Physics M10A, Exam #4, Ch. 9 – Ch. 10

Directions:
There 20 multiple choice questions. Each question is 5 points, and the exam is 100 points. The exam is open book, open notes. However, you may not discuss the exam with your classmates.
Submit your work for each question in Canvas.
𝜏 = π‘ŸπΉ sin πœƒ, 𝜏 = π‘ŸπΉβŠ₯, 𝜏 = π‘ŸβŠ₯𝐹, 𝑔 = 9.8 π‘š/𝑠𝑒𝑐2 ,
πœ” = πœ”π‘– + 𝛼𝑑, πœƒ = πœƒπ‘– + πœ”π‘– 𝑑 + 1
2 𝛼𝑑2, (πœ”)2 = (πœ”π‘– )2 + 2π›ΌΞ”πœƒ, 𝑣 = π‘Ÿπœ”,
π‘Žπ‘‘ = π‘Ÿπ›Ό, 𝜏 = π‘ŸπΉ, πœπ‘›π‘’π‘‘ = 𝐼𝛼, 𝐾𝐸 = 1
2 π‘šπ‘£2 + 1
2 πΌπœ”2, π‘ˆπ‘” = π‘šπ‘”β„Ž, 𝐿 = πΌπœ”

Problems 1–2 refer to the following situation. A beam, resting on two pivots, has a length of 𝐿 = 6.00 π‘š and mass 𝑀 = 90.0 π‘˜π‘”. The pivot under the left end exerts a normal force 𝑁1 on the beam, and the second pivot placed a distance β„“ = 4.00 π‘š from the left end, exerts a normal force 𝑁2. A woman of mass π‘š = 55.0 π‘˜π‘” is a distance π‘₯ = 2.00 π‘š from the left end.
The system is in equilibrium.

 

Problem 1:
What is the normal force 𝑁1?

  • a. 190 N
  • b. 290 N
  • c. 390 N
  • d. 490 N

Problem 2:

What is the normal force 𝑁2?

  • a. 831 N
  • b. 931 N
  • c. 1031 N
  • d. 1131 N
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. 𝐹𝐢 is the force exerted by the food being chewed against the jawbone, 𝑇 is the force of tension in the masseter, and 𝑅 is the force exerted by the socket on the mandible. We will consider 𝐹𝐢 = 50.0 𝑁. The system is in equilibrium.

Problem 3:
What is 𝑇, 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 𝑅, 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 β€œwith his back” (see figure below on the left) rather than β€œwith his knees” 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 12.0∘. The system is in equilibrium.

Problem 5:
What is the tension 𝑇?

  • 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, 𝑅π‘₯?

  • 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, 𝑅𝑦?

  • a. βˆ’42.5 𝑁
  • b. βˆ’32.5 𝑁
  • c. βˆ’22.5 𝑁
  • d. βˆ’12.5 𝑁
Problems 8–13 refer to the following situation. Two forces 𝐹1 and 𝐹2 are acting on a rod, as shown in the figure below. The rod is pivoted about point 𝑂, on the left end of the rod.

Problem 8:
What is the torque due to 𝐹1, about point 𝑂?

  • a. βˆ’10.0 𝑁 β‹… π‘š
  • b. βˆ’20.0 𝑁 β‹… π‘š
  • c. βˆ’30.0 𝑁 β‹… π‘š
  • d. βˆ’40.0 𝑁 β‹… π‘š

Problem 9:
What is the torque due to 𝐹2, about point 𝑂?

  • a. 12.0 𝑁 β‹… π‘š
  • b. 22.0 𝑁 β‹… π‘š
  • c. 32.0 𝑁 β‹… π‘š
  • d. 42.0 𝑁 β‹… π‘š

Problem 10:
What is the net torque about point 𝑂?

  • a. βˆ’18.0 𝑁 β‹… π‘š
  • b. βˆ’28.0 𝑁 β‹… π‘š
  • c. βˆ’38.0 𝑁 β‹… π‘š
  • d. βˆ’48.0 𝑁 β‹… π‘š

Problem 11:
What is the moment of inertia of the rod about the point 𝑂? The formula for the moment of inertia of a rod, about one of its ends, is 𝐼 = π‘šπΏ2/3, where π‘š is the mass of the rod and 𝐿 is the length of the rod. The rod has a mass of 1.5 π‘˜π‘” and a length of 5.00 m.

  • a. 8.5 π‘˜π‘” β‹… π‘š2
  • b. 10.5 π‘˜π‘” β‹… π‘š2
  • c. 12.5 π‘˜π‘” β‹… π‘š2
  • d. 14.5 π‘˜π‘” β‹… π‘š2

Problem 12:
What is the angular acceleration of the rod?

  • a. βˆ’1.24 π‘Ÿπ‘Žπ‘‘/𝑠𝑒𝑐2
  • b. βˆ’2.24 π‘Ÿπ‘Žπ‘‘/𝑠𝑒𝑐2
  • c. βˆ’3.24 π‘Ÿπ‘Žπ‘‘/𝑠𝑒𝑐2
  • d. βˆ’4.24 π‘Ÿπ‘Žπ‘‘/𝑠𝑒𝑐2

Problem 13:
The rod is initially at rest (πœ”π‘– = 0 π‘Ÿπ‘Žπ‘‘/𝑠𝑒𝑐). Let’s take the initial position of the rod to be πœƒ0 = 0 π‘Ÿπ‘Žπ‘‘. How long does it take the rod to rotate to the angle πœƒ = βˆ’πœ‹/2 ?

  • a. 0.88 sec
  • b. 1.18 sec
  • c. 1.48 sec
  • d. 1.78 sec
Problems 14–17 refer to the following situation. A uniform solid disk with mass 40.0 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is stationary. The disk is initially at rest, and then a constant force 𝐹 = 30.0 𝑁 is applied tangent to the rim of the disk. The moment of inertia of a solid disk is 𝐼 = 𝑀𝑅2/2, where 𝑀 is the mass of the disk and 𝑅 is the radius of the disk.

Problem 14:
What is the net torque acting on the disk?

  • a. 2.00 𝑁 β‹… π‘š
  • b. 4.00 𝑁 β‹… π‘š
  • c. 6.00 𝑁 β‹… π‘š
  • d. 8.00 𝑁 β‹… π‘š

Problem 15:
What is the angular acceleration of the disk?

  • a. 6.50 π‘Ÿπ‘Žπ‘‘/𝑠𝑒𝑐2
  • b. 7.50 π‘Ÿπ‘Žπ‘‘/𝑠𝑒𝑐2
  • c. 8.50 π‘Ÿπ‘Žπ‘‘/𝑠𝑒𝑐2
  • d. 9.50 π‘Ÿπ‘Žπ‘‘/𝑠𝑒𝑐2

Problem 16:
What is the angular speed of the disk after 0.400 sec?

  • a. 1.00 π‘Ÿπ‘Žπ‘‘/𝑠𝑒𝑐
  • b. 2.00 π‘Ÿπ‘Žπ‘‘/𝑠𝑒𝑐
  • c. 3.00 π‘Ÿπ‘Žπ‘‘/𝑠𝑒𝑐
  • d. 4.00 π‘Ÿπ‘Žπ‘‘/𝑠𝑒𝑐

Problem 17:
Let’s define the initial angular position of the disk as πœƒ0 = 0 π‘Ÿπ‘Žπ‘‘. What is the angular position of the disk, πœƒ, 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 𝐼 = 𝑀𝑅2/2, where 𝑀 is the mass of the disk and 𝑅 is the radius of the disk. Treat the piece of putty as particle. The moment of inertia of a particle is 𝐼 = π‘šπ‘Ÿ2, where π‘š is the mass of the particle and π‘Ÿ 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 π‘˜π‘” β‹… π‘š2
  • b. 6.00 π‘˜π‘” β‹… π‘š2
  • c. 7.00 π‘˜π‘” β‹… π‘š2
  • d. 8.00 π‘˜π‘” β‹… π‘š2

Problem 19:
What is the moment of inertia of the disk and the piece of putty?

  • a. 5.20 π‘˜π‘” β‹… π‘š2
  • b. 6.20 π‘˜π‘” β‹… π‘š2
  • c. 7.20 π‘˜π‘” β‹… π‘š2
  • d. 8.20 π‘˜π‘” β‹… π‘š2

Problem 20:
What is the angular speed of the system after the piece of putty is dropped on the disk?Β Β  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