So if ω 0 increases 3fold, the stopping angle increases 3 2 = 9fold.


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1 Name: MULTIPLE CHOICE: Questions 111 are 5 points each. 1. A safety device brings the blade of a power mower from an angular speed of ω 1 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 ω 3 that is three times as great, ω 3 = 3ω 1? a. 1/9 revolution b. 1/3 revolution c. 1 3 revolution d. 3 revolutions e revolutions f revolutions ω ω = αδθ 0 ω0 Δθ = α So if ω 0 increases 3fold, the stopping angle increases 3 = 9fold.. 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. 1/1 ML b. 1/3 ML c. 1/ ML d. /3 ML e. 3/4 ML f. 5/6 ML g. ML h. 5/4 ML i. 3/ ML j. ML Parallelaxis theorem: I = I CM + Md, where d = distance from axis to center of mass. Here d = L/ so Md = ½ ML and I CM = 1/1 M(L ) = 1/6 ML, so I = (1/ + 1/6) ML = /3 ML. L PHYS 110 Exam 3 1 of 7
2 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 /3 translational. b. /3 rotational and 1/3 translational. c. 1/5 rotational and 4/5 translational. d. 4/5 rotational and 1/5 translational. e. /5 rotational and 3/5 translational. f. 3/5 rotational and /5 translational. g. /7 rotational and 5/7 translational. h. 5/7 rotational and /7 translational. i. 1/ rotational and 1/ translational. Total kinetic energy is translational plus rotational, K = 1/ mv + 1/ Iω. Because it is rolling without slipping, v = ωr and ω = v/r. Because it is a uniform solid sphere, I = /5 mr. So we have K = ½ mv + ½ (/5 mr )ω = ½ mv + 1/5 mv = 7/10 mv. 1/ = 5/10 and 1/5 = /10, so the rotational kinetic energy is /7 of the total and the translational kinetic energy is 5/7 of the total. 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? In an incompressible fluid, pressure p = p 0 + ρgh, a. The pressure increases. where p 0 is pressure at the top, ρ is the fluid b. The pressure does not change. density, g is gravitational field strength, and h is depth. The deeper h the float gets, the greater the c. The pressure decreases. pressure on it. 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. 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. 6. Water flows through a pipe with a narrowing radius, as shown in the diagram. Where is the volume flow rate of the water the fastest? a. At point A. b. At point B. c. It is the same at A and B. By continuity of flow, the volume flow rate is the same at all points in unbranched flow. A B PHYS 110 Exam 3 of 7
3 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. 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 crosssectional area. Where A is smaller (point B), v must be faster. 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. By the Bernoulli equation, p A + ρgy A + ½ ρv A = p B + ρgy B + ½ ρv B. The heights y A and y B are the same and v B > v A, so p B < p A. 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? Gravitational force is a. 5/40 the gravitational pull of Earth s Moon. Gm 1 m /r. Both forces have b. 1/(5 40) the gravitational pull of Earth s Moon. the same G and m 1 (the mass c. 5/40 the gravitational pull of Earth s Moon. of the sun). The smaller mass d. 1/(5 40 dimishes the force by a factor ) the gravitational pull of Earth s Moon. of 5, and the greater distance e. 1/(5 40 ) the gravitational pull of Earth s Moon. dimishes the force by a factor f. 40/5 the gravitational pull of Earth s Moon. of 40. g. 1/(5 40) the gravitational pull of Earth s Moon. h. 5 /40 the gravitational pull of Earth s Moon. i. 5 /40 the gravitational pull of Earth s Moon. k. 40 /5 the gravitational pull of Earth s Moon. PHYS 110 Exam 3 3 of 7
4 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 B 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? PHYS 110 Exam 3 4 of 7
5 FREE RESPONSE: Questions 114 are 15 points each. Show all your work. 1. 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.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. L 1 = L I 1 ω 1 = I ω ω = ω 1 I 1 /I Now our task is to find the ratio of moments of inertia I 1 /I. We treat each astronaut as a point particle, each of which has a moment of inertia I = mr. 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) : I = (7/5) I 1. ω = ω 1 (5/7) = (.00 rad/s)(5/49) = 1.0 rad/s. PHYS 110 Exam 3 5 of 7
6 13. The Moon averages m from Earth and orbits Earth with a period of 7.3 days ( s). The Moon s mass is kg and the Earth s mass is 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.) The circumference of the orbit is πr and it travels this distance in time T, so its orbital speed is v = πr/t. Its kinetic energy then is 1/ mv = m(πr/t). m(πr/t) = ( kg)[π( m)/( s)] = J. b. How much additional translational kinetic energy would the Moon need to gain to escape from Earth orbit? 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 = ( Nm /kg )( kg)( kg)/( m) = J Difference = J J = J. This is the origin of the expression orbit is halfway to the stars. PHYS 110 Exam 3 6 of 7
7 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/m 3. a. What is the water level in the rain barrel? p 1 + ρgy 1 + ½ ρv 1 = p + ρgy + ½ ρv We ll take position 1 as the water line and position as the hole at the bottom. Since both are exposed to the atmosphere, p 1 = p. The height of the water line, which we want to find, we will call h, and h = y 1 y. If the water level does not significantly change, v 1 = 0. Then Oh, that s Torricelli s theorem. p 1 p + ρg(y 1 y ) + 0 = ½ ρv ρgh = ½ ρv v = gh v = gh h = v /(g) h = (3.50 m/s) /(19.6 m/s ) = (1.5/19.6) m = 0.65 m. b. How fast is the barrel losing water in liters/second? (1000 liters = 1 m 3 ) Mass flow rate dv/dt = va = v (πr) = (3.50 m/s)(π)( m) = m 3 /s = ( m 3 /s)(1000 L/m 3 ) = L/s. PHYS 110 Exam 3 7 of 7
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