Problems 3–4 refer to the following situation. The chewing muscle, the masseter, is one of the strongest in the human body. It is attached to the mandible (lower jawbone) as shown in the Figure below on the left. The jawbone is pivoted about a socket just in front of the auditory
canal. The forces acting on the jawbone are equivalent to those acting on the curved bar in Figure below on the right. πΉπΆ is the force exerted by the food being chewed against the jawbone, π is the force of tension in the masseter, and π
is the force exerted by the socket on the mandible. We will consider πΉπΆ = 50.0 π. The system is in equilibrium.
Problem 3:
What is π, the force of tension in the masseter?
a. 57.1 N b. 107.1 N c. 157.1 N d. 207.1 N
Problem 4:
What is π
, the force exerted by the socket on the mandible?
a. 57.1 N b. 107.1 N c. 157.1 N d. 207.1 N
Problems 5–7 refer to the following situation. A person bending forward to lift a load βwith his backβ (see figure below on the left) rather than βwith his kneesβ can be injured by large forces exerted on the muscles and vertebrae. The spine pivots mainly at the fifth lumbar vertebra, with the principal supporting force provided by the erector spinalis muscle in the back. To see the magnitude of the forces involved, and to understand why back problems are common among humans, consider the model shown in the figure below on the right of a person bending forward to lift a 200 N object. The spine and upper body are represented as a uniform horizontal rod of length 45.0 cm and weight 350 N, pivoted at the base of the spine.
The erector spinalis muscle, attached at a point two thirds of the way up the spine (a distance of 30.0 cm from the spine), maintains the position of the back. The angle between the spine and this muscle is 12.0β. The system is in equilibrium.
Problem 5:
What is the tension π?
a. 2705.5 N b. 2905.5 N c. 3105.5 N d. 3305.5 N
Problem 6:
What is the x–component of the force on the spine, π
π₯?
a. 2146.4 N b. 2646.4 N c. 3146.4 N d. 3646.1 N
Problem 7:
What is the y–component of the force on the spine, π
π¦?
a. β42.5 π b. β32.5 π c. β22.5 π d. β12.5 π
Problem 8:
What is the torque due to πΉ1, about point π?
a. β10.0 π β
π b. β20.0 π β
π c. β30.0 π β
π d. β40.0 π β
π
Problem 9:
What is the torque due to πΉ2, about point π?
a. 12.0 π β
π b. 22.0 π β
π c. 32.0 π β
π d. 42.0 π β
π
Problem 10:
What is the net torque about point π?
a. β18.0 π β
π b. β28.0 π β
π c. β38.0 π β
π d. β48.0 π β
π
Problem 11:
What is the moment of inertia of the rod about the point π? The formula for the moment of
inertia of a rod, about one of its ends, is πΌ = ππΏ2/3, where π is the mass of the rod and πΏ is
the length of the rod. The rod has a mass of 1.5 ππ and a length of 5.00 m.
a. 8.5 ππ β
π2 b. 10.5 ππ β
π2 c. 12.5 ππ β
π2 d. 14.5 ππ β
π2
Problem 12:
What is the angular acceleration of the rod?
a. β1.24 πππ/π ππ2 b. β2.24 πππ/π ππ2 c. β3.24 πππ/π ππ2 d. β4.24 πππ/π ππ2
Problem 13:
The rod is initially at rest (ππ = 0 πππ/π ππ). Letβs take the initial position of the rod to be
π0 = 0 πππ. How long does it take the rod to rotate to the angle π = βπ/2 ?
a. 0.88 sec b. 1.18 sec c. 1.48 sec d. 1.78 sec
Problem 14:
What is the net torque acting on the disk?
a. 2.00 π β
π b. 4.00 π β
π c. 6.00 π β
π d. 8.00 π β
π
Problem 15:
What is the angular acceleration of the disk?
a. 6.50 πππ/π ππ2 b. 7.50 πππ/π ππ2 c. 8.50 πππ/π ππ2 d. 9.50 πππ/π ππ2
Problem 16:
What is the angular speed of the disk after 0.400 sec?
a. 1.00 πππ/π ππ b. 2.00 πππ/π ππ c. 3.00 πππ/π ππ d. 4.00 πππ/π ππ
Problem 17:
Letβs define the initial angular position of the disk as π0 = 0 πππ. What is the angular position of the disk, π, after 0.400 sec?
a. 0.200 rad b. 0.400 rad c. 0.600 rad d. 0.800 rad
Problems 18–20 refer to the following situation. A solid, horizontal cylinder of mass 10.0 kg and
radius 1.00 m rotates with an angular speed of 7.00 rad/sec about a fixed vertical axis through
its center. A 0.250 kg piece of putty is dropped vertically onto the cylinder at a point 0.900 m
from the center of rotation and sticks to the cylinder. The moment of inertia of a solid disk is
πΌ = ππ
2/2, where π is the mass of the disk and π
is the radius of the disk. Treat the piece of
putty as particle. The moment of inertia of a particle is πΌ = ππ2, where π is the mass of the
particle and π is the distance of the particle from the axis of rotation.
Problem 18:
What is the moment of inertia of the disk?
a. 5.00 ππ β
π2 b. 6.00 ππ β
π2 c. 7.00 ππ β
π2 d. 8.00 ππ β
π2
Problem 19:
What is the moment of inertia of the disk and the piece of putty?
a. 5.20 ππ β
π2 b. 6.20 ππ β
π2 c. 7.20 ππ β
π2 d. 8.20 ππ β
π2
Problem 20:
What is the angular speed of the system after the piece of putty is dropped on the disk?
Hint: Use conservation of angular momentum.
a. 5.73 rad/sec b. 6.73 rad/sec c. 7.73 rad/sec d. 8.83 rad/sec