Physics 1303,1403 - Spring 1997

Homework Assignment #6

Due:
10 April (Sections 002 and 802)
11 April (Sections 001 and 801)

READING

Chapters 11 and 12.

QUESTIONS

Chapter 11 - 1, 2, 12, 13, 19.
Chapter 12 - 1, 5, 6, 7, 8, 9, 10, 13, 14, 15.

PROBLEMS

Chapter 11 - Review Problem, 2, 3a, 4, 7, 10, 20, 27, 29, 39, 49, 56, 57, 63, 64.
Chapter 12 - 6, 7, 13, 17, 21, 33, 43, 58.

ANSWERS

Q 11-1) The torque depends on the location of the center of rotation.
Q 11-2) A.(BxC) is a scalar.
(A.B) is a scalar, so it cannot be crossed into the vector C. The cross product acts on two vectors to produce a third.
Q 11-12) Sphere wins, solid cylinder next, hoop last.
Q 11-13) The turntable begins to rotate in the opposite direction, not necessarily with the same angular speed. (Conservation of Angular Momentum.)
Q 11-19) Roll them down an incline; the solid one will win.
Q 12-1) No.
Q 12-5) Done in lecture with a map of Texas.
Q 12-6) Over the leg that is touching the ground.
Q 12-7) Two equal forces to the right: one acting on the top of a wheel, the other acting at the bottom of the wheel. There is a net force to the right, but zero net torque.
Q 12-8) Two forces of the same magnitude: one acting to the right on the top of a wheel, the other acting to the left at the bottom of the wheel. The forces cancel, but the torques add.
Q 12-9) No.
Q 12-10) The tall crate will topple first -- its center of gravity will move from a position over the base first.
Q 12-13) Any example in which the net FORCE vanishes, but the net TORQUE does not.
Q 12-14)
  1. could be rotating at a CONSTANT angular speed omega.
  2. could be translating with a CONSTANT linear speed v.

Q 12-15) Rough ground, frictionless wall.
P 11-Review)
  1. 7/3 m d2
  2. mgd
  3. (9 g)/(21 d) counterclockwise
  4. 2/7 g up
  5. mgd
  6. Sqrt[(6 g)/(7 d)]
  7. 7/3 m d2 Sqrt[(6 g)/(7 d)]
  8. Sqrt(2 g d/21)

P 11-2) 1.05 x 104 J
P 11-3a) adisk = 2/3 g sin(theta); ahoop = 1/2 g sin(theta)
P 11-4) vdisk = Sqrt(4 g h/3) vhoop = Sqrt(g h) (The disk wins.)
P 11-7)
  1. -17 k
  2. 70.5 degrees = 1.2304571 rad

P 11-10)
  1. -7 N.m (k)
  2. +11 N.m (k)

P 11-20)
  1. 2.10 x 1010 kg.m2/s
  2. No, L is a constant.

P 11-27)
  1. omegaf = I1/(I1 + I2) omegao
  2. Kf/Ki = I1/(I1 + I2) which is < 1.

P 11-29)
  1. 0.36 rad/s counterclockwise
  2. 100 J

P 11-39)
  1. T = Mg/3
  2. a = 2g/3
  3. v = Sqrt(4gh/3)

P 11-49)
  1. vo ro/ r
  2. m (vo ro)2/r3
  3. m vo2/2 (ro2/r2 - 1)
  4. v = 4.50 m/s; T = 10.1 N; W = 0.450 J

P 11-56)
  1. va = Sqrt(10 g h/7)
  2. vb = Sqrt(2 g h) faster!
  3. ta/tb = 1.18

P 11-57)
  1. Proof required
  2. Proof required
  3. v = Sqrt[(8 F d)/(3 M)]

P 11-63)
F1 - spool rolls to the right
F2 - spool spins but does not move left or right
F3 - spool rolls to the left
F4 - spool rolls to the left
P 11-64) Proof required
P 12-6) 3010 N
P 12-7) Proof required
P 12-13) 0.789 L
P 12-17) 2R/5
P 12-21) 3L/4
P 12-33) NA = 5.98 x 105 N; NB = 4.80 x 105 N;
P 12-43) T = 5080 N; Fx = 4770 N; Fy = 8260 N
P 12-58) 1.09 m

Please report any corrections to Professor Scalise.

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