PH201 Chapter 6 – Solutions 6.2. Set Up: Since the stone travels in a circular path, its acceleration is
directed toward the
center of the circle. The only horizontal force on the stone is the tension of the string. Set the tension in the string
equal to its maximum value.
Solve: (a) The free-body diagram for the stone is given in the figure below. In the diagram the stone is at a point
to the right of the center of the path.
(b)
gives
*6.3. Set Up: Each hand travels in a circle of radius 0.750 m and has mass (0.0125)(52 kg) = 0.65 kg and
weight 6.4 N. The period for each hand is
Let
path.
(a) The free-body diagram for one hand is given in the figure below.
wrist. This force has both horizontal and vertical components.
be toward the center of the circular
is the force exerted on the hand by the
(b)
gives
(c)
Reflect:
The horizontal force from the wrist is 12 times the weight of the hand.
The wrist must also exert a vertical force on the hand equal to its weight. *6.5. Set Up: The person moves in a circle of radius
acceleration of the person is
for one revolution is the period
The
directed horizontally to the left in the figure in the problem. The time
Solve: (a) The free-body diagram is given in the figure below.
(b)
gives
is the force applied to the seat by the rod.
and
gives
Combining
these two equations gives
Then the period is
(c) The net force is proportional to m so in
the mass divides out and the angle for a given rate of
rotation is independent of the mass of the passengers.
Reflect: The person moves in a horizontal circle so the acceleration is horizontal. The net inward force required
for circular motion is produced by a component of the force exerted on the seat by the rod.
6.6. Set Up: Apply
to the button. The button moves in a circle, so it has acceleration
The
situation is equivalent to that of Example 6.3.
Solve: (a)
Expressing v in terms of the period T,
so
A platform speed of 40.0
rev/min corresponds to a period of 1.50 s, so
(b) For the same coefficient of static friction, the maximum radius is proportional to the square of the period
(longer periods mean slower speeds, so the button may be moved farther out) and so is inversely proportional to
the square of the speed. Thus, at the higher speed, the maximum radius is (0.150 m)
Reflect:
The maximum radial acceleration that friction can give is
rate T is smaller so R must be smaller to keep
At the faster rotation
the same.
6.10. Set Up: The ball has acceleration
directed toward the center of the circular path. When the
ball is at the bottom of the swing, its acceleration is upward. Take
upward, in the direction of the
acceleration. The bowling ball has mass
Solve: (a)
upward.
(b) The free-body diagram is given in the figure below.
gives
6.13. Set Up:
A
angle is
In (c) use coordinates where
Solve: (a)
of a full rotation, so in
is upward, in the direction of
a hand travels through a distance of
at the bottom of the swing.
and
(b) The free-body diagram is shown in the figure below. F is the force exerted by the blood vessel.
(c)
gives
and
upward
(d) When the arm hangs vertically and at rest,
so
Reflect: The acceleration of the hand is only about 20% of g, so the increase in the force on the blood drop
when the arm swings is about 20%.
6.18. Set Up: The net force on each astronaut is the gravity force exerted by the other astronaut. Call the
astronauts A and B, where
and
Solve: (a) The free-body diagram for astronaut A is given in Figure (a) and for astronaut B in Figure (b) below.
for A gives
and
And for B,
(b)
gives
so
and
and
(c) Their accelerations would increase as they moved closer and the gravitational attraction between them
increased.
*6.19. Set Up: The gravitational force between two objects is
and the mass of the sun is
The mass of the earth is
The distance from the earth to the sun is
and the distance from the earth to the moon is
Solve:
It is more accurate to say the moon orbits the sun.
Reflect: The sun is farther away but has a much greater mass than the earth.
6.22. Set Up: Use coordinates where
is to the right. Each gravitational force is attractive, so is toward the
mass exerting it. Treat the masses as uniform spheres, so the gravitational force is the same as for point masses
with the same center-to-center distances. The free-body diagrams for (a) and (b) are given in Figures (a) and (b)
below.
Solve: (a)
The net force on A is
to the right.
(b)
The net force on A is
to the left.
6.26. Set Up: The earth has radius
Let the earth’s mass be
Solve: At the surface of the earth your weight is
and your mass be m.
At height h above the earth’s surface,
so
and
6.32. Set Up: (a) Although equation 6.7 was derived for the earth it can be applied to the asteroid, which we
assume
to
be
spherically
symmetric;
thus,
we
have
Also,
we
know
that
and
(b) The astronaut’s weight on the asteroid can be calculated using
(c) We note that
is nearly constant for a fall of 1.0 meter so we can use the equations of motion for constant
acceleration: use equation 2.12
(d) Assuming that
with the positive x-axis upward,
and
is nearly constant over the height of the jump, we can use the equations of motion for
constant acceleration: use equation 2.13 (
). At the top of the astronaut’s jump we have
and we will assume that the astronaut jumps with the same initial velocity,
on both the earth and the
asteroid.
Solve: (a)
(b)
(c) We can solve equation 2.12 for time:
(d) We can solve equation 2.13 for
and use it to compare what happens on the asteroid to what happens on
earth:
Thus, we have
Reflect: Contrary to our assumption, the acceleration of the astronaut does not remain perfectly constant during
this incredible 560 meter jump. Using the inverse-square law, we can show that the astronaut’s acceleration
decreases by only about 6%—and so we have probably underestimated the height of the astronaut’s jump
slightly.
*6.37. Set Up: The rotational period of the earth is
The radius of the satellite’s path is
where h is its height above the surface of the earth and
applying
and r are related by
to the motion of the satellite.
Solve: (a)
(b)-(c) The free-body diagram for the satellite is given in Figure (a) below.
and
(d)
so
gives
and
The sketch of the earth and orbit of the satellite is given in Figure (b) above.
Reflect: The orbital period is proportional to
altitude above the earth’s surface.
To achieve a period as large as 24 h, the satellite is at a large
*6.43. Set Up:
Solve:
The distance a point on the rim travels in 1 revolution is
so
In one minute a point on the rim travels a distance
minute is
The station must turn at 2.55 rpm.
The number of revolutions in one
6.45. Set Up:
Use coordinates where +y is upward. At the maximum height,
Solve: At the surface of Io,
so
Now solve for
when
and
gives
6.46. Set Up: Your friend slides toward you when the friction force on her isn’t sufficient to make her turn as
you turn. Set
and apply
to your friend. Set her acceleration equal to that of the car,
Solve: (a) In the absence of friction your friend moves in a straight line, so turn to the right, toward her.
(b) The free-body diagram for your friend is given in the figure below.
gives
gives
and
Reflect: The larger the coefficient of friction, the smaller must be the radius of the turn.
*6.47. Set Up: The person moves in a horizontal circle of radius
to its maximum value,
The person has an acceleration
directed toward the center of the circle. The period is
Set the static friction force equal
and the person has speed
Solve: (a) The free-body diagram is given in the figure below. The diagram is for when the wall is to the right of
her, so the center of the cylinder is to the left.
(b)
gives
gives
gives
so
Combining these two equations
and
(c) The mass m of the person divides out of the equation for
the answer to (b) does not depend on the mass
of the person.
Reflect: The greater the rotation rate the larger the normal force exerted by the wall and the larger the friction
force. Therefore, for smaller
the rotation rate must be larger.
6.54. Set Up: The block moves in a horizontal circle of radius
string makes an angle
Each
with the vertical.
so
The block has acceleration
directed to the left in the figure in the problem.
Solve: (a) The free-body diagram for the block is given in the figure below.
gives
(b)
gives
Reflect: The tension in the upper string must be greater than the tension in the lower string so that together they
produce an upward component of force that balances the weight of the block.