Name:
MULTIPLE CHOICE: Questions 1-11 are 5 points each.
1. A safety device brings the blade of a power mower from an angular speed of ω1 to rest in 1.00 revolution. At the same constant angular acceleration, how many revolutions would it take the blade to come to rest from an initial angular speed ω3 that is three times as great, ω3 = 3ω1?
a. 1/9 revolution b. 1/3 revolution c.
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1 3 revolution d.
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3 revolutions e. 3.00 revolutions f. 9.00 revolutions
2. What is the moment of inertia of a square plate of mass M and length L along a side when it is rotated about an axis that is perpendicular to the plane of the plate and through a corner of the square?
a. 1/12 ML2 b. 1/3 ML2 c. 1/2 ML2 d. 2/3 ML2 e. 3/4 ML2 f. 5/6 ML2 g. ML2 h. 5/4 ML2 i. 3/2 ML2 j. 2 ML2
L α
θ ω
θ α ω ω
2 2
2 0 2 0 2
= − Δ
Δ
=
−
So if ω0 increases 3-fold, the stopping angle increases 32 = 9-fold.
Parallel-axis theorem: I = ICM + Md2, where d = distance from axis to center of mass. Here d = L/ 2 so Md2 = ½ ML2 and ICM = 1/12 M(2L2) = 1/6 ML2, so I = (1/2 + 1/6) ML2 = 2/3 ML2.
3. A uniform solid sphere rolls without slipping along a level surface. What fraction of its total kinetic energy is rotational, and what fraction is translational?
a. 1/3 rotational and 2/3 translational.
b. 2/3 rotational and 1/3 translational.
c. 1/5 rotational and 4/5 translational.
d. 4/5 rotational and 1/5 translational.
e. 2/5 rotational and 3/5 translational.
f. 3/5 rotational and 2/5 translational.
g. 2/7 rotational and 5/7 translational.
h. 5/7 rotational and 2/7 translational.
i. 1/2 rotational and 1/2 translational.
4. A fish bites at a baited hook and swims downward, pulling the fishing line and float down with it. As the fish pulls the float deeper below the surface, how does the inward pressure on the float change?
a. The pressure increases.
b. The pressure does not change.
c. The pressure decreases.
5. In the scenario of the previous question, how does the upward force of buoyancy acting on the submerged float change as the fish pulls the float deeper under water?
a. The buoyancy force increases.
b. The buoyancy force does not change.
c. The buoyancy force decreases.
Total kinetic energy is translational plus rotational, K = 1/2 mv2 + 1/2 Iω2. Because it is rolling without slipping, v = ωr and ω = v/r. Because it is a uniform solid sphere, I = 2/5 mr2. So we have K = ½ mv2 +
½ (2/5 mr2)ω2 = ½ mv2 + 1/5 mv2 = 7/10 mv2. 1/2 = 5/10 and 1/5 = 2/10, so the rotational kinetic energy is 2/7 of the total and the translational kinetic energy is 5/7 of the total.
In an incompressible fluid, pressure p = p0 + ρgh, where p0 is pressure at the top, ρ is the fluid density, g is gravitational field strength, and h is depth. The deeper h the float gets, the greater the pressure on it.
The buoyancy force is ρgV, where ρ is the fluid density, g is the gravitational field strength, and V is submerged volume. In an incompressible fluid, none of these
quantities depends on depth.
7. In the scenario of the previous question, where is the speed of the water the fastest?
a. At point A.
b. At point B.
c. It is the same at A and B.
8. In the scenario of the previous question, where is the pressure of the water the greatest?
a. At point A.
b. At point B.
c. It is the same at A and B.
9. The Kuiper Belt Object known as Pluto is about 1/5 the mass of Earth’s Moon and averages about 40 times farther from the Sun than Earth’s Moon. Compared to the average gravitational pull Earth’s Moon exerts on the sun, how strong is the gravitational pull that Pluto exerts on the sun?
a. 5/40 the gravitational pull of Earth’s Moon.
b. 1/(5·40) the gravitational pull of Earth’s Moon.
c. 5/402 the gravitational pull of Earth’s Moon.
d. 1/(52·402) the gravitational pull of Earth’s Moon.
e. 1/(5·402) the gravitational pull of Earth’s Moon.
f. 40/5 the gravitational pull of Earth’s Moon.
g. 1/(52·40) the gravitational pull of Earth’s Moon.
h. 52/40 the gravitational pull of Earth’s Moon.
i. 52/402 the gravitational pull of Earth’s Moon.
k. 402/5 the gravitational pull of Earth’s Moon.
The volume flow rate is the same at all points, and it is equal to vA, where v is the fluid speed and A the cross-sectional area.
Where A is smaller (point B), v must be faster.
By the Bernoulli equation, pA + ρgyA + ½ ρvA
2 = pB + ρgyB + ½ ρvB 2. The heights yA and yB are the same and vB > vA, so pB < pA.
Gravitational force is Gm1m2/r2. Both forces have the same G and m1 (the mass of the sun). The smaller mass dimishes the force by a factor of 5, and the greater distance dimishes the force by a factor of 402.
10. A giant windmill with 3 blades of length L is turning in the wind. Point A is at the tip of the blade while point B is halfway out from the axis of rotation. Which of the following is true?
a. The angular speed of A is twice that of B, but they have the same linear speed.
b. The angular speed of B is twice that of A, but they have the same linear speed.
c. The linear speed of A is twice that of B, but they have the same angular speed.
d. The linear speed of B is twice that of A, but they have the same angular speed.
e. They both have the same angular and linear speeds.
11. A disk is rotating counter clockwise, and its angular speed is increasing. Which figure below best shows the direction of the net force at the right edge of the disk?
A B
FREE RESPONSE: Questions 12-14 are 15 points each. Show all your work.
12. Dr. Ryan Stone and Lt. Matt Kowalsi, stranded in orbit, are connected by a cable 5.00 m long. They rotate about their mutual center of mass at a rate of 2.00 rad/s. The cable slips and they drift apart until Lt. Kowalski catches the cable again when they are 7.00 m apart. What is their new rate of rotation about their center of mass?
Dr. Stone’s mass is 50.0 kg and Lt. Kowalski’s mass is 80.0 kg. You may treat each astronaut as a point particle. The mass of the cable can be neglected.
This is a conservation of angular momentum problem.
L1 = L2
I1ω1 = I2ω2
ω2 = ω1 I1/I2
Now our task is to find the ratio of moments of inertia I1/I2. We treat each astronaut as a point particle, each of which has a moment of inertia I = mr2. Their masses do not change, but their distances from the center of mass increase by a factor of 7/5.
Thus their moments of inertia increase by a factor of (7/5)2: I2 = (7/5)2 I1. ω2 = ω1(5/7)2 = (2.00 rad/s)(25/49) = 1.02 rad/s.
13. The Moon averages 3.84×108 m from Earth and orbits Earth with a period of 27.3 days (2.36×106 s). The Moon’s mass is 7.35×1022 kg and the Earth’s mass is 5.97×1024 kg.
a. What is the Moon’s orbital kinetic energy? (Its kinetic energy of traveling around the Earth, ignoring energy of rotation about its axis.)
b. How much additional translational kinetic energy would the Moon need to gain to escape from Earth orbit?
The circumference of the orbit is 2πr and it travels this distance in time T, so its orbital speed is v = 2πr/T. Its kinetic energy then is 1/2 mv2 = 2m(πr/T)2. 2m(πr/T)2 = 2(7.35×1022 kg)[π(3.84×108 m)/(2.36×106 s)]2 = 3.84×1028 J.
The Moon’s gravitational potential energy is –GMm/r, where G is the gravitational constant, M is Earth’s mass, and m is the Moon’s mass. The total kinetic energy needed to escape Earth would be +GMm/r; the additional energy needed would be this amount minus the translational kinetic energy the Moon already has.
U = –GMm/r = (6.67×10–11 Nm2/kg2)(5.97×1024 kg)(7.35×1022 kg)/(3.84×108 m)
= –7.62×1028 J
14. A rain barrel springs a leak at the very bottom. The hole is perfectly circular and 3.00 mm in diameter. The stream of water coming out has an initial speed of 3.50 m/s.
Meanwhile, the water level in the barrel is not significantly changed. Recall that water has a density of 1000 kg/m3.
a. What is the water level in the rain barrel?
b. How fast is the barrel losing water in liters/second? (1000 liters = 1 m3) p1 + ρgy1 + ½ ρv1
2 = p2 + ρgy2 + ½ ρv2 2
We’ll take position 1 as the water line and position 2 as the hole at the bottom. Since both are exposed to the atmosphere, p1 = p2. The height of the water line, which we want to find, we will call h, and h = y1 – y2. If the water level does not significantly change, v1 = 0. Then
p1 – p2 + ρg(y1 – y2) + 0 = ½ ρv2 2
ρgh = ½ ρv2 2
v2
2 = 2gh Oh, that’s Torricelli’s theorem.
v22 = 2gh h = v2
2/(2g)
h = (3.50 m/s)2/(19.6 m/s2) = (12.25/19.6) m = 0.625 m.
Mass flow rate dV/dt = vA = v (πr2) = (3.50 m/s)(π)(0.0015 m)2 = 2.474×10–5 m3/s
= (2.474×10–5 m3/s)(1000 L/m3) = 0.02474 L/s.
