Answers for Problem Set 4

Last revised 1999/09/27

HRW pp. 72-80 # 69, 91
Sheet # V, VI, VII
HRW pp. 99-107 # 10, 11

Chapter 4

4-69.

a_c = 4 pi² R cos(latitude) / (1 day)² = 0.03 m / sec²,
where R is the radius of the earth.

4-91.

(a) The speed of our galaxy as seen from Galaxy A is 0.35 c.

(b) The speed of Galaxy B as seen from Galaxy A is 0.62 c, not 0.70 c .

Sheet # V.

(a) The man should row at an angle given by

sin (theta) = - v_rowing / ( v_walking + v_water)

provided that v_walking > ( v_rowing² - v_water² ) / v_water ; otherwise sin(theta)=- v_water / v_rowing .

The angle is defined relative to a line directed straight across the river; positive angles point downstream.

(b) The trip takes

T = w [ v_walking + v_water + v_rowing sin (theta) ] / [ v_rowing v_walking cos (theta) ]

= w sqrt [ (1 / v_rowing)² (1 + v_water / v_walking )² - (1 / v_walking)² ]

where w is the width of the river.

Sheet # VI.

(a) t = sqrt ( 2 pi r / b )

(b), (c) v = sqrt ( 2 b pi R ) downward

(d)
a_c = - 2 pi b
a_tangential = - b

(e)
The magnitude of the total acceleration is a = b sqrt( 4 pi² + 1 )
The angle is given by tan (theta) = pi + [1 / (2 pi)] , measured from the positive x-axis.

Sheet # VII.

(a) The path will be an ellipse.

(b) v = omega [ x₀ cos (omega t) 1_x - y₀ sin (omega t) 1_y ]

(c) a = - (omega)² r

(d) The velocity is a maximum or minimum whenever x = 0 or y = 0

Chapter 5

5-10.
a = (1/m) [ F₁ cos(theta₁) + F₂ - F₃ cos(theta₃) ]1_x + (1/m) [ F₁ sin(theta₁) - F₃ sin(theta₃) ]1_y

= 0.86 m/sec² 1_x - 0.16 m/sec² 1_y

= 0.88 m/sec², 11^o below the x axis

[numbers from an previous edition of HRW]

5-11. [HRW answers]

(a) a = 0 [Forces in -x direction]

(b) a = 0.83 m/sec² 1_x [Forces in the -x direction]

(c) a = 0 [Forces at 34^o above and below the -x direction]