − (Rω
h cos ωt) î − (Rω sin ωt) iĵ
= −ω 2 (R cos ωt) î + (R sin ωt) ĵ
=
−ω 2~r.
So, the acceleration vector points back along the displacement vector, from the
point, back to the origin. The magnitude of ~a = a = ω 2 R. But, v = ωR, and so
2
v2
v
2
R
=
,
a=ω R=
R2
R
(d) The net force acting on the particle is just F~ = m~a, and so
h
i
F~ = m~a = −mω 2 (R cos ωt) î + (R sin ωt) ĵ ,
which points along the acceleration (i.e., back towards the origin) with a magnitude F = ma = mRω 2 = .8 × 4 × (2π)2 = 126 N.
10
10. After a parachutist jumps from an airplane (but before he pulls the rip cord to open his
parachute), a downward speed of up to 180 km/h can be reached. When the parachute
is finally opened, the drag force is increased by about a factor of 10, and this can create
a large jolt on the jumper. Suppose this jumper falls at 180 km/h before opening his
chute.
(a) Determine the parachutist’s acceleration when the chute is just opened, assuming
his mass is 60 kg.
(b) If rapid accelerations greater than 5.0g can harm the structure of the human body,
is this a safe practice?
————————————————————————————————————
Solution
Applications of Newton’s Laws 521
(a) From Newton’s laws, the sum of the forces
is his net acceleration. So, after his chute
opened, The free-body
Picture the Problem
diagram showsXthe
Fy = drag
Fd − Fgforce
= 10bv 2 F
−dmg = maopen ,
y
r
Fd = 10bv 2 ˆj
exerted by the air and the
So
10bv 2
gravitational force Fg exerted
− g.
aopen = by the
m
Earth acting on
onhethe
parachutist
Before
opened
the chute the drag force
just after his chute
was onlyhas
Fd =opened.
bv 2 . So, We
X
can apply Newton’s
2nd law to2 the
Fy = Fd − Fg = bv − mg = maclosed .
parachutist to obtain
an expression
If the parachutist
was traveling
for his acceleration
as a function
of at termivelocity,evaluate
vT , before he
pulled the cord,
his speed andnal then
this
then his acceleration aclosed = 0, and so
expression for bv==mg/v
v t . T2 .
(a) Apply
r
Fg = −mgˆj
Plugging this back into the open acceleration equation gives
!
2
2
2
∑ Fy = ma y to the a = 10bv −10
bv − mgv= ma
g = g 10
− chute
1 .
open
m
vT
open
parachutist immediately after the
chute opens: So, his acceleration depends
Solving
on how fast he was traveling before he pulled the
chute, which makes sense. If he was traveling at terminal velocity, vT , then
!
2
v
T
b 1) 2= 9g.
for achute open yields: aopen = g 10
− 1 = g(10 −
(1)
vTa chute open = 10 v − g
m
So, he’s subjected to an acceleration of 9g.
(b) This
is 4g greater than what is needed2 to harm the body.
Before the chute
opened:
bv − mg = ma y
So, this is not a very
safe practice (for a variety of reasons!).
Under terminal speed conditions,
ay = 0 and:
11bv 2
t
− mg = 0 ⇒
b
g
= 2
m vt