Assignment #1

Section 11-2

Due: Wed. Jan. 9

1-1. You drop a 3.50 kg rock from a height of 1.25 m on Mars whose mass is 6.419x10 ²³kg and mean equatorial radius is 3393 km {on the Moon whose mass is 7.349x10 ²²kg and mean radius is 1738 km}.

(A) At what rate will the rock accelerate downward ?
(B) How long will the rock take to reach the ground ?

1-2. At what distance above the Earth’s surface will the acceleration of gravity drop by 10.0% {20.0%} of value at the Earth’s surface of 9.80 m/s ². The equatorial radius of the Earth is 6378 km and its mass is 5.974x10 ²⁴kg.

1-3. The mass of the Moon is 7.349x10 ²²kg {Sun is 1.99x10 ³⁰kg} and its mean distance from the Earth is 3.84x10 ⁸m {1.49x10 ¹¹m}.

(A) What is the magnitude of the gravitational force between the Earth and the Moon {Sun}?
(B) At what distance from the center of the Earth (between the Earth and the Moon {Sun}) will the total gravitational force of the Earth and the Moon {Sun} on another object vanish (i.e. at what point will the gravitational force of Earth and the Moon be equal)?

{Ans: #1}

{1-1.} (A) 1.62 m/s 2 (B) 1.24 s

{1-2}. 753 km

{1-3.} (A) 3.56x1022N (B) 2.58x108m

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Assignment #2

Section 11-3

Due: Fri. Jan. 11

2-1. An 5.90 kg object is launched straight up at 9.50 km/s {8.10 km/s} from the surface of the Earth at the Earth’s North Pole (polar radius 6356 km).

(A) What is the initial Mechanical Energy of the object relative to a base line at infinity ?
(B) How high will such an object rise above the Earth’s surface neglecting air resistance ?
BONUS:
(C) If object were launched straight up from the Earth’s Equator (6378 km), how high would the object rise above the Earth’s surface ? Hint use the conservation of angular moment to find the object’s final tangential velocity when it reaches its maximum altitude and assume that the change in the Earth’s angular momentum in negligible.

2-2. Determine the velocity that an air molecule would need to escape from the surface of the Moon {Mercury m = 7.28x10 ²³kg, R = 2400 km} ? Both the Moon and Mercury rotates slowly so you can neglect any rotational effects.

2-3. If an object of 3.00 kg {11.0 mg} object falls (from rest) from a great distance (infinity) into a black hole, calculate the kinetic energy of the object when it reaches the event horizon, R = 2GM/c².

{Ans: #2}

{2-1.} (A) -176 MJ (B) 6970 km

{2-2.} 6.36 km/s

{2-3.} 4.95x1011J

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Assignment #3

Notes on Kepler’s 1 st Law

Due: Mon. Jan. 14

3-1. A moon of Neptunian is in a nearly circular orbit at 73,000 miles {150,000 miles} from the planet's center. If Neptune has a mass of 1.03x10 ²⁶kg,

(A) at what speed in km/s is it moving in its orbit about Neptune ?
(B) how long (in days) does this moon take to orbit Neptune once ?

3-2. Mars’ closest moon Phobos ( m = 9.60x10 ¹⁵kg) {fictitious moon ( m = 2.70x10 ¹⁵kg)} orbits Mars once every 7.66 hours {30.3 hr}. When Phobos {the fictitious moon} is at its closest, it is 9,210 km {18,000 km} from Mars’ center. Recall that the mass of Mars is 6.419x10 ²³kg.

(A) What is semimajor axis of the moon in its orbit about Mars ?
(B) What is the slowest speed at which the moon will be moving ?
(C) What is the mechanical energy of the moon ?

{ANS: #3}

3-1. (A) v = 5.34 km/s (B) T = 3.29 days

3-2. (A) 23,500 km (B) 1.07 km/s (C) -2.47x1021J

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Assignment #4

Notes Kepler’s 2nd Law

Date Due: Tue. Jan. 15

4. A rock is launched from the surface of the Earth (assume R = 6378 km} at a speed of 2.60 km/s {7.80 km/s} at an angle of 32.0 ⁰ relative to local horizontal. Neglecting air resistance and any rotational motion of the Earth:

(A) Determine the semimajor axis of the rock’s orbit about the center of the Earth.
(B) Determine the eccentricity of the rocks orbit about the center of the Earth.
(C) What is the maximum altitude the rock will reach above the Earth?
(D) How far will the rock hit from it launch point measured along the surface of the Earth?

{ANS:#4}

(A) a = 6210 km

(B) e = .530

(C) h = 3130 km

(D) s = 12,400 km

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Assignment #5

Section 14-1

Due: Wed. Jan. 16

5. Each blade of an electric shaver moves back and forth in simple harmonic motion over a distance of 3.30 mm {2.50 mm}; making 1500 {1800} complete oscillations every 15.0 seconds.

(A) What is the angular frequency of the end of the blade ?
(B) What is the maximum acceleration {maximum speed} with which the end of the blade moves?
(C) Find the equation of motion of the end of the blade if t = 0 corresponds to the moment when the end of the blade is at its center of motion (i.e., at x = 0) and moving in the negative {positive} x-direction along the surface of the shaver head.
(D) What is the speed of the end of the blade and it direction of motion, T/3 seconds after the starting time t = 0, i.e., when the blade is one third of the way through its period ?
BONUS:
(E) What is distance of the end of the blade from its starting position (along the surface) when the blade's speed is one third its maximum speed ?

{ANS:} #5

(A) 754 rad/s

(C) x(t) = (1.25 mm) sin[(754 rad/s) t]

(D -47.1 cm/s to left

(E) ±1.18 mm

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Assignment #6

Section 14-2

Due: Fri. Jan.18

6. A 1.80 kg {36.0 kg} block (on a frictionless, horizontal surface) is attached to a massless, horizontal spring with a spring constant of 29.0 N/cm. The block is pushed to the left so that the spring is compressed 12.0 cm and then the block is given an initial speed of 4.70 m/s {53.0 cm/s} to the right.

(A) What is the total mechanical energy of the block-spring system ?
(B) What is the amplitude of the block's oscillation ?
(C) What is period with which the block oscillates ?
(D) Determine the equation of motion of the block’s velocity.
BONUS:
(E) what is the block kinetic energy when the block's acceleration is ±108 m/s ² {±6.40 m/s ²} ?

{ANS:} #6

(A) 25.9 J

(B) 13.4 cm

(C) 700 ms

(D) v(t) = (1.20 m/s) cos[(8.98 rad/s) t - 1.11 rad] or v(t) = (1.20 m/s) sin[(8.98 rad/s) t +.460 rad]

(E) 16.8 J

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Assignment #7

Section 14-3

Due: Tue. Jan. 22

7. A physical pendulum is constructed from a 396 cm {2.60 m} long, thin, vertical rod that is pivoted 83.0 cm from its top. The mass of the rod is 3.90 kg. The rod is pulled to the side through an angle of 3.80 ^o and released at time t = 0.

(A) What is the moment of inertia of the rod about this pivot point ?
(B) What is the period of oscillation of the rod?
(C) What is the angular speed of the rod when the rod reaches the bottom of its swing, i.e., when the rod is vertical ?
(D) How long will it take for the rod to be at an angle of 2.50 ^o from vertical after it is released ?
BONUS: (for those who enjoy a good calculus problem)
(E) Express the period as function of the pivot distance from the rod's center of mass, call it x, and determine the value of the pivot distance x at which the period of the rod will be a minimum.

{ANS:} #7

(A) I = 3.06 kg-m 2

(B) 2.59 s

(C) -.161 rad/s

(D) .352 s

(E) x = 75.1 cm from center of rod

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