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Homework 8 Problems: Rotational Motion
1. A machine part rotates at an angular speed of 0.06 rad/s; its angular speed is then
increased to 2.2 rad/s at an angular acceleration of 0.70 rad/s2.
a. [5 points] Find the angle through which the part rotates before reaching this final
speed.
b. [5 points] In general, if both the initial and final angular speed are doubled at the
same angular acceleration, by what factor is the angular displacement changed?
Why?
Answer:
(a) To determine the final angular position, we use rotational kinematics equation 3
(
)
(
(
)
)
(b) Based on rotational kinematics equation 3, we see that
is proportional to
.
Therefore, doubling the final and initial speed (while keeping the angular acceleration
constant) will quadruple the angular displacement.
2. A car initially traveling at 29.0 m/s undergoes a constant negative acceleration of
magnitude 1.75 m/s2 after its brakes are applied.
a. [7 points] How many revolutions does each tire make before the car comes to a
stop, assuming the car does not skid and the tires have radii of 0.330 m?
b. [3 points] What is the angular speed of the wheels when the car has traveled half
the total distance?
Answer:
(a) To determine the amount of revolutions each tire makes, we need to convert linear
quantities to angular quantities. The final angular speed is given by
The angular acceleration is given by
To determine the number of revolutions made, we use kinematics equation 2
(
)
(
)
(b) If the car travels half the total distance, this implies that the wheels will rotate half the
total revolutions. Assuming the same deceleration, the angular speed of the wheels when
the car has traveled half the total distance is given by
(
√
(
)
√(
)
)
(
)(
)
3. A pail of water is rotated in a vertical circle of radius 1.00 m.
a. [3 points] What two external forces act on the water in the pail? Which of the two
forces is most important in causing the water to move in a circle?
b. [4 points] What is the pail’s minimum speed at the top of the circle if no water is
to spill out?
c. [3 points] If the pail with the speed found in part (b) were to suddenly disappear at
the top of the circle, describe the subsequent motion of the water. Would it differ
from the motion of a projectile?
Answer:
(a) For a pail of water moving in a circle, there are two forces that operate on the water: the
weight of the water itself (i.e. the gravitational force on the water) and the normal force
from the pail acting on the water. The normal force of the bottom of the pail is most
important in causing the water to move in a circle since the normal force always points
towards the center of the circle for a pail of water moving in a vertical circle.
(b) At the top of the circle, the gravitational force pulls the water toward the center of the
circular path. If the water is just on the verge of spilling out, this means that the normal
force between the pail and the water is zero. Thus, the only force on the water must be the
gravitational force
Solving for the minimum speed gives
√
√(
)(
)
(c) If the pail was removed, then there is no longer a centripetal force that causes the water to
move in a circular path. Therefore, the water will initially move in the direction of the
tangential velocity at that point. Since the water is in the air, the water will travel like a
projectile with a horizontal initial velocity of 3.13 m/s.
4. Because of Earth’s rotation about its axis, a point on the equator has a centripetal
acceleration of 0.0340 m/s2, whereas a point at the poles has no centripetal acceleration.
a. [7 points] Show that, at the equator, the gravitational force on an object (i.e. the
object’s true weight) must exceed the object’s apparent weight.
b. [3 points] What are the apparent weights of a 75.0-kg person at the equator and at
the poles? (Assume Earth is a uniform sphere and take g = 9.800 m/s2.)
Answer:
(a) The apparent weight of an object (as measured on a scale) is the magnitude of the normal
force. Taking the earth as uniform sphere, Newton’s 2nd law in the radial direction gives
∑
Here, ω is the angular rotation speed of the earth. When an object is at the poles, the
distance from the object to the axis of rotation is zero and thus there is no centripetal
acceleration. This implies that the object’s apparent weight is equal to the gravitational
force on the object (i.e. the object’s true weight). However, when an object is at the
equator, the distance from the object to the axis of rotation is no longer zero and thus
there is a centripetal acceleration. This indicates that there is an imbalance of the forces in
the radial direction, giving an apparent weight of
(
This implies that
the object’s apparent weight.
)
and thus, the gravitational force on the object exceeds
This can also be explained via the centrifugal force. For a person at the pole, the
centrifugal force is zero and thus, the apparent weight is the same as the person’s true
weight. For a person at the equator, the centrifugal force points outward (i.e. away from
Earth’s surface) and this will counterbalance the person’s true weight. Consequently, the
apparent weight will be less than the true weight of the person.
(b) For a 75.0-kg person, the apparent weight of the man at the poles is
(
)(
)
whereas the apparent weight of the man at the equator is
(
)
(
)(
)
5. One end of a cord is fixed and a small 0.500-kg object is attached to the other end, where
it swings in a section of a vertical circle of radius 2.00 m, as shown below. When θ =
20.0°, the speed of the object is 8.00 m/s.
a. [4 points] Find the tension in the string,
b. [3 points] Find the tangential and radial components of acceleration. Find the total
acceleration.
c. [3 points] Is your answer changed if the object is swinging down toward its lowest
point instead of swinging up? Explain your answer.
Answer:
(a) Here the tension force serves as the centripetal force necessary for this object to move in
a circular path. Using a coordinate system in which the x-axis points in the tangential
direction and the y-axis points in the radial direction, we can apply Newton’s 2nd law to
determine the tension. Using Newton’s 2nd law in the y-direction, we have
Solving for the tension force
(
)
(
gives
(
)
)(
(
)
(
)
)
(b) The centripetal acceleration can be calculated from the object’s speed when θ = 20.0°:
(
)
The tangential acceleration can be calculated from Newton’s 2nd law in the x-direction
Solving for the tangential acceleration gives
(
)
The total acceleration is given by
√
√(
)
(
)
(c) A schematic diagram of this problem is given below
If the object is swinging down instead of up, this would imply that the tangential
acceleration would be positive. However, the magnitude of the tangential acceleration
would be the same since the angle θ is the same. This would also imply that the speed is
the same, which further implies that the centripetal acceleration is the same. Thus, the
total acceleration would remain the same.
Bonus: [7 points] A 0.50-kg ball that is tied to the end of a 1.5-m light cord is revolved in a
horizontal plane with the cord making a 30⁰ angle with the vertical.
a. Determine the ball’s speed.
b. If, instead the ball is revolved so that its speed is 4.0 m/s, what angle does the cord make
with the vertical?
c. If the cord can withstand a maximum tension of 9.8 N, what is the highest speed at which
the ball can move?
Answer:
(a) The ball is subject to two forces: the gravitational force and the tension force. Here, the
horizontal component of the tension force produces a centripetal acceleration, which acts
towards the center of the circle. Using Newton’s 2nd law in the radial and vertical
direction we have
∑
∑
Eliminating T in both equations gives
If the height from the hand to the plane of the ball is h, we can also write
Setting both equations equal to each other gives
√
√
Note that the radius of the horizontal circle can be written as
from the hand to the plane of the ball h can be written as
√
√
√
(
)(
)
(
(
and the height
. This gives us
)
)
(b) If the ball now has a tangential velocity of 4.0 m/s, then we can write the tangential
velocity as a function of the radius of the circle
(
)
Solving this quadratic equation gives
.
(c) If the cord can withstand a maximum tension of 9.8 N, then the angle of the cord with the
vertical is given by
(
(
(
)
)(
)
Therefore, the maximum velocity is given as
√
√
√
(
)(
)
(
(
)
)
)