Download Physics 201 Homework

Physics 201 Homework 4
Jan 30, 2013
1. There is a clever kitchen gadget for drying lettuce leaves after you wash them.
It consists of a cylindrical container mounted so that it can be rotated about its
axis by turning a hand crank. The outer wall of the cylinder is perforated with
small holes. You put the wet leaves in the container and turn the crank to spin
off the water. The radius of the container is 0.12 meters. When the cylinder is
rotating at 2.0 revolutions per second, what is the magnitude of the centripetal
acceleration at the outer wall?
19 m/s2
Solution
When we are told the cylinder is rotating at 2.0 revolutions per second, we are
being told the period. This is because the period is the time it takes to make one
revolution. It is the reciprocal of the number we are given. The number of cycles
per second is the frequency of the motion. So, in symbols we have:
T = 1/f = 1/(2.0) = 0.50
Now that we have the period we can use the alternate formula for centripetal
acceleration, a = 4π 2 r/T 2 :
a=
4π 2 (0.12)
= 18.950
(0.5)2
In fact, we can combine our formulas in this problem and get a second alternative
formula for centripetal acceleration:
a = 4π 2 rf 2
But it is so simple to get the period directly, that this formula is not quite as
useful.
2. A jet flying at 123 m/s banks to make a horizontal circular turn. The radius of
the turn is 3810 meters, and the mass of the jet is 2.00 × 105 kilograms. Calculate
the magnitude of the necessary lifting force.
Solution
Notice that we are not told the angle involved. So, let’s tread a bit carefully here.
The forces involved are weight and lift. The motion is horizontal, so we should
align our coordinate system as such. The net force in the horizontal is nothing
but the horizontal component of the lift:
Fnet,x = Lx
and the net force in the vertical is
Fnet,y = Ly − W
Since the motion is in the horizontal, there is no net force in the vertical. Thus,
Ly − mg = 0
So,
Ly = (2.00 × 105 )(9.8) = 1.9600 × 106
But we don’t know the angle involved. Well, what about the horizontal motion?
We do know the horizontal net force is responsible for the centripetal acceleration,
F = mv 2 /r. Thus,
Lx = mv 2 /r = (2.00 × 105 )(123)2 /(3810) = 7.9417 × 105
1
2.11 × 106 meters
Now we know the two components of the lifting force:
Lx = 7.9417 × 105
Ly = 1.9600 × 106
q
The magnitude of this vector is given by mag =
~ L2x + L2y , so:
L=
p
(7.9417 × 105 )2 + (1.9600 × 106 )2 = 2.1148 × 106
3. Mars has a mass of 6.46 × 1023 kilograms and a radius of 3.39 × 106 meters.
(a) What is the acceleration due to gravity on Mars? (b) How much would a
65-kilogram person weigh on this planet?
(a) 3.75 m/s2 ; (b) 244 newtons
Solution
(a) This is an application of the Newton’s law of gravity, Fgrav = GM m/r2 , and
Newton’s second law of motion, Fnet = ma. When calculating the weight, we
have Fnet = Fgrav which allows us to combine these two.
ma = GM m/r2
Cancelling m gives us the general formula for the acceleration due to gravity.
a = GM/r2
Notice that when we plug in the mass and the radius of the earth, we get the
number we expect:
a = (6.672 × 10−11 )
5.974 × 1024
= 9.8016
(6.378 × 106 )2
For Mars, we get:
a = (6.672 × 10−11 )
6.46 × 1023
= 3.7517
(3.39 × 106 )2
Or, about 38% of that on earth.
(b) The weight on Mars is still the same formula, but different numbers:
W = mg = (65)(3.7515) = 243.86
4. A car travels at a constant speed around a circular track whose radius is
2600 meters. The car goes once around the track in 360 seconds. What is the
magnitude of the centripetal acceleration of the car?
Solution
The centripetal acceleration required to maintain uniform circular motion is given
by a = v 2 /r. The v in this equation is the speed around the circle. We are told the
period is 360 seconds, but not the speed. The missing piece is the distance. But
the distance is simply the circumference of the circle which is given by C = 2πr:
C = 2π(2600) = 16336
This is the distance covered in one cycle. Divide by the time to get the average
speed:
(16336)
d
= 45.379
v= =
t
(360)
Finally, we can plug this into the formula for centripetal acceleration:
a=
v2
(45.379)2
=
= 0.79200
r
(2600)
2
0.79 m/s2
A different (but similar) way to do this is to use an alternate formula for centripetal acceleration:
a = 4π 2 r/T 2
This formula is the combination of v = 2πr/T and a = v 2 /r, so using it is
essentially the same as the previous solution:
a=
(4π 2 )(2600)
= 0.79200
(360)2
5. In a skating stunt known as crack-the-whip, a number of skaters hold hands
and form a straight line. They try to skate so that the line rotates about the
skater at one end, who acts as the pivot. The skater farthest out has a mass of
80.0 kilograms and is 6.10 meters from the pivot. He is skating at a speed of 6.80
m/s. Determine the magnitude of the centripetal force that acts on him.
606 newtons
Solution
The centripetal force required to maintain the motion is given by Newton’s second
law: Fnet = ma, but in this case the a is given by a = v 2 /r. So,
F =
(80.0)(6.80)2
= 606.43
(6.10)
In this case the centripetal force is provided by the tension between the skaters’
hands.
6. The moon orbits the earth at a distance of 3.85×108 meters. Assume that this
distance is between the centers of the earth and the moon and that the mass of
the earth is 5.98 × 1024 kilograms. Find the period for the moon’s motion around
the earth. Express the answer in days and compare it to the length of a month.
27.5 days
Solution
The centripetal force is provided by gravity. The magnitude of the force is given
by Fgrav = GM m/r2 . Since we are interested in the period, we can combine this
with the second formula for centripetal acceleration, Fcent = 4π 2 mr/T 2 . Thus,
GM m
4π 2 mr
=
r2
T2
Which simplifies to
GM T 2 = 4π 2 r3
This is sometimes called the 1-2-3 law and is also known as Kepler’s third law of
planetary motion. Historically, this formula was one of the main clues Newton
had in uncovering his law of gravity. We can use it to solve our problem:
(6.674 × 10−11 )(5.98 × 1024 )(T )2 = 4π 2 (3.85 × 108 )3 =⇒ T = 2.3759 × 106
This is in seconds. There are 86,400 seconds in one day, so the period in days is
T = 2.3759 × 106 s ×
1 dy
= 27.498
86400 s
Which is just short of one full month.
7. The National Aeronautics and Space Administration (NASA) studies the
physiological effects of large accelerations on astronauts. Some of these studies
use a machine known as a centrifuge. This machine consists of a long arm, to one
end of which is attached a chamber in which the astronaut sits. The other end of
the arm is connected to an axis about which the arm and chamber can be rotated.
The astronaut moves on a circular path, much like a model airplane flying in a
circle on a guideline. The chamber is located 15 meters from the center of the
3
33 m/s
circle. At what speed must the chamber move so that an astronaut is subjected
to 7.5 times the acceleration due to gravity?
Solution
This will involve the formula a = v 2 /r. We know r is equal to 15 meters and the
acceleration is 7.5 times g. Thus,
(7.5)(9.8) =
(v)2
=⇒ v = 33.204
15
8. The hammer throw is a track-and-field event in which a 7.3-kilogram ball (the
“hammer”) is whirled around in a circle several times and released. It then moves
upward on the familiar curving path of projectile motion and eventually returns
to earth some distance away. The world record for this distance is 86.75 meters,
achieved in 1986 by Yuriy Sedykh. Ignore air resistance and the fact that the ball
is released above the ground rather than at ground level. Furthermore, assume
that the ball is whirled on a circle that has a radius of 1.8 meters and that its
velocity at the instant of release is directed 41◦ above the horizontal. Find the
magnitude of the centripetal force acting on the ball just prior to the moment of
release.
3500 newtons
Solution
Okay. Let’s start with the end: the centripetal force. Its formula is F = mv 2 /r.
Now we know m and r. So we need to find v. What we have remaining is the range
of the hammer. We can extract this from the range formula, R = (v 2 /g) sin 2θ.
Notice that we know R and θ, so we can solve for the initial velocity. Thus,
(86.75) =
(v)2
sin 2(41◦ ) =⇒ v = 29.300
(9.80)
We can take this velocity and calculate the centripetal force:
F =
(7.3)(29.300)2
= 3481.7
1.8
9. A car moving at 5.0 m/s tries to round a corner in a circular arc of 8.0 meters
radius. The roadway is flat. How large must the coefficient of fricton be between
the wheels and roadway if the car is not to skid?
Solution
There are three forces in this problem: weight, support (from the roadway) and
friction. The free-body diagram is in Figure 1.
Top View
Side View
N
F = µN
Fcent
r
W = mg
Figure 1: Free body diagram
Since there is no motion up and down, the support force is equal to the weight,
mg. The formula for the friction involves this support force, so we have
F = µN = µmg
4
0.32
On the other hand, this friction provides the centripetal force Fcent . The formula
for the centripetal force is
Fcent = mv 2 /r
So,
µmg = mv 2 /r =⇒ µ = v 2 /rg
We have
µ=
(5.0)2
= 0.31888
(8.0)(9.8)
10. A satellite orbits the Earth at a height of 200 kilometers. Find (a) the speed
of the satellite and (b) the time taken (in minutes) to complete one revolution.
Solution
(a) The radius of the satellite orbit is the 200 kilometers plus the radius of the
earth 6371 kilometers. After converting to meters, we have:
r = 6.571 × 106
Since the force of gravity F = GM m/r2 provides the centripetal force Fcent =
mv 2 /r, we can cancel the mass m to yield:
GM m
mv 2
=
2
r
r
r
GM
=⇒ v =
r
For our problem we have
r
v=
(6.673 × 10−11 )(5.974 × 1024 )
= 7789
6.571 × 106
(b) Now that we have the velocity we can use the formula v = 2πr/T
7798 =
(2π)(6.571 × 106 )
=⇒ T = 5301
T
But this answer is in seconds. Divide by 60 to convert to minutes:
T = 88.34 min
We could also use Kepler’s 3rd law GM T 2 = 4πr3 to answer part (b) without
part (a).
5
(a) 7.8 km/s; (b) 88 minutes