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Homework 9 Problems – Rotational Dynamics
1. [10 points] A uniform plank of length 2.00 m and mass 30.0 kg is supported by three
ropes, as indicated by the blue vectors in the figure below. Find the tension in each rope
when a 700-N person is d = 0.500 m from the left end.
Answer: In order for the man to remain stationary, the system must be in mechanical
equilibrium, which implies that ⃗
and ⃗
. Defining the axis of rotation at the left
end of the plank, we have
(
)
)(
(
Solving for
(
)
)(
)(
)
gives
(
)(
)
(
(
)(
)(
)
)
Since the system is in mechanical equilibrium, the net force on the system must also be balanced
⃗
⃗⃗
⃗⃗
⃗⃗
⃗
⃗
In the x-direction, we have
∑
(
)
In the y-direction, we have
(
∑
(
)(
)
)(
)
2. [10 points] The arm in the figure below weighs 41.5 N. The force of gravity acting on the
arm acts through point A. Determine the magnitudes of the tension force Ft in the deltoid
muscle and the force Fs exerted by the shoulder on the humerus (upper-arm bone) to hold
the arm in the position shown.
Answer: In order for the arm to remain in its position, it must be in mechanical equilibrium,
which implies that ⃗
and ⃗
. Defining the axis of rotation at point O, we have
(
Solving for
)(
)
(
)(
)
gives
(
(
)(
)
)
Since the system is in mechanical equilibrium, the net force on the system must also be balanced.
In the x-direction, we have
∑
In the y-direction, we have
∑
The magnitude of
is given by
√
3. [10 points] Use conservation of mechanical energy to determine the angular speed of the
spool shown below after the 3.00 kg bucket has fallen 4.00 m, starting from rest. The
light string attached to the bucket is wrapped around the spool and does not slip as it
unwinds.
Answer: Mechanical energy conservation states that (
starts from rest, we have
(
)
. Since the object
)
The spool can be modeled as a solid cylinder and thus its moment of inertia is
Therefore, we have
(
)
(
)
Simplifying the above expression gives
(
)
Solving for the final angular speed gives
√
√[
(
(
)(
)(
)
)(
(
)
)(
)
]
.
4. [10 points] An Atwood’s machine consists of blocks of masses m1 = 10.0 kg and m2 =
20.0 kg attached by a cord running over a pulley. The pulley is a solid cylinder with mass
M = 8.00 kg and radius r = 0.200 m. The block of mass m2 is allowed to drop, and the
cord turns the pulley without slipping.
a. Why must the tension T2 be greater than the tension T1?
b. What is the acceleration of the system, assuming the pulley axis is frictionless?
c. Find the tensions T1 and T2.
Answer:
(a) In order to induce an angular acceleration, which leads to an acceleration of the system as
shown in the figure above, there must be a net clockwise torque applied on the pulley.
This can only occur if is greater than .
(b) To determine the acceleration of the system, we use Newton’s 2nd law on block m1
(
∑
)
( )
Using Newton’s 2nd law on block m2 gives
(
∑
)
( )
Using the rotational form of Newton’s 2nd law for the pulley (assuming that the pulley
can be modeled as a solid cylinder), we have
∑
(
Substituting (1) and (2) into (3) gives
)
( )
(
)
(
)
Solving for the acceleration gives us
(
)
(
(
)
The tension forces are given by
(
)
(
)(
)
(
)
(
)(
)
)(
)
5. [10 points] A 0.005-kg bullet traveling horizontally 1.00 x 103 m/s enters an 18.0-kg
door, embedding itself 10.0 cm from the side opposite the hinges. The 1.00-m-wide door
is free to swing on its hinges.
a. Before it hits the door, does the bullet have angular momentum relative to the
door’s axis of rotation? Defend your answer.
b. Is mechanical energy conserved in this collision? Defend your answer.
c. At what angular speed does the door swing open immediately after the collision?
Assume that the door has the same moment of inertia as a rod with an axis at one
end.
d. Calculate the energy of the door-bullet system and determine whether it is less
than or equal to the kinetic energy of the bullet before the collision.
Answer:
(a) Relative to the axis of rotation of the door, the bullet has angular momentum since the
angular velocity is defined as the product of the linear velocity of the bullet and the
distance from the axis of rotation. Also, the bullet is moving perpendicular to the door,
implying that the lever arm and the linear momentum vectors are perpendicular to one
another.
(b) Mechanical energy will not be conserved in this collision since there will be kinetic
energy loss due to the collision. In other words, this is a perfectly inelastic collision.
(c) To determine the angular speed of the door immediately after the collision, we use
conservation of angular momentum
(
)
(
)
The bullet can be modeled as a particle which indicates that
(where
is the distance from the axis of rotation to the bullet) and the door can be modeled as a
rod with a rotation axis at one end, which indicates that
. Therefore,
conservation of angular can be written as
(
)
(
(
)
)(
)(
((
)(
)
)
(
)(
) )
(d) The rotational kinetic energy of the bullet-door system after the collision is given as
(
)
(
((
)
)(
)
(
)(
) )(
The linear kinetic energy of the bullet before the collision is given as
(
)(
)
)
Bonus: [6 points] A rigid, massless rod has three particles with equal masses attached to it as
shown below. The rod is free to rotate in a vertical plane about a frictionless axle perpendicular
to the rod through the point P and is released from rest in the horizontal position at time t = 0.
Assuming that m = 2.0 kg and d = 2.0 cm,
a. Find the moment of inertia, torque, and angular acceleration of the system at t = 0.
b. Find the maximum kinetic energy of the system and the maximum angular speed reached
by the rod.
c. Find the maximum angular momentum of the system
Answer:
(a) The moment of inertia for the system at
[(
The torque of the system at
)
is given by
( )
(
(
) ]
)(
)
is given by
(
)
(
)(
)(
)
The angular acceleration of the system is given by Newton’s 2nd law
(b) If we neglect air resistance, the total mechanical energy is conserved. The maximum
kinetic energy will occur when the rod is in the vertical position. Thus, the potential
energy relative to point P is given by
( )
( )
(
( )
)(
)(
)
Thus, conservation of mechanical energy states that the change in potential energy will be
the same as the change in total kinetic energy. Thus, as the system starts from rest, the
maximum kinetic energy will be
. The maximum angular speed will be
√
(
)
(c) The maximum angular momentum of the system is given by
(
)(
)