Physics 1303,1403 - Spring 1998
Homework Assignment #6
Due: 9 April
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
These are the ANSWERS only, not the SOLUTIONS. It is not sufficient to
copy these and turn them in as homework. You must show your work.
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)
- could be rotating at a CONSTANT angular speed omega.
- could be translating with a CONSTANT linear speed v.
Q 12-15) Rough ground, frictionless wall.
P 11-Review)
- 7/3 m d2
- mgd
- (9 g)/(21 d) counterclockwise
- 2/7 g up
- mgd
- Sqrt[(6 g)/(7 d)]
- 7/3 m d2 Sqrt[(6 g)/(7 d)]
- 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)
- -17 k
- 70.5 degrees = 1.2304571 rad
P 11-10)
- -7 N.m (k)
- +11 N.m (k)
P 11-20)
- 2.10 x 1010 kg.m2/s
- No, L is a constant.
P 11-27)
- omegaf =
I1/(I1 + I2) omegao
- Kf/Ki =
I1/(I1 + I2) which is < 1.
P 11-29)
- 0.36 rad/s counterclockwise
- 100 J
P 11-39)
- T = Mg/3
- a = 2g/3
- v = Sqrt(4gh/3)
P 11-49)
- vo ro/ r
- m (vo ro)2/r3
- m vo2/2
(ro2/r2 - 1)
- v = 4.50 m/s; T = 10.1 N; W = 0.450 J
P 11-56)
- va = Sqrt(10 g h/7)
- vb = Sqrt(2 g h) faster!
- ta/tb = 1.18
P 11-57)
- Proof required
- Proof required
- 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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