Here's another conversion to know, this time it's about angles. The most natural unit of angular measure is not the degree, but the radian. We can generate 1 radian in the following manner:

1) take a circle
2) remove its radius
3) squash the radius up against the perimeter of the circle
4) the angle thus formed is the radian

Notice that the radian measure is the same for large and small circles.

Identify the following central angles:

38.	39.	40.
41.	42.	43.
38. Ø = 2 m/2 m Ø = 1 rad	39. Ø = 4 m/2 m Ø = 2 rad	40. Ø = 1 m/2 m Ø = .5 rad
41. Ø = 9 m/6 m Ø = 1.5 rad	42. Ø = 5 m/25 m Ø = .2 rad	43. Ø = s/r

The last relationship, Ø = s/r, should be memorized. (Actually, memorized is a poor term. Most relationships in physics are so reasonable that memory becomes automatic once the principle is used a little.)

Now we will convert from radians to degrees and vice-versa. Recall that the circumference of a circle is given by C = 2(3.14)r. That is, 2(3.14) radii can be wrapped neatly around a circle.

2(3.14) r's stretched out	2(3.14) r's wrapped around circle

Note that the interior angle defined by the end points of this wrapping is exactly one complete circle, or 360°. But in our new improved notation, it is 2(3.14) radians. Thus we may say that 360° = 2(3.14) radians.

44. Convert 1.6 radians to degrees.

(1.6 rad)(360°/2(3.14) rad) = 91.7°

45. Convert to radians or degrees, whichever is appropriate.
a) 38°
b) 0.22 rad
c) 3.14 rad
d) 422°

a) (38°)(2(3.14) rad/360°) = .66 rad	b) (.22 rad)(360°/2(3.14) rad) = 12.6°
c) (3.14 rad)(360°/2(3.14) rad) = 180°	d) (422°)(2(3.14) rad/360°) = 7.37 rad

46. Convert 2.0 complete revolutions to radians.

(2.0 rev)(2(3.14) rad/1 rev) = 4(3.14) rad

47. Convert 0.3 revolutions to radians and 1.12 x 10³ rev to rad.

(.3 rev)(2(3.14) rad/1 rev) = 1.88 rev (1.12 * 103 rev)(2(3.14) rad/1 rev) = 7.04 * 103 rad

48. Convert 63.6 radians and 4.4 x 10³ radians to revolutions.

(63.6 rad)(1 rev/2(3.14) rad) = 10.1 rev (4.4 * 10-3rad)(1 rev/2(3.14) rad) = 7.0 * 10-4 rev

49. A motor shaft turns at 28,000 rev/min. Convert this to (a) rad/sec and (b) deg/sec.

a) (28000 rev/min)(2(3.14) rad/1 rev)(1 min/60 sec) = 2.93 * 103 rad/sec

b) (28000 rev/min)(360°/1 rev)(1 min/60 sec) = 1.68 * 105 °/sec

50. Find the central angle in degrees.

a)

b)

a) Ø = (4/6 rad)(360°/2(3.14) rad) Ø = 38.2°

b) Ø = [(12.1/3.6) rad](360°/2(3.14) rad) Ø = 193°

51. Earth's moon has a diameter of 3476 km, and is located 3.84 x 10⁵ km from Earth. What is its apparent size in degrees, as seen from Earth?

Ø = [(3476/3.84 * 105) rad](360°/2(3.14) rad) Ø = .52°

52. The sun has the same apparent size as the moon. (That's why solar eclipses come out so nicely.) If the sun is 9.3 x 10⁷ miles away, what is its diameter? You will need to use problem 51 in this.

Ø = s/r s = rØ s = (9.3 * 107 mi)[(3476/3.84 * 105) rad] s = 8.4 * 105 mi That is, the diameter of the sun is its distance times its apparent size in radius.

53. A merry-go-round rotates at 0.077 rad/sec. How long will it take to complete one full revolution?

We have rad/sec, but we want sec/rev. Thus (sec/rad) * (rad/rev) will give us our answer. (1 sec/.077 rad)(2(3.14) rad/1 rev) = 81.6 sec/rev

54. A wheel of 80 cm diameter spins at 142 rad/sec.
a) How far does a point on the rim travel in 1 second?
b) How fast is a point 6 cm from the center traveling?

a) In 1 second it rotates: (142 rad/sec)(1 sec) = 142 rad So the rim travels s = rØ: (80/2 cm)(142 rad) = 5680 cm

b) v = s/t v = rØ/t v = [(6 cm)(142)]/(1 sec) v = 852 cm/sec

55*. Find the rate of angular motion, in rad/sec, of Earth about the sun.

It takes 365 days for the earth to make one complete orbit. angular motion = (1 rev/365 days)(2(3.14) rad/1 rev)(1 day/24 hr)(1 hr/60 min)(1 min/60 sec) angular motion = 2.0 * 10-7 rad/sec

56*. A beetle takes a joy ride on a pendulum. The string of the pendulum is 183 cm long. If the beetle rides through a swing of 40°, how far has he traveled along the path of the pendulum? (Your answer will be in centimeters.)

First we need the angle in radians through which the pendulum has swung: (40°)(2(3.14) rad/360°) = .70 rad s/r = Ø s/r = .70 rad s = (.70)r s = (.70)(183 cm) s = 128 cm

57*. If the aforementioned (problem 57) beetle swings through 0.15 radians, how far has he traveled?

s = rØ s = (183 cm)(.15) s = 27.5 cm

58*. If the pendulum (problem 57) at some instant is swinging at 1.4 rad/sec, how fast is the beetle traveling in cm/sec?

In one second the beetle travels 1.4 rad, so: Ø = s/r 1.4 = s/183 cm s = (1.4)(183 cm) s = 256 cm Thus the speed is 256 cm/sec

59*. Convert the following into linear speed of the beetle (yes, still from problem 57).
a) 0.88 rad/sec
b) 2.4 x 10^-5 rad/sec
c) 38 °/sec
d) 127 °/sec

a) (.88 rad/sec)(183 cm) = 161 cm/sec	b) (2.4 * 10-5 rad/sec)(183 cm) = 4.39 * 10-3 cm/sec
c) (38 °/sec)(2(3.14) rad/360°)(183 cm) = 121 cm/sec	d) (127 °/sec)(2(3.14) rad/360°)(183 cm) = 406 cm/sec

Note that since a radian is dimensionless, it can be dropped in and out of any unit.

60*. A small wheel, with radius of 1.6 cm, drives a large wheel of 14.1 cm radius by their circumferences being pressed together. If the small wheel turns at 480 rad/sec, what does the big one turn at?

The rim of the small wheel moves at: v = (480 rad/sec)(1.6 cm) v = 768 cm/sec The rim of the large wheel is forced to move at the same speed, as the rims are in contact. Thus... w = (768 cm/sec)/(14.1 cm) w = 54 rad/sec

61*. A bucket is pulled from a well by a hand-powered winch. The handle on the winch has a radius of 86 cm, the axle has a radius of 11 cm. If the handle is turned at 80 cm/sec, how fast does the bucket rise?

w = velocity/radius w = (80 cm/sec)/(86 cm) w = .930 rad/sec velocity = w * r v = (.930 rad/sec)(11 cm) v = 10.2 cm/sec

I have introduced the radians and rotational velocities because they are concepts that need some time to be thoroughly internalized. We will return to them periodically until they come out full blast in Circular and Rotational motion. Be sure you have made sense out of the last few problems, as they will help immensely later in the course.

Another concept that's good to be exposed to early is density. The Greek letter p (pronounced "ro") is used to represent density.

62. The density of a block of clay is 1.4 gm/cm³. What is the mass of 288 cm³ of clay? Hint: Let the units be your guide.

m = (1.4 gm/cm3)(288 cm3) m = 403 gm

63. A block of metal measures 3.6 cm x 4.2 cm x 6.7 cm. What is its density if the mass is 346 gm?

p = mass/volume p = 346 gm/[(3.6 cm)(4.2 cm)(6.7 cm)] p = 3.42 gm/cm3

64. A sphere whose radius is 5.7 cm has a mass of 426 gm. What is its density? (Recall that for a sphere, V = 4/3 r³.)

p = mass/volume p = m/[(4/3)(3.14)(r3)] p = (426 gm)/[(4/3)(3.14)(5.7)3] p = 0.55 gm/cm3

65. A brass ball, density 5.6 gm/cm³, has a radius of 6.4 cm. What is its mass?

p = mass/volume m = p * V m = (5.6 gm/cm3)[4/3(3.14)](6.4 cm)3 m = 6.15 * 103 gm

66. A 15 gm marble has a density of 1.6 gm/cm³. What is its radius?

p = m/V V = m/p V = (15 gm)/(1.6 gm/cm3) V = 9.38 cm3 But 9.38 cm3 = [4/3(3.14)r3] r = 1.31 cm

67. A cylinder, 5.0 cm in diameter, and 13 cm high, has a mass of 2.64 kg. What is its density in gm/cm³?

For a cylinder, V = (3.14)r2h (area of base times height) p = m/V p = [(2.64 kg)(1000 gm/1 kg)]/[(3.14)(2.5 cm)2(13 cm)] p = 10.3 gm/cm3

68*. A 422 gm cylinder is 2.2 inches high and has a density of 3.4 gm/cm³. What is its radius in cm? (Be careful of the mixed unit systems.)

First get everything into the same system: (2.2 in)(2.54 cm/1 in) = 5.59 cm Next: p = m/V p = m/[(3.14)r2h] r = [m/[(3.14)hp]]1/2 r = [(422 gm)/[(3.14)(5.59 cm)(3.4 gm/cm3)]]1/2 r = 2.66 cm

69*. A piece of pipe has an outer radius of 4.7 cm and an inner radius of 2.6 cm. What is the mass of a 35 cm length of the pipe if its density is 8.4 gm/cm³?

To get the volume of the pipe, calculate the volume of the outer cylinder and subtract the volume of the hole. m = pV m = (8.4 gm/cm3)[(3.14)(4.7 cm)2(35 cm) - (3.14)(2.6 cm)2(35 cm)] m = 1.42 * 104 gm

70*. A hollow sphere has an inner radius of 7.6 cm, an outer radius of 11.4 cm, and a density of 3.8 gm/cm³. What is its mass?

As before: m = pV m = (3.8 gm/cm3)[(4/3)(3.14)(11.4 cm)3 - (4/3)(3.14)(7.6 cm)3] m = 1.66 * 104 gm

I hope you noticed that in #70 and #71 that the terms could be simplified.

V = (3.14)r₁²h - (3.14)r₂²h = (3.14)h(r₁² - r₂²)
V = (4/3)(3.14)r₁³ - (4/3)(3.14)r₂³ = (4/3)(3.14)(r₁³ - r₂³)

71*. A cylinder, 15.0 cm long and 2.5 cm in radius, is made of two different metals bonded to make a single bar. If the densities are 4.1 gm/cm³ and 6.3 gm/cm³, what length of the lighter metal is needed? The total mass is 1500 gm.

m = p1V1 + p2V2 m = (4.1 gm/cm3)[x * (3.14) * (2.52 cm2)] + (6.3 gm/cm3)(15 - x)[(3.14) * (2.52 cm2)] m = 1500 gm This messy algebraic equation can be simplified to: 80.5x + 1855 - 123.70x = 1500 x = 8.22 cm

72**. A hollow sphere has an outer radius of 21.6 cm and masses 6.8 x 10⁴ gm. When the sphere is filled with liquid whose density is 4.8 gm/cm³ the total mass rises to 16.5 x 10⁴ gm. What is the density of the material making up the hollow sphere?

For the inner sphere, m = pV The mass increase was: (16.5 * 104 - 6.8 * 104 gm) = (4.8 gm/cm3)(4/3(3.14)r3) r = 16.9 cm So, for the outer sphere: (6.8 * 104 gm) = p[4/3(3.14)(21.63 - 16.93) cm3] p = 3.09 gm/cm3

One last skill you will need for physics is the ability to manipulate expressions without using numbers. While the geometry of the next few problems is not used much in the rest of the course, the algebraic methods will be pervasive.

73. Express the perimeter of each figure in terms of the given dimensions.

a)

b)

c)

a) P = 2(s + 2s) P = 2(3s) P = 6s

b) P = 2s + s + [s2 + (2s)2]1/2 P = 3s + (5s2)1/2 P = [3 + 51/2]s

c) P = 2(3.14)r

74. Now for some slightly harder ones:

a)

b)

c)

a) For 1/2 circle: P1 = 1/2[2(3.14)r] P1 = (3.14)r For base: P2 = 2r Total: P = P1 + P2 P = 2r + (3.14)r P = [2 + (3.14)]r

b) Slant side: (s2 + s2)1/2 = (2)1/2s Total: P = 2(2s) + 2(s) + (2)1/2s P = [6 + (2)1/2]s

c) P = 2r + 4r + 2r + r + (3.14)r + r P = [10 + (3.14)]r

75. Find the surface area of each object:

a)

b)

c)

a) Top and Bottom (sides #1): A1 = 2(s * 2s) A1 = 4s2 Front and Back (sides #2): A2 = 2(s * 2s) A2 = 4s2 Ends (sides #3): A3 = 2(s * s) A3 = 2s2 A = A1 + A2 + A3 A = 10s2

b) A1 = 2[(3.14)r2] A2 = [2(3.14)r]h A = A1 + A2 A = [2(3.14)r2 + 2(3.14)rh] A = [2(3.14)r(r + h)]

c) A = (2r)2 - 1/4(3.14)r2 A = [4 - (3.14/4)]r2

I hope you've noticed the units. For perimeter, which is a linear measure, the expressions are all in the first power of length. For area, they are in the second power of length, for volume, the third power. There are no exceptions to this rule. Examples: 4 (3.14)r2 must be an area. (3.14)r2h must be a volume. 2L + 2W must be a perimeter.

76. Find the volumes of the figures.

a)

b)

c)

a) V = 1/2(b * h) * l V = (bhl)/2

b) V = [(3.14)r2]h V = (3.14)r2h

c) V = [s2 - (3.14)(s/4)2]t V = s2t[1 - (3.14/16)]

Remember that for cylinders, whether they are circular cylinders, triangular cylinders, or anything else, their volume is given by the area of the base times the height.

77*. Find the areas and volumes of the figures below.

a)

b)

c)

a) A1 = 2[1/2(3.14)r2] A2 = 1/2[2(3.14)r](2r) A2 = 2(3.14)r2 A3 = (2r)(2r) A3 = 4r2 A = A1 + A2 + A3 A = [4 + 3(3.14)]r2 V = [1/2(3.14)r2](2r) V = (3.14)r3

b) A1 = 2[(2r)2 - 1/4[(3.14)r2]] A1 = [8 - (3.14/2))r2 A2 = 2(hr) A2 = 2hr A3 = 1/4[2(3.14)r]h A3 = [(3.14)rh]/2 A4 = 2(2r * h) A4 = 4rh A = [6 + (3.14/2)]rh + [8 - (3.14/2)]r2 V = [(2r)2 - 1/4(3.14)r2]h V = [4 - (3.14/4)]r2h

c) A1 = 1/2[4(3.14)r22] A1 = 2(3.14)r22 A2 = 1/2[4(3.14)r12] A2 = 2(3.14)r12 A3 = (3.14)r22 - (3.14)r12 A3 = (3.14)(r22 - r12) A = 2(3.14)(r22 + r12) + (3.14) (r22 - r12) V = 1/2[(4/3)(3.14)r23] - 1/2[(4/3)(3.14)r13] V = [2(3.14)/3][r23 - r13]

78. Karen Pheden can run around the unusual track shown in time t. What is her average velocity? (Express v in terms of r and t.)

velocity = distance/time velocity = [1/2(2)(3.14)r + 2r]/t velocity = [(3.14) + 2]r/t

79*. C. D. Twain can drive his car at a constant velocity v over any race track in the world. How long will it take him to complete the one at left?

velocity = distance/time time = distance/velocity time = [4r = 4[1/4(2)(3.14)r]]/v time = [4 + 2(3.14)]r/v

79. The object shown at right has a density r. What is its mass?

p = m/V m = pV m = p[(3.14)r22 - (3.14)r12]h m = (3.14)ph(r22 - r12)

80*. If the quarter cylinder shown were to be melted down and poured into a spherical shape, what would the radius of the completed sphere be?

Let R = radius of final sphere: 4/3(3.14)R3 = 1/4(3.14)r2h R3 = 3/16r2h R = (3/16r2h)1/3

81** What size cube will have the same surface area as the hemisphere shown?

The surface area of the object is: A = 1/2[4(3.14)r2] + (3.14)r2 * 3(3.14)r2 The surface of a cube of edge S is: A = 6S6 Thus, 3(3.14)r2 = 6S6 S = [(3.14)/2]1/2r