PHY 131 Second Midterm, Fall 2001

Moments of inertia:

Solid sphere, mass M, radius R: 2/5 MR2

Solid cylinder mass M, radius R: 1/2 MR2

Thin walled hollow cylinder mass M, radius R: MR2

Thin walled sphere mass M, radius R: 2/3 MR2

Rod of length l, rotating around an axis in center: 1/12 Ml2

Thin solid rectangle, length l, width w, rotating around an axis along an edge parallel to its length: 1/3 Mw2


 

  1. A 6.0m (5.0m) long ladder is on ice, leaning against a vertical. It is tied to the wall by a horizontal rope, running 1.1m above the ground. The angle between the wall and the ladder is 35o. The ladder has a weight of 410N, and it has a uniform mass distribution along its length. The friction between the wall and the ladder and between the floor and the ladder is negligible. What is the tension in the rope?

The force F at the wall is horizontal, and its magnitude is unknown. At the bottom of the ladder the force is vertical, and its magnitude is 410N (since there is no other vertical force to balance the weight). Distance from the top of the ladder to the rope is d=(cos35o)(6.0m)-1.1m. The torque around the top of the ladder is: (sin35o)(3.0m) (410N)-(sin35o)(6.0m)410N-Fd. This must be zero for equilibrium. We get F=(sin35o)(3.0m) (410N)/d=185N (196N)
  1. The liquid in the open tube manometer in the Figure is mercury (alcohol), and y1=4.00cm and y2=37.00cm. The atmospheric pressure is 0.170atm (1atm = 1.013 x 105Pa). What is the absolute pressure in the gas tank? (Use the enclosed tables.)
The pressure is changing as p= rgh, where h is the distance from the top surface of the liquid. The density of mercury is 13.6 x 103 kg/m3 and h=y2-y1=0.33m. The pressure is p = p0 + rgh = 6.12 x 104Pa (1.98x104Pa).
  1. One end of a horizontal rubber tube is fastened to a fixed support, the other end passes over a pulley, and supports a weight of 5.0kg (8.0kg). The free length between the support and the pulley is 12.0m long, and the mass of the tube in this length is 0.90kg A transverse wave runs along the tube. How long does it take to go from one end to the other end?
The mass per unit length is m = 0.9kg/12.0m = 0.075 kg/m. The tension is T = 5.0kg x 9.81m/s2=49N, therefore the speed of the wave is v = (T/m)1/2 = (49/0.075)1/2=25.6m/s. It takes t=12.0/25.6=0.47s (0.37s) to go along the full length.
  1. A longitudinal wave of frequency 4200Hz is propagating in liquid alcohol (mercury). What is the wavelength? (All data necessary to solve this problem can be found in the tables provided with the exam.)
The velocity of longitudinal waves in bulk materials is v=(B/r)1/2. We find B= 1.0 x 109N/m2 and r= 0.79 x 103kg/m3 for alcohol. The speed of sound is v=1125m/s. The wavelength is l=v/f=0.26m (0.10m).
  1. (This problem refers to a laboratory experiment you have performed, but some parameters are different.) A circular table rotates around a frictionless vertical axis with an angular velocity of 1.20rad/sec. A solid disk of uniform thickness, initially at rest, is dropped on the table from a small height, so that the rim of the disk matches the rim of the table. The mass of the disk is 1.00kg, and the radius is 0.10m. After the collision, the table and the disk rotates together with an angular velocity of 0.90rad/sec. What is the moment of inertia of the table?
  1. 0.0050 kgm2
  2. 0.015 kgm2
  3. 0.020 kgm2
  4. 0.050 kgm2
  5. 0.15 kgm2
  6. 0.20 kgm2
Use the conservation of angular momentum. Initially the table has L = wI and the disk has none. When they rotate together, the total moment of inertia is I + 1/2 MR2 and the angular momentum is L = w' (I + 1/2 MR2). Therefore wI = w' (I + 1/2 MR2), and I = w'/(w - w') 1/2 MR2 = 0.015kgm2.
  1. A mass of 1.62kg hangs from a vertical spring, and it finds an equilibrium when the string is stretched by 0.315m. We pull the mass down by an additional 0.130m, and release it. How long does it take for the mass to reach the equilibrium position again?


    2. A. 0.18s

    B. 0.20s

    C. 0.28s

    D. 0.56s

    E. 0.72s

    F. 1.12s

The spring constant is k=mg/x=50.4N/m. The period of the oscillation is T=2p(m/k)1/2 =1.126s. It takes one quarter of the period to get from the maximum to the equilibrium: t=0.282s.
  1. A solid uniform cylinder of mass m and radius r rolls down a ramp of angle q=30o. What is the magnitude of the acceleration of the center of the cylinder, if the motion is without sliding? (The moment of inertia for a solid cylinder is 1/2 mr2.)

  1. 4.95 m/s2
  2. 4.05m/s2
  3. 3.27m/s2
  4. 1.89m/s2
  5. 0.98m/s2
  6. something else
The torque equation yields (1/2) mr2 (a/r) = Fr, where F is the force between the slope and the cylinder. The linear motion is described by mg sin q - F =ma. The acceleration is (2/3) g sin q = 3.27m/s2.
  1. A 0.45kg hockey puck heading east with a velocity of 4.2m/s collides head-on with a 0.90kg hockey puck, initially at rest. Assuming perfectly elastic collision, what will be the velocity (magnitude and direction) of the 0.45kg puck after the collision? (in a "head-on collision" all motion of the objects is along the same straight line.)


    2. A. 1.4m/s east

    B. 1.4m/s west

    C. 2.1 m/s east

    D. 2.1m/s west

    E. 4.2m/s east

    F. 4.2m/s west
     

Use conservation of energy and momentum. Momentum conservation yields: m1v1+0=m1v1'+m2v2'. Combine this with kinetic energy conservation and obtain v1-v2=v2'-v1', as discussed in the book and lecture. With numbers: (0.45)(4.2)= (0.45)v1'+(0.90)v2' and 4.2=v2'-v1'. Express v2'=v1'+4.2, and get (0.45)(4.2)= (0.45) (v1') +(0.90)(v1'+4.2) =(0.45)(v1')+(0.90)(4.2)+(0.90)v1'. Therefore -1.89=(0.135)v1' and v1'=-1.4m/s. The negative sign means that it moves west.
  1. According to official stats, US Open winner Venus Williams can serve a ball with a speed of 120miles/hour. What is the average force between the ball and her racket, if the ball is in touch with the racket for 0.03s, and the mass of the ball is 0.057kg? (Note: This is and estimate, and the result is rounded to one significant digit.)
  1. 0.09N
  2. 0.6N
  3. 6N
  4. 100N
  5. 200N
  6. 900N
The impulse is mv, and the force is F=mv/t=(0.057kg)(1600m)/(3600s)/(0.03s)~100N