# Physics 1303 - Summer 1998

# Homework Assignment #6

**Due:**Monday 27 July 1998

### 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 d
^{2}
- mgd
- (9 g)/(21 d) counterclockwise
- 2/7 g up
- mgd
- Sqrt[(6 g)/(7 d)]
- 7/3 m d
^{2} Sqrt[(6 g)/(7 d)]
- Sqrt(2 g d/21)

**P 11-2)** 1.05 x 10^{4} J

**P 11-3a)** a_{disk} = 2/3 g sin(theta);
a_{hoop} = 1/2 g sin(theta)

**P 11-4)** v_{disk} = Sqrt(4 g h/3)
v_{hoop} = 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 10
^{10} kg.m^{2}/s
- No, L is a constant.

**P 11-27)**
- omega
_{f} =
I_{1}/(I_{1} + I_{2}) omega_{o}
- K
_{f}/K_{i} =
I_{1}/(I_{1} + I_{2}) 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)**
- v
_{o} r_{o}/ r
- m (v
_{o} r_{o})^{2}/r^{3}
- m v
_{o}^{2}/2
(r_{o}^{2}/r^{2} - 1)
- v = 4.50 m/s; T = 10.1 N; W = 0.450 J

**P 11-56)**
- v
_{a} = Sqrt(10 g h/7)
- v
_{b} = Sqrt(2 g h) faster!
- t
_{a}/t_{b} = 1.18

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

**P 11-63)**

F_{1} - spool rolls to the right

F_{2} - spool spins but does not move left or right

F_{3} - spool rolls to the left

F_{4} - 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)** N_{A} = 5.98 x 10^{5} N;
N_{B} = 4.80 x 10^{5} N;

**P 12-43)** T = 5080 N; F_{x} = 4770 N; F_{y} = 8260 N

**P 12-58)** 1.09 m

Please report any corrections to
Professor Scalise.

Back to the Physics 1303 Home Page