43. A person sits on a freely spinning lab stool that has no friction in its axle. When this person extends her arms,

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1 43. A person sits on a freely spinning lab stool that has no friction in its axle. When this person extends her arms, A) her moment of inertia increases and her rotational kinetic energy remains the same. B) her moment of inertia increases and her rotational kinetic energy decreases. C) her moment of inertia decreases and her rotational kinetic energy decreases. D) her moment of inertia increases and her rotational kinetic energy increases. E) her moment of inertia decreases and her rotational kinetic energy increases. The moment of inertia of a mass point with mass m a distance r away from the rotation axis is I = mr 2. The total moment of inertia of a solid object is the sum over all mass points of the solid object. When the person extends her arms, she moves some mass further away from the rotation axis and her moment of inertia increases. Rotational kinetic energy is E kin = ½ I ω 2 and angular momentum L is L = I ω. Angular momentum L is conserved when the person extends her arms. Therefore, E kin = ½ L ω = ½ L 2 / I. Since I increases, E kin decreases. 44. A solid uniform disk of diameter 3.20 m and mass 42 kg rolls without slipping to the bottom of a hill, starting from rest. If the angular speed of the disk is 4.27 rad/s at the bottom, how high did it start on the hill? A) 4.28 m B) 3.14 m C) 3.57 m D) 2.68 m E) 1.34 m mgy mv Iω 2 = mgy mv Iω 2 0 mg( y y 0 ) = 1 2 mv Iω 2 v = ωr; I = 1 2 mr 2 mg( y y 0 ) = 3 4 mr 2 ω 2 h = ( y y 0 ) = 3r 2 ω 2 g = 3.57 m

2 45. A wheel having a moment of inertia of 5.00 kg m 2 starts from rest and accelerates under a constant torque of 3.00 N m for 8.00 s. What is the wheel's angular velocity at the end of 8.00 s? A) 4.80 rad/s B) 3.26 rad/s C) 6.45 rad/s D) 5.67 rad/s E) 8.92 rad/s τ = Iα α = τ I = ω = ω 0 + αt = 0.6 rad/s Nm = rad/s kg m ( ) 8.00 s ( ) = 4.80 rad/s 46. From what height above the surface of the earth should an object be dropped to initially experience an acceleration of 0.54g? A) 2300 km B) 2900 km C) 5400 km D) 1700 km E) 8700 km g E = G m E R g(h) = G m E 2 E R E + h ( ) 2 h = Gm E g(h) R E = m 47. An astronaut goes out for a "space-walk" at a distance above the earth equal to the radius of the earth. What is her acceleration due to gravity at that point? A) g/ 2 B) g/4 C) g D) g/2 E) zero g(2r E ) = G m E ( ) 2 = G m E 2R E 4R E 2 = g 4

3 48. A satellite of mass M takes time T to orbit a planet. If the satellite had twice as much mass, the time for it to orbit the planet at the same altitude would be A) T/4 B) T C) T/2 D) 2T E) 4T The orbit period T does not depend on the mass of the satellite, but on the mass M of the planet: T = 4π 2 r 3 GM 49. Suppose you pull a simple pendulum to one side y an angle of 5 o, let go and measure the period of oscillation to be 6 seconds. Then you stop the pendulum, pull the pendulum to an angle of 10 o, and let go. The resulting oscillation will have a period of A) 24 seconds B) 12 seconds C) 3 seconds D) 1.5 seconds E) 6 seconds The period of a pendulum is T = 2π L / g and does not depend on the amplitude.

4 50. A mass on a spring undergoes simple harmonic motion. When the mass is at its maximum distance from the equilibrium position, which of the following statements about it are true? A) Its acceleration is zero. B) Its speed is zero. C) Its elastic potential energy is zero. D) Its kinetic energy is a maximum. E) Its total mechanical energy is zero. When the mass is at its maximum distance from the equilibrium position, either when the spring is maximally stretched on compressed, the magnitude of the acceleration is maximal, the speed is zero, the elastic potential energy is maximal, and the kinetic energy is zero. The total mechanic energy is constant and non-zero for a mass undergoing oscillation. 51. An object is undergoing simple harmonic motion of amplitude 2.3 m. If the maximum velocity of the object is 10 m/s, what is the object's angular frequency? A) 4.3 rad/s B) 4.8 rad/s C) 3.5 rad/s D) 4.0 rad/s E) 2.9 rad/s v max = Aω ω = v max A = 10 m/s 2.3 m = 4.3 rad/s

5 52. Consider a wave on a string moving to the right as shown in the figure. What is the direction of the velocity of a particle of string at point P? A) B) C) D) E) zero The wave on the string is a transverse wave. Therefore particles on the string move perpendicular (up-down) to the wave velocity (to the right). At the instant shown in the figure, the wave crest to the left of point P is moving towards point P and the particle of string is moving away (upward) from the string. 53. What is the frequency of the fundamental mode of vibration of a steel piano wire stretched to a tension of 440 N? The wire is m long and has a mass of 5.60 g. A) 366 Hz B) 181 Hz C) 234 Hz D) 312 Hz E) 517 Hz v = F T m / L = ( 440 N) kg ( ) / ( m) ( 217 m/s) 2( m) λ 0 = 2L f 0 = v / λ 0 = v 2L = = 217 m/s = 181 Hz

6 54. What is the ratio of the intensities of two sounds with intensity levels of 70 db and 30 db? A) 10 : 1 B) 100 : 1 C) 1,000 : 1 D) 10,000 : 1 E) 100,000 : 1 β = (10 db)log I I 0 I = I 0 10β /(10 db) I 70 = 10(70 db)/(10 db) 107 = (30 db)/(10 db) I = = 10, A siren emitting sound of frequency 1000 Hz approaches a stationary observer at onehalf the speed of sound. The observer hears a frequency of A) 750 Hz B) 2000 Hz C) 1333 Hz D) 1500 Hz E) None of the above f ( 1000 Hz) f ' = 1 v = source 1 0.5v = 2000 Hz v v

7 56. Six seconds after a brilliant flash of lightning, thunder shakes the house. Approximately how far was the lightning strike from the house? A) much closer than one kilometer B) about one kilometer away C) about two kilometers away D) much farther away than two kilometers E) It is impossible to say. d = vt = 343 m/s ( ) 6 s ( ) = 2,000 m 57. The figure shows a standing wave in a resonating pipe that is open at one end and closed at the other end. The length of the pipe is L. The wave length of the wave is A) 3 L/4 B) 4 L/3 C) L/4 D) L/2 E) 3 L/2 L = 3 4 λ λ = 4L A ball of lead (density 11.3 g/cm 3 ) is placed in a tub of mercury (density 13.6 g/cm 3 ). Which answer best describes the result? A) The lead ball will float with about 83% of its volume above the surface of the mercury. B) The lead ball will float with about 17% of its volume above the surface of the mercury. C) The lead ball will float with about 69% of its volume above the surface of the mercury. D) The lead will sink to the bottom of the mercury. E) The lead ball will float with its top exactly even with the surface of the mercury. V subm V total = ρ Pb ρ Hg = 3 ( 11.3 g/cm ) = 0.83 = 83% ( 13.6 g/cm 3 )

8 59. What buoyant force does a 0.60-kg solid gold crown experience when it is immersed in water? The density of gold is kg/m 3 and that of water is 1000 kg/m 3. A) 0.30 N B) N C) N D) N E) N F B = m disp g = ρ water V crown g ( ) = ρ water m crown ( ρ gold ) ( ) g = 0.60 kg ( 1000 kg/m 3 ) 19,300 kg/m 3 ( 9.8 m/s2 ) = 0.3 N ( ) 60. A cubical block of stone is lowered at a steady rate into the ocean by a crane, always keeping the top and bottom faces horizontal. Which one of the following graphs best describes the buoyant force B on this block as a function of time t if the block just enters the water at time t = 0 s? Answer B. The buoyant force is equal to the weight of the displaced water. The volume and thus the weight of the displaced water will increase proportional to time until the block is completely submersed. After that the submerged volume stays constant and therefore the buoyant force will also not change anymore.

9 61. What force does the water exert (in addition to that due to atmospheric pressure) on a submarine window of radius 44.0 cm at a depth of 9400 m in sea water? The density of sea water is 1025 kg/m3. A) N B) N C) N D) N E) N P P atm = ρgh = 1025 kg/m 3 ( ) 9.8 m/s 2 ( ) 9400 m ( ) = 94 MPa F = ( P P atm ) A = ( 94 MPa)π ( 0.44 m) 2 = N 62. Ideal incompressible fluid flows through a 4.0-cm-diameter pipe at 1.0 m/s. There is a 2.0 cm diameter constriction in the line. What is the speed of the fluid in this constriction? A) 0.25 m/s B) 0.50 m/s C) 2.0 m/s D) 4.0 m/s E) 16 m/s 2 2 A A 1 v 1 = A 0 v 0 v 1 = v 0 πr 0 = v A 1 πr = v r cm 2 0 = ( 1.0 m/s) = 4.0 m/s 1 r cm 63. Water flows out of a large reservoir through an open pipe, as shown in the figure. What is the speed of the water as it comes out of the pipe? A) 8.9 m/s B) 9.9 m/s C) 13 m/s D) 14 m/s E) 16 m/s P top ρv 2 top + ρgy top = P outlet ρv 2 outlet P top = P outlet ; v top 0; h = y top y outlet ρgh = 1 2 ρv 2 outlet + ρgy outlet v outlet = 2gh = 2( 9.8 m/s 2 )( 8.0 m) = 13 m

10 Physics 111 Exam 3 - KEY 43 B 53 B 63 C 44 C 54 D 45 A 55 B 46 A 56 C 47 B 57 B 48 B 58 B 49 E 59 A 50 B 60 B 51 A 61 A 52 A 62 D

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