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# Practice Exam 4 for General Physics 1 PHYS113

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Quiz instructions You should have both an exam and a machine-gradeable answer card (Scantron). On the other side of this page is an official “physics fact sheet”. Detach it if you wish. Not all of the information on the “fact sheet” will be useful. Before you start: ● Fill in the bubbles for your Wildcat ID number. Scantron cards without ID number bubbles filled out won't be counted. ● Fill in the the test version number (written in big letters before the first question), and the answer sheet number (1). ● Also print your name, the course, and the date in the box at the top of the card. ● Also write your name on the first and last pages of the exam. (If you detach this page, write your name on the next page, and also the last page.) ● Take a deep breath. Relax and pay attention. It's really not that bad. Please note: ● Your score on the multiple choice questions will be determined by your answers on the answer card. However, it is a good idea to mark your answers on the exam sheet too, for your own reference. ● Do not fold or roll the card. Unreadable cards won't be counted. ● Mark your responses on the answer card by filling in the bubbles completely with a #2 pencil. ● Do not fill in more than one response for each question. ● If you erase an answer, make sure you erase it completely. ● If you need a new card, please ask. When you are done: ● Recheck your filled-in circles against your exam sheet. ● Be sure that you have answered all questions, that you have all of the pages of the exam, and that both the question number and your response match agree with your exam sheet. You should see the words “The End” after the last question. ● Turn in both the Scantron card and your exam at the front of the room. ● Your exam will be returned to you after all exams are scanned and checked and the written part is graded. ● Please write your name on your exam. Constants SI prefixes Gravity at Earth's surface g = 9.80 m/s , down -12 12 -11 3 -1 -2 pico 10-9 p tera 10 9 T Gravitational constant G = 6.674 × 10 5 kg 2 nano 10 -6 n giga 10 6 G Atmospheric pressure 1 atm = 1.013×10 N/m micro 10 μ mega 10 M Boltzmann's constant k = 1.38×102J/K milli 10-3 m kilo 10 3 k Ideal gas constant R = 8.31 J/(mole·K) centi 10 -2 c hecto 10 2 h Avagadro's number NA= 6.02×1023 Speed of light c = 2.998 × 10 m/s Trigonometry sin θ = “opposite” / “hypotenuse” Kinematics, dynamics, energy, momentum cos θ = “adjacent” / “hypotenuse” Δ ⃗ ω= Δθ hypotenuse = opposite + adjacent ⃗= Δt Δt Δω t hy a= Δ⃗v α= Δt s po e n Δt o use ⃗ θ(t)=θ +ω t+ α1 t2 p θ ∑ F=ma ⃗ 0 0 2 const o adjacent 1 2 2 2 ⃗(t)=⃗0+v⃗0+ 2 ⃗const ω =ω +0αΔθ 2 2 ∑ τ=I α Impulses and collisions v =v0+2aΔ x Impulse: F Δ t=Δ ⃗ 1 2 KE rotationω 2 KE linear mv 2 Collision: ∑ ⃗pinitial ⃗final 2 W=τΔθ W=F dcosθ Fd Simple harmonic motion and waves p=mv L=I ω ⃗ ⃗ Period: T=2π inertia √ restoring Some kinds of potential energy Uniform gravity: PE Gmgy Spring: F=k x Pendulum: F=mg x/L Spring: PE = kx 2 Frequency: f =1/T S 2 Oscillator: v0/ A=2π f Waves: v= f λ Geometry of circles and spheres Standing: f =n(v/2L),any n 360 degrees = 2π radians = 1 full circle n Circumference of a circle: 2πr f nn(v/4L),odd n Rolling distance = (radius)*(angle in radians) Small angles: (linear size of object) ≈ (arc length) Area of a sphere: 4πr Other definitions and relations Density: ρ=m/V Pressure: P=F/ A Circular or rotational motion v2 2 In a fluid: Δ P=ρ gΔ y v=rω, acentrip =rω Buoyancy: F Bρ fluidg r 2 τ=r Fsinθ, I= ∑ mr Gravity (sphere): F =G m 1 2 G r2 Static equilibrium Friction: F Fμ F N ax=a y0 → ∑ F x ∑ F y0 Ideal gas: PV=nRT=NkT 3 α=0 → ∑ τ=0 Avg transl. KE: KE= kT 2 Name: Date: Exam 4 practice, version 2 6 5 1. A 3.0×10 kg asteroid and a smaller 1.8×10 kg asteroid orbit around each other, keeping a constant distance 7000 m. How far is the center of mass from the 3.0×10 kg asteroid? +03 6 6 5 A. 6.6×10 m, calculated from (0+(3.0×10 kg)*(7000 m))/((3.0×10 kg)+(1.8×10 kg)) B. 7.4×10 +03m, calculated from ((3.0×10 kg)*(7000 m)+(1.8×10 kg)*(7000 m))/(3.0×10 kg) C. 7×10 +03m, calculated from ((3.0×10 kg)*(7000 m)+(1.8×10 kg)*(7000 m))/((3.0×10 kg)+ 5 (1.8×10 kg)) D. 4×10 +02m, calculated from (0+(1.8×10 kg)*(7000 m))/((3.0×10 kg)+(1.8×10 kg)) 2. An astronaut standing on the higher mass asteroid uses rope and pully system to pull the lower mass asteroid closer. Does the higher mass asteroid move closer to the center of mass? A. No B. Yes page 1 of 9 A bicycle, with crank length indicated. 3. As a bicycle travels down a road, its 0.67 m diameter wheels turn 3.7 revolutions/second How fast is the bicycle moving? A. 7.8 m/s B. 2.5 m/s C. 0.57 m/s D. 3.7 m/s 4. The bicyclist exerts a force of 645 N on one pedal of the bike. The distance of the pedals from the crank axle is 0.16 m What is the magnitude of the torque if the force is exerted perpendicular to the crank axle to the pedal? (See figure.) +03 A. 4×10 m*N, calculated from (645 N)/(0.16 m) B. 0.00025 m*N, calculated from (0.16 m)/(645 N) C. 1×10 +02m*N, calculated from (0.16 m)*(645 N) D. 6.4×10 +02m*N, calculated from (645 N) 5. For the same force and crank length, what is the torque when the angle between the force and the crank arm is 120 degrees? (See figure.) A. -52 m*N B. -0.00012 m*N C. 0 m*N D. 5.6×10 +02m*N E. 89 m*N 6. Suppose the bicyclist were apply the same downward force to both pedals at the same time. What would the total torque be? A. 5.6×10 +02m*N, calculated from (645 N)*sin((120 degrees)*pi/180) +02 B. 1.8×10 m*N, calculated from 2*(0.16 m)*(645 N)*sin((120 degrees)*pi/180) C. -52 m*N, calculated from (0.16 m)*(645 N)*cos((120 degrees)*pi/180) D. -0.00012 m*N, calculated from (0.16 m)/(645 N)*cos((120 degrees)*pi/180) E. 0 m*N, calculated from 0 page 2 of 9 7. A centrifuge whirls 3 samples of mass 0.083 kg at a distance 0.14 m from the axis of rotation. What is the total moment of inertia of the 3 samples? 2 A. 0.035 kg*m B. 0.0049 kg*m 2 C. 0.0024 kg*m 2 2 D. 0.00054 kg*m 8. Suppose it takes 2.7 s for a centrifuge to come up to an angular speed of 135 radians/s when a torque 1.2 m*N is applied. What is the angular acceleration of the centrifuge? 2 A. 50 radians/s , calculated from (135 radians/s)/(2.7 s) B. 0.14 radians/s , calculated from 1/(2.7 s) C. 0.37 radians/s, calculated from 1/(2.7 s) D. 1.8×10 +04radians/s , calculated from (135 radians/s) 9. What is the TOTAL moment of inertia of the centrifuge plus samples? (It is greater than that of the samples due to the mass of the rotor in the centrifuge.) A. 1.6×10 +02kg*m 2 2 B. 0.024 kg*m C. 3.2 kg*m 2 +02 2 D. 4.4×10 kg*m 10. What torque is required to bring the centrifuge to rest from an angular speed of 135 radians/s* in 1.15 s? A. 2 m*N, calculated from (2.7 s)/((1.15 s)*(1.2 m*N)) B. 2.6 m*N, calculated from (1.15 s)*(2.7 s)/(1.2 m*N) C. 0.35 m*N, calculated from (1.15 s)/((2.7 s)*(1.2 m*N)) D. 2.8 m*N, calculated from (1.2 m*N)*(2.7 s)/(1.15 s) page 3 of 9 Drawing of a tetherball. 11. A 0.5 kg "tether ball" rotates on a string of length 2.5 m a distance 1.4 m from the pole. (See figure.) What is the moment of inertia of the tether ball? Ignore the mass of the string. A. 5.8 kg*m /s B. 0.7 kg*m C. 1.3 kg*m/s D. 3.1 kg*m 2 2 E. 0.98 kg*m 12. What is the torque on the tether ball if air resistance exerts a force 1.8 N on it? A. 2.5 m*N, calculated from (1.4 m)*(1.8 N) B. 0.78 m*N, calculated from (1.4 m)/(1.8 N) C. 1.8 m*N, calculated from (1.8 N) D. 1.3 m*N, calculated from (1.8 N)/(1.4 m) page 4 of 9 13. What is the angular acceleration of the tether ball due to air resistance? [Reminder, units are important. Remember Mars Climate Orbiter!] 2 A. 2.6 radians/s B. 2.6 radians/s C. 3.6 m/s 2 D. 2.6 rpm/s E. 2.6 radians/s 3 14. An almost horizontal clothes line pulls the top of a 3.7 m tall pole with a horizontal component of force 242 N A "guy wire" on the opposite side of the pole has an angle 53 deg from horizontal. (See figure.) What is the tension in the "guy wire" such that the pole experiences no torque about its base? A. 4.8×10 +02 N, calculated from 2*(242 N) +02 B. 2.4×10 N, calculated from (242 N) C. 3.0×10 +02 N, calculated from (242 N)/sin((53 deg)*pi/180) +02 D. 4.0×10 N, calculated from (242 N)/cos((53 deg)*pi/180) 15. Does the weight of the pole exert a torque about the base of the pole? A. Yes B. No page 5 of 9 Cartoon of a person climbing a wall using a rope. 16. A 55 kg climber stands perpendicular to a vertical wall, hanging on to a rope with makes an angle 65 deg from horizontal. (See figure.) The climber's center of gravity is 1.08 m from the wall, and her shoulders are 1.62 m from the wall. What is the tension in the rope? [Hint: use the feet of the climber as the center of rotation when calculating torque.] +02 A. 3.6×10 N B. 4.0×10+02N +02 C. 8.1×10 N D. 5.4×10 +02N 17. What force horizontal force does the climber feel in her legs? (I.e., what is the normal force from the wall on the climber?) A. 1.7×10 +02N, calculated from (1.08 m)*(55 kg)*(9.80 m/s )/((1.62 m)*tan((65 deg)*pi/180)) +02 2 B. 5.4×10 N, calculated from (55 kg)*(9.80 m/s ) C. 8.1×10+02N, calculated from (55 kg)*(9.80 m/s )*(1.62 m)/(1.08 m) +02 2 D. 3.6×10 N, calculated from (55 kg)*(9.80 m/s )*(1.08 m)/(1.62 m) 18. What force of friction on the climber's feet is required? A. 3.6×10 +02N B. 1.8×10+02N C. 8.1×10+02N D. 5.4×10 +02N page 6 of 9 19. In which direction does the force of friction act on the climber's feet? A. into the wall B. out from the wall C. down D. up 20. If the climber slowly walks up the wall and continues to stand perpendicular to the wall, the rope becomes more and more horizontal. Does the tension in the rope increase, decrease, or stay the same? A. Increase B. Stay the same C. Decrease 21. If the climber were to hang straight down from the rope, would the tension in the rope be greater than, less than, or the same as it is when the climber stands perpendicular to the wall? A. Less B. Greater C. The same page 7 of 9 Written answer question A 60 kg bowler holds a 7.0 kg bowling ball 0.55 m in front of her. She must lean back for balance. [Note: your solution will be graded for the correctness of the work shown as well as on the final answer. Show and explain all work! Be sure to include proper units in all answers!] (i) Draw a diagram showing the bowler and the forces acting on her. (ii) Calculate how far back (horizontal distance) the bowler's center of gravity must be relative to her feet. The End page 8 of 9 Quiz instructions You should have both an exam and a machine-gradeable answer card (Scantron). On the other side of this page is an official “physics fact sheet”. Detach it if you wish. Not all of the information on the “fact sheet” will be useful. Before you start: ● Fill in the bubbles for your Wildcat ID number. Scantron cards without ID number bubbles filled out won't be counted. ● Fill in the the test version number (written in big letters before the first question), and the answer sheet number (1). ● Also print your name, the course, and the date in the box at the top of the card. ● Also write your name on the first and last pages of the exam. (If you detach this page, write your name on the next page, and also the last page.) ● Take a deep breath. Relax and pay attention. It's really not that bad. Please note: ● Your score on the multiple choice questions will be determined by your answers on the answer card. However, it is a good idea to mark your answers on the exam sheet too, for your own reference. ● Do not fold or roll the card. Unreadable cards won't be counted. ● Mark your responses on the answer card by filling in the bubbles completely with a #2 pencil. ● Do not fill in more than one response for each question. ● If you erase an answer, make sure you erase it completely. ● If you need a new card, please ask. When you are done: ● Recheck your filled-in circles against your exam sheet. ● Be sure that you have answered all questions, that you have all of the pages of the exam, and that both the question number and your response match agree with your exam sheet. You should see the words “The End” after the last question. ● Turn in both the Scantron card and your exam at the front of the room. ● Your exam will be returned to you after all exams are scanned and checked and the written part is graded. ● Please write your name on your exam. Constants SI prefixes Gravity at Earth's surface g = 9.80 m/s , down -12 12 -11 3 -1 -2 pico 10-9 p tera 10 9 T Gravitational constant G = 6.674 × 10 5 kg 2 nano 10 -6 n giga 10 6 G Atmospheric pressure 1 atm = 1.013×10 N/m micro 10 μ mega 10 M Boltzmann's constant k = 1.38×102J/K milli 10-3 m kilo 10 3 k Ideal gas constant R = 8.31 J/(mole·K) centi 10 -2 c hecto 10 2 h Avagadro's number NA= 6.02×1023 Speed of light c = 2.998 × 10 m/s Trigonometry sin θ = “opposite” / “hypotenuse” Kinematics, dynamics, energy, momentum cos θ = “adjacent” / “hypotenuse” Δ ⃗ ω= Δθ hypotenuse = opposite + adjacent ⃗= Δt Δt Δω t hy a= Δ⃗v α= Δt s po e n Δt o use ⃗ θ(t)=θ +ω t+ α1 t2 p θ ∑ F=ma ⃗ 0 0 2 const o adjacent 1 2 2 2 ⃗(t)=⃗0+v⃗0+ 2 ⃗const ω =ω +0αΔθ 2 2 ∑ τ=I α Impulses and collisions v =v0+2aΔ x Impulse: F Δ t=Δ ⃗ 1 2 KE rotationω 2 KE linear mv 2 Collision: ∑ ⃗pinitial ⃗final 2 W=τΔθ W=F dcosθ Fd Simple harmonic motion and waves p=mv L=I ω ⃗ ⃗ Period: T=2π inertia √ restoring Some kinds of potential energy Uniform gravity: PE Gmgy Spring: F=k x Pendulum: F=mg x/L Spring: PE = kx 2 Frequency: f =1/T S 2 Oscillator: v0/ A=2π f Waves: v= f λ Geometry of circles and spheres Standing: f =n(v/2L),any n 360 degrees = 2π radians = 1 full circle n Circumference of a circle: 2πr f nn(v/4L),odd n Rolling distance = (radius)*(angle in radians) Small angles: (linear size of object) ≈ (arc length) Area of a sphere: 4πr Other definitions and relations Density: ρ=m/V Pressure: P=F/ A Circular or rotational motion v2 2 In a fluid: Δ P=ρ gΔ y v=rω, acentrip =rω Buoyancy: F Bρ fluidg r 2 τ=r Fsinθ, I= ∑ mr Gravity (sphere): F =G m 1 2 G r2 Static equilibrium Friction: F Fμ F N ax=a y0 → ∑ F x ∑ F y0 Ideal gas: PV=nRT=NkT 3 α=0 → ∑ τ=0 Avg transl. KE: KE= kT 2 Name: Date: Exam 4 practice, version 3 6 5 1. A 3.0×10 kg asteroid and a smaller 1.6×10 kg asteroid orbit around each other, keeping a constant distance 6000 m. How far is the center of mass from the 3.0×10 kg asteroid? +03 6 6 5 A. 5.7×10 m, calculated from (0+(3.0×10 kg)*(6000 m))/((3.0×10 kg)+(1.6×10 kg)) B. 6.3×10 +03m, calculated from ((3.0×10 kg)*(6000 m)+(1.6×10 kg)*(6000 m))/(3.0×10 kg) C. 3×10 +02m, calculated from (0+(1.6×10 kg)*(6000 m))/((3.0×10 kg)+(1.6×10 kg)) +03 6 5 6 D. 6×10 m, calculated from ((3.0×10 kg)*(6000 m)+(1.6×10 kg)*(6000 m))/((3.0×10 kg)+ (1.6×10 kg)) 2. An astronaut standing on the higher mass asteroid uses rope and pully system to pull the lower mass asteroid closer. Does the higher mass asteroid move closer to the center of mass? A. Yes B. No page 1 of 9 A bicycle, with crank length indicated. 3. As a bicycle travels down a road, its 0.65 m diameter wheels turn 3.9 revolutions/second How fast is the bicycle moving? A. 3.9 m/s, calculated from (3.9 revolutions/second) B. 0.52 m/s, calculated from pi*(0.65 m)/(3.9 revolutions/second) C. 8 m/s, calculated from pi*(0.65 m)*(3.9 revolutions/second) D. 2.5 m/s, calculated from (0.65 m)*(3.9 revolutions/second) 4. The bicyclist exerts a force of 575 N on one pedal of the bike. The distance of the pedals from the crank axle is 0.15 m What is the magnitude of the torque if the force is exerted perpendicular to the crank axle to the pedal? (See figure.) A. 86 m*N, calculated from (0.15 m)*(575 N) B. 0.00026 m*N, calculated from (0.15 m)/(575 N) C. 5.8×10 +02m*N, calculated from (575 N) +03 D. 3.8×10 m*N, calculated from (575 N)/(0.15 m) 5. For the same force and crank length, what is the torque when the angle between the force and the crank arm is 80 degrees? (See figure.) A. 15 m*N, calculated from (0.15 m)*(575 N)*cos((80 degrees)*pi/180) B. 85 m*N, calculated from (0.15 m)*(575 N)*sin((80 degrees)*pi/180) C. 5.7×10 +02m*N, calculated from (575 N)*sin((80 degrees)*pi/180) D. 4.5×10 -05m*N, calculated from (0.15 m)/(575 N)*cos((80 degrees)*pi/180) E. 0 m*N, calculated from 0 6. Suppose the bicyclist were apply the same downward force to both pedals at the same time. What would the total torque be? A. 4.5×10 -05m*N, calculated from (0.15 m)/(575 N)*cos((80 degrees)*pi/180) +02 B. 5.7×10 m*N, calculated from (575 N)*sin((80 degrees)*pi/180) C. 1.7×10 +02m*N, calculated from 2*(0.15 m)*(575 N)*sin((80 degrees)*pi/180) D. 0 m*N, calculated from 0 E. 15 m*N, calculated from (0.15 m)*(575 N)*cos((80 degrees)*pi/180) page 2 of 9 7. A centrifuge whirls 4 samples of mass 0.087 kg at a distance 0.15 m from the axis of rotation. What is the total moment of inertia of the 4 samples? A. 0.052 kg*m , calculated from (4)*(0.087 kg)*(0.15 m) 2 2 B. 0.00049 kg*m , calculated from (0.087 kg)*(0.15 m) /(4) C. 0.0039 kg*m , calculated from (4)*(0.087 kg)*(0.15 m) /2 2 2 2 D. 0.0078 kg*m , calculated from (4)*(0.087 kg)*(0.15 m) 8. Suppose it takes 2.7 s for a centrifuge to come up to an angular speed of 145 radians/s when a torque 1.3 m*N is applied. What is the angular acceleration of the centrifuge? 2 2 A. 0.14 radians/s , calculated from 1/(2.7 s) B. 54 radians/s , calculated from (145 radians/s)/(2.7 s) C. 0.37 radians/s, calculated from 1/(2.7 s) D. 2.1×10 +04 radians/s , calculated from (145 radians/s)2 9. What is the TOTAL moment of inertia of the centrifuge plus samples? (It is greater than that of the samples due to the mass of the rotor in the centrifuge.) A. 1.9×10 +02 kg*m , calculated from (1.3 m*N)*(145 radians/s) B. 5.1×10 +02 kg*m , calculated from (1.3 m*N)*(145 radians/s)*(2.7 s) 2 C. 0.024 kg*m , calculated from (1.3 m*N)/((145 radians/s)/(2.7 s)) D. 3.5 kg*m , calculated from (1.3 m*N)*(2.7 s) 10. What torque is required to bring the centrifuge to rest from an angular speed of 145 radians/s* in 1.05 s? A. 0.3 m*N, calculated from (1.05 s)/((2.7 s)*(1.3 m*N)) B. 3.3 m*N, calculated from (1.3 m*N)*(2.7 s)/(1.05 s) C. 2 m*N, calculated from (2.7 s)/((1.05 s)*(1.3 m*N)) D. 2.2 m*N, calculated from (1.05 s)*(2.7 s)/(1.3 m*N) page 3 of 9 Drawing of a tetherball. 11. A 0.8 kg "tether ball" rotates on a string of length 2.1 m a distance 1.3 m from the pole. (See figure.) What is the moment of inertia of the tether ball? Ignore the mass of the string. A. 2.1 kg*m/s, calculated from (0.8 kg)*(2.6 m/s) B. 7.1 kg*m /s, calculated from (0.8 kg)*(2.1 m) *((2.6 m/s)/(1.3 m)) C. 3.5 kg*m , calculated from (0.8 kg)*(2.1 m) 2 D. 1 kg*m, calculated from (0.8 kg)*(1.3 m) E. 1.4 kg*m , calculated from (0.8 kg)*(1.3 m) 2 12. What is the torque on the tether ball if air resistance exerts a force 1.1 N on it? A. 0.85 m*N, calculated from (1.1 N)/(1.3 m) B. 1.2 m*N, calculated from (1.3 m)/(1.1 N) C. 1.4 m*N, calculated from (1.3 m)*(1.1 N) D. 1.1 m*N, calculated from (1.1 N) page 4 of 9 13. What is the angular acceleration of the tether ball due to air resistance? [Reminder, units are important. Remember Mars Climate Orbiter!] 2 A. 1.1 rpm/s, calculated from (1.3 m)*(1.1 N)/((0.8 kg)*(1.3 m) ) B. 1.1 radians/s , calculated from (1.3 m)*(1.1 N)/((0.8 kg)*(1.3 m) ) 2 C. 1.4 m/s , calculated from (1.1 N)/(0.8 kg) D. 1.1 radians/s , calculated from (1.3 m)*(1.1 N)/((0.8 kg)*(1.3 m) ) 2 E. 1.1 radians/s, calculated from (1.3 m)*(1.1 N)/((0.8 kg)*(1.3 m) ) 14. An almost horizontal telephone wire pulls the top of a 4.6 m tall pole with a horizontal component of force 169 N A "guy wire" on the opposite side of the pole has an angle 51 deg from horizontal. (See figure.) What is the tension in the "guy wire" such that the pole experiences no torque about its base? A. 3.4×10 +02 N, calculated from 2*(169 N) +02 B. 2.2×10 N, calculated from (169 N)/sin((51 deg)*pi/180) C. 2.7×10 +02 N, calculated from (169 N)/cos((51 deg)*pi/180) +02 D. 1.7×10 N, calculated from (169 N) 15. Does the weight of the pole exert a torque about the base of the pole? A. Yes B. No page 5 of 9 Cartoon of a person climbing a wall using a rope. 16. A 66 kg climber stands perpendicular to a vertical wall, hanging on to a rope with makes an angle 60 deg from horizontal. (See figure.) The climber's center of gravity is 1.04 m from the wall, and his shoulders are 1.56 m from the wall. What is the tension in the rope? [Hint: use the feet of the climber as the center of rotation when calculating torque.] +02 2 A. 4.3×10 N, calculated from (66 kg)*(9.80 m/s )*(1.04 m)/(1.56 m) B. 5.0×10 +02 N, calculated from (1.04 m)*(66 kg)*(9.80 m/s )/((1.56 m)*sin((60 deg)*pi/180)) +02 2 C. 6.5×10 N, calculated from (66 kg)*(9.80 m/s ) D. 9.7×10 +02 N, calculated from (66 kg)*(9.80 m/s )*(1.56 m)/(1.04 m) 17. What force horizontal force does the climber feel in his legs? (I.e., what is the normal force from the wall on the climber?) A. 4.3×10 +02 N, calculated from (66 kg)*(9.80 m/s )*(1.04 m)/(1.56 m) B. 9.7×10 +02 N, calculated from (66 kg)*(9.80 m/s )*(1.56 m)/(1.04 m) +02 2 C. 2.5×10 N, calculated from (1.04 m)*(66 kg)*(9.80 m/s )/((1.56 m)*tan((60 deg)*pi/180)) D. 6.5×10 +02 N, calculated from (66 kg)*(9.80 m/s ) 18. What force of friction on the climber's feet is required? A. 9.7×10 +02 N, calculated from (66 kg)*(9.80 m/s )*(1.56 m)/(1.04 m) +02 2 B. 4.3×10 N, calculated from (66 kg)*(9.80 m/s )*(1.04 m)/(1.56 m) C. 6.5×10 +02 N, calculated from (66 kg)*(9.80 m/s ) +02 2 2 D. 2.2×10 N, calculated from (66 kg)*(9.80 m/s ) - (1.04 m)*(66 kg)*(9.80 m/s )/(1.56 m) page 6 of 9 19. In which direction does the force of friction act on the climber's feet? A. up B. out from the wall C. into the wall D. down 20. If the climber slowly walks up the wall and continues to stand perpendicular to the wall, the rope becomes more and more horizontal. Does the tension in the rope increase, decrease, or stay the same? A. Increase B. Stay the same C. Decrease 21. If the climber were to hang straight down from the rope, would the tension in the rope be greater than, less than, or the same as it is when the climber stands perpendicular to the wall? A. The same B. Greater C. Less page 7 of 9 Written answer question A 60 kg bowler holds a 7.0 kg bowling ball 0.55 m in front of her. She must lean back for balance. [Note: your solution will be graded for the correctness of the work shown as well as on the final answer. Show and explain all work! Be sure to include proper units in all answers!] (i) Draw a diagram showing the bowler and the forces acting on her. (ii) Calculate how far back (horizontal distance) the bowler's center of gravity must be relative to her feet. The End page 8 of 9 Quiz instructions You should have both an exam and a machine-gradeable answer card (Scantron). On the other side of this page is an official “physics fact sheet”. Detach it if you wish. Not all of the information on the “fact sheet” will be useful. Before you start: ● Fill in the bubbles for your Wildcat ID number. Scantron cards without ID number bubbles filled out won't be counted. ● Fill in the the test version number (written in big letters before the first question), and the answer sheet number (1). ● Also print your name, the course, and the date in the box at the top of the card. ● Also write your name on the first and last pages of the exam. (If you detach this page, write your name on the next page, and also the last page.) ● Take a deep breath. Relax and pay attention. It's really not that bad. Please note: ● Your score on the multiple choice questions will be determined by your answers on the answer card. However, it is a good idea to mark your answers on the exam sheet too, for your own reference. ● Do not fold or roll the card. Unreadable cards won't be counted. ● Mark your responses on the answer card by filling in the bubbles completely with a #2 pencil. ● Do not fill in more than one response for each question. ● If you erase an answer, make sure you erase it completely. ● If you need a new card, please ask. When you are done: ● Recheck your filled-in circles against your exam sheet. ● Be sure that you have answered all questions, that you have all of the pages of the exam, and that both the question number and your response match agree with your exam sheet. You should see the words “The End” after the last question. ● Turn in both the Scantron card and your exam at the front of the room. ● Your exam will be returned to you after all exams are scanned and checked and the written part is graded. ● Please write your name on your exam. Constants SI prefixes Gravity at Earth's surface g = 9.80 m/s , down -12 12 -11 3 -1 -2 pico 10-9 p tera 10 9 T Gravitational constant G = 6.674 × 10 5 kg 2 nano 10 -6 n giga 10 6 G Atmospheric pressure 1 atm = 1.013×10 N/m micro 10 μ mega 10 M Boltzmann's constant k = 1.38×102J/K milli 10-3 m kilo 10 3 k Ideal gas constant R = 8.31 J/(mole·K) centi 10 -2 c hecto 10 2 h Avagadro's number NA= 6.02×1023 Speed of light c = 2.998 × 10 m/s Trigonometry sin θ = “opposite” / “hypotenuse” Kinematics, dynamics, energy, momentum cos θ = “adjacent” / “hypotenuse” Δ ⃗ ω= Δθ hypotenuse = opposite + adjacent ⃗= Δt Δt Δω t hy a= Δ⃗v α= Δt s po e n Δt o use ⃗ θ(t)=θ +ω t+ α1 t2 p θ ∑ F=ma ⃗ 0 0 2 const o adjacent 1 2 2 2 ⃗(t)=⃗0+v⃗0+ 2 ⃗const ω =ω +0αΔθ 2 2 ∑ τ=I α Impulses and collisions v =v0+2aΔ x Impulse: F Δ t=Δ ⃗ 1 2 KE rotationω 2 KE linear mv 2 Collision: ∑ ⃗pinitial ⃗final 2 W=τΔθ W=F dcosθ Fd Simple harmonic motion and waves p=mv L=I ω ⃗ ⃗ Period: T=2π inertia √ restoring Some kinds of potential energy Uniform gravity: PE Gmgy Spring: F=k x Pendulum: F=mg x/L Spring: PE = kx 2 Frequency: f =1/T S 2 Oscillator: v0/ A=2π f Waves: v= f λ Geometry of circles and spheres Standing: f =n(v/2L),any n 360 degrees = 2π radians = 1 full circle n Circumference of a circle: 2πr f nn(v/4L),odd n Rolling distance = (radius)*(angle in radians) Small angles: (linear size of object) ≈ (arc length) Area of a sphere: 4πr Other definitions and relations Density: ρ=m/V Pressure: P=F/ A Circular or rotational motion v2 2 In a fluid: Δ P=ρ gΔ y v=rω, acentrip =rω Buoyancy: F Bρ fluidg r 2 τ=r Fsinθ, I= ∑ mr Gravity (sphere): F =G m 1 2 G r2 Static equilibrium Friction: F Fμ F N ax=a y0 → ∑ F x ∑ F y0 Ideal gas: PV=nRT=NkT 3 α=0 → ∑ τ=0 Avg transl. KE: KE= kT 2 Name: Date: Exam 4 practice, version 1 6 5 1. A 4.0×10 kg asteroid and a smaller 1.8×10 kg asteroid orbit around each other, keeping a constant distance 6000 m. How far is the center of mass from the 4.0×10 kg asteroid? +03 6 5 6 A. 6×10 m, calculated from ((4.0×10 kg)*(6000 m)+(1.8×10 kg)*(6000 m))/((4.0×10 kg)+ (1.8×10 kg)) B. 2.6×10 +02m, calculated from (0+(1.8×10 kg)*(6000 m))/((4.0×10 kg)+(1.8×10 kg)) +03 6 5 6 C. 6.3×10 m, calculated from ((4.0×10 kg)*(6000 m)+(1.8×10 kg)*(6000 m))/(4.0×10 kg) D. 5.7×10 +03m, calculated from (0+(4.0×10 kg)*(6000 m))/((4.0×10 kg)+(1.8×10 kg)) 2. An astronaut standing on the higher mass asteroid uses rope and pully system to pull the lower mass asteroid closer. Does the higher mass asteroid move closer to the center of mass? A. Yes B. No page 1 of 9 A bicycle, with crank length indicated. 3. As a bicycle travels down a road, its 0.67 m diameter wheels turn 4.1 revolutions/second How fast is the bicycle moving? A. 8.6 m/s, calculated from pi*(0.67 m)*(4.1 revolutions/second) B. 4.1 m/s, calculated from (4.1 revolutions/second) C. 0.51 m/s, calculated from pi*(0.67 m)/(4.1 revolutions/second) D. 2.7 m/s, calculated from (0.67 m)*(4.1 revolutions/second) 4. The bicyclist exerts a force of 465 N on one pedal of the bike. The distance of the pedals from the crank axle is 0.18 m What is the magnitude of the torque if the force is exerted perpendicular to the crank axle to the pedal? (See figure.) A. 84 m*N B. 2.6×10+03 m*N C. 4.6×10+02 m*N D. 0.00039 m*N 5. For the same force and crank length, what is the torque when the angle between the force and the crank arm is 100 degrees? (See figure.) A. 4.6×10 +02 m*N, calculated from (465 N)*sin((100 degrees)*pi/180) -05 B. -6.7×10 m*N, calculated from (0.18 m)/(465 N)*cos((100 degrees)*pi/180) C. 82 m*N, calculated from (0.18 m)*(465 N)*sin((100 degrees)*pi/180) D. 0 m*N, calculated from 0 E. -15 m*N, calculated from (0.18 m)*(465 N)*cos((100 degrees)*pi/180) 6. Suppose the bicyclist were apply the same downward force to both pedals at the same time. What would the total torque be? A. -15 m*N B. 0 m*N C. 1.6×10+02 m*N D. -6.7×10 -05m*N +02 E. 4.6×10 m*N page 2 of 9 7. A centrifuge whirls 4 samples of mass 0.081 kg at a distance 0.13 m from the axis of rotation. What is the total moment of inertia of the 4 samples? 2 2 A. 0.0055 kg*m , calculated from (4)*(0.081 kg)*(0.13 m) B. 0.042 kg*m , calculated from (4)*(0.081 kg)*(0.13 m) C. 0.00034 kg*m , calculated from (0.081 kg)*(0.13 m) /(4) 2 2 D. 0.0027 kg*m , calculated from (4)*(0.081 kg)*(0.13 m) /2 8. Suppose it takes 2.4 s for a centrifuge to come up to an angular speed of 165 radians/s when a torque 1.3 m*N is applied. What is the angular acceleration of the centrifuge? +04 2 A. 2.7×10 radians/s B. 69 radians/s 2 C. 0.42 radians/s D. 0.17 radians/s 2 9. What is the TOTAL moment of inertia of the centrifuge plus samples? (It is greater than that of the samples due to the mass of the rotor in the centrifuge.) A. 5.1×10 +02 kg*m , calculated from (1.3 m*N)*(165 radians/s)*(2.4 s) 2 B. 0.019 kg*m , calculated from (1.3 m*N)/((165 radians/s)/(2.4 s)) C. 2.1×10 +02 kg*m , calculated from (1.3 m*N)*(165 radians/s) 2 D. 3.1 kg*m , calculated from (1.3 m*N)*(2.4 s) 10. What torque is required to bring the centrifuge to rest from an angular speed of 165 radians/s* in 0.85 s? A. 0.27 m*N B. 2.2 m*N C. 1.6 m*N D. 3.7 m*N page 3 of 9 Drawing of a tetherball. 11. A 0.6 kg "tether ball" rotates on a string of length 2.1 m a distance 1.4 m from the pole. (See figure.) What is the moment of inertia of the tether ball? Ignore the mass of the string. 2 2 A. 4.9 kg*m /s, calculated from (0.6 kg)*(2.1 m) *((2.6 m/s)/(1.4 m)) B. 0.84 kg*m, calculated from (0.6 kg)*(1.4 m) C. 2.6 kg*m , calculated from (0.6 kg)*(2.1 m) 2 D. 1.2 kg*m , calculated from (0.6 kg)*(1.4 m) 2 E. 1.6 kg*m/s, calculated from (0.6 kg)*(2.6 m/s) 12. What is the torque on the tether ball if air resistance exerts a force 1.3 N on it? A. 1.8 m*N B. 1.1 m*N C. 1.3 m*N D. 0.93 m*N page 4 of 9 13. What is the angular acceleration of the tether ball due to air resistance? [Reminder, units are important. Remember Mars Climate Orbiter!] 3 2 A. 1.5 radians/s , calculated from (1.4 m)*(1.3 N)/((0.6 kg)*(1.4 m) ) B. 2.2 m/s , calculated from (1.3 N)/(0.6 kg) 2 2 C. 1.5 radians/s , calculated from (1.4 m)*(1.3 N)/((0.6 kg)*(1.4 m) ) D. 1.5 radians/s, calculated from (1.4 m)*(1.3 N)/((0.6 kg)*(1.4 m) ) 2 E. 1.5 rpm/s, calculated from (1.4 m)*(1.3 N)/((0.6 kg)*(1.4 m) ) 14. An almost horizontal telephone wire pulls the top of a 3.9 m tall pole with a horizontal component of force 180 N A "guy wire" on the opposite side of the pole has an angle 53 deg from horizontal. (See figure.) What is the tension in the "guy wire" such that the pole experiences no torque about its base? A. 1.8×10+02 N +02 B. 2.3×10 N C. 3.0×10+02 N +02 D. 3.6×10 N 15. Does the weight of the pole exert a torque about the base of the pole? A. Yes B. No page 5 of 9 Cartoon of a person climbing a wall using a rope. 16. A 67 kg climber stands perpendicular to a vertical wall, hanging on to a rope with makes an angle 69 deg from horizontal. (See figure.) The climber's center of gravity is 1.00 m from the wall, and her shoulders are 1.50 m from the wall. What is the tension in the rope? [Hint: use the feet of the climber as the center of rotation when calculating torque.] +02 2 A. 4.7×10 N, calculated from (1.00 m)*(67 kg)*(9.80 m/s )/((1.50 m)*sin((69 deg)*pi/180)) B. 4.4×10 +02N, calculated from (67 kg)*(9.80 m/s )*(1.00 m)/(1.50 m) +02 2 C. 9.8×10 N, calculated from (67 kg)*(9.80 m/s )*(1.50 m)/(1.00 m) D. 6.6×10 +02N, calculated from (67 kg)*(9.80 m/s ) 17. What force horizontal force does the climber feel in her legs? (I.e., what is the normal force from the wall on the climber?) A. 6.6×10 +02N B. 1.7×10 +02N +02 C. 4.4×10 N D. 9.8×10 +02N 18. What force of friction on the climber's feet is required? A. 4.4×10 +02N, calculated from (67 kg)*(9.80 m/s )*(1.00 m)/(1.50 m) +02 2 B. 6.6×10 N, calculated from (67 kg)*(9.80 m/s ) C. 9.8×10 +02N, calculated from (67 kg)*(9.80 m/s )*(1.50 m)/(1.00 m) +02 2 2 D. 2.2×10 N, calculated from (67 kg)*(9.80 m/s ) - (1.00 m)*(67 kg)*(9.80 m/s )/(1.50 m) page 6 of 9 19. In which direction does the force of friction act on the climber's feet? A. down B. into the wall C. out from the wall D. up 20. If the climber slowly walks up the wall and continues to stand perpendicular to the wall, the rope becomes more and more horizontal. Does the tension in the rope increase, decrease, or stay the same? A. Increase B. Stay the same C. Decrease 21. If the climber were to hang straight down from the rope, would the tension in the rope be greater than, less than, or the same as it is when the climber stands perpendicular to the wall? A. The same B. Less C. Greater page 7 of 9 Written answer question A 60 kg bowler holds a 7.0 kg bowling ball 0.55 m in front of her. She must lean back for balance. [Note: your solution will be graded for the correctness of the work shown as well as on the final answer. Show and explain all work! Be sure to include proper units in all answers!] (i) Draw a diagram showing the bowler and the forces acting on her. (ii) Calculate how far back (horizontal distance) the bowler's center of gravity must be relative to her feet. The End page 8 of 9 Quiz instructions You should have both an exam and a machine-gradeable answer card (Scantron). On the other side of this page is an official “physics fact sheet”. Detach it if you wish. Not all of the information on the “fact sheet” will be useful. Before you start: ● Fill in the bubbles for your Wildcat ID number. Scantron cards without ID number bubbles filled out won't be counted. ● Fill in the the test version number (written in big letters before the first question), and the answer sheet number (1). ● Also print your name, the course, and the date in the box at the top of the card. ● Also write your name on the first and last pages of the exam. (If you detach this page, write your name on the next page, and also the last page.) ● Take a deep breath. Relax and pay attention. It's really not that bad. Please note: ● Your score on the multiple choice questions will be determined by your answers on the answer card. However, it is a good idea to mark your answers on the exam sheet too, for your own reference. ● Do not fold or roll the card. Unreadable cards won't be counted. ● Mark your responses on the answer card by filling in the bubbles completely with a #2 pencil. ● Do not fill in more than one response for each question. ● If you erase an answer, make sure you erase it completely. ● If you need a new card, please ask. When you are done: ● Recheck your filled-in circles against your exam sheet. ● Be sure that you have answered all questions, that you have all of the pages of the exam, and that both the question number and your response match agree with your exam sheet. You should see the words “The End” after the last question. ● Turn in both the Scantron card and your exam at the front of the room. ● Your exam will be returned to you after all exams are scanned and checked and the written part is graded. ● Please write your name on your exam. Constants SI prefixes Gravity at Earth's surface g = 9.80 m/s , down -12 12 -11 3 -1 -2 pico 10-9 p tera 10 9 T Gravitational constant G = 6.674 × 10 5 kg 2 nano 10 -6 n giga 10 6 G Atmospheric pressure 1 atm = 1.013×10 N/m micro 10 μ mega 10 M Boltzmann's constant k = 1.38×102J/K milli 10-3 m kilo 10 3 k Ideal gas constant R = 8.31 J/(mole·K) centi 10 -2 c hecto 10 2 h Avagadro's number NA= 6.02×1023 Speed of light c = 2.998 × 10 m/s Trigonometry sin θ = “opposite” / “hypotenuse” Kinematics, dynamics, energy, momentum cos θ = “adjacent” / “hypotenuse” Δ ⃗ ω= Δθ hypotenuse = opposite + adjacent ⃗= Δt Δt Δω t hy a= Δ⃗v α= Δt s po e n Δt o use ⃗ θ(t)=θ +ω t+ α1 t2 p θ ∑ F=ma ⃗ 0 0 2 const o adjacent 1 2 2 2 ⃗(t)=⃗0+v⃗0+ 2 ⃗const ω =ω +0αΔθ 2 2 ∑ τ=I α Impulses and collisions v =v0+2aΔ x Impulse: F Δ t=Δ ⃗ 1 2 KE rotationω 2 KE linear mv 2 Collision: ∑ ⃗pinitial ⃗final 2 W=τΔθ W=F dcosθ Fd Simple harmonic motion and waves p=mv L=I ω ⃗ ⃗ Period: T=2π inertia √ restoring Some kinds of potential energy Uniform gravity: PE Gmgy Spring: F=k x Pendulum: F=mg x/L Spring: PE = kx 2 Frequency: f =1/T S 2 Oscillator: v0/ A=2π f Waves: v= f λ Geometry of circles and spheres Standing: f =n(v/2L),any n 360 degrees = 2π radians = 1 full circle n Circumference of a circle: 2πr f nn(v/4L),odd n Rolling distance = (radius)*(angle in radians) Small angles: (linear size of object) ≈ (arc length) Area of a sphere: 4πr Other definitions and relations Density: ρ=m/V Pressure: P=F/ A Circular or rotational motion v2 2 In a fluid: Δ P=ρ gΔ y v=rω, acentrip =rω Buoyancy: F Bρ fluidg r 2 τ=r Fsinθ, I= ∑ mr Gravity (sphere): F =G m 1 2 G r2 Static equilibrium Friction: F Fμ F N ax=a y0 → ∑ F x ∑ F y0 Ideal gas: PV=nRT=NkT 3 α=0 → ∑ τ=0 Avg transl. KE: KE= kT 2 Name: Date: Exam 4 practice, version 4 6 5 1. A 3.0×10 kg asteroid and a smaller 1.6×10 kg asteroid orbit around each other, keeping a constant distance 7000 m. How far is the center of mass from the 3.0×10 kg asteroid? +03 6 6 5 A. 6.6×10 m, calculated from (0+(3.0×10 kg)*(7000 m))/((3.0×10 kg)+(1.6×10 kg)) B. 3.5×10 +02m, calculated from (0+(1.6×10 kg)*(7000 m))/((3.0×10 kg)+(1.6×10 kg)) C. 7.4×10 +03m, calculated from ((3.0×10 kg)*(7000 m)+(1.6×10 kg)*(7000 m))/(3.0×10 kg)6 +03 6 5 6 D. 7×10 m, calculated from ((3.0×10 kg)*(7000 m)+(1.6×10 kg)*(7000 m))/((3.0×10 kg)+ (1.6×10 kg)) 2. An astronaut standing on the higher mass asteroid uses rope and pully system to pull the lower mass asteroid closer. Does the higher mass asteroid move closer to the center of mass? A. Yes B. No page 1 of 9 A bicycle, with crank length indicated. 3. As a bicycle travels down a road, its 0.69 m diameter wheels turn 3.9 revolutions/second How fast is the bicycle moving? A. 2.7 m/s, calculated from (0.69 m)*(3.9 revolutions/second) B. 8.5 m/s, calculated from pi*(0.69 m)*(3.9 revolutions/second) C. 3.9 m/s, calculated from (3.9 revolutions/second) D. 0.56 m/s, calculated from pi*(0.69 m)/(3.9 revolutions/second) 4. The bicyclist exerts a force of 595 N on one pedal of the bike. The distance of the pedals from the crank axle is 0.18 m What is the magnitude of the torque if the force is exerted perpendicular to the crank axle to the pedal? (See figure.) +03 A. 3.3×10 m*N, calculated from (595 N)/(0.18 m) B. 0.0003 m*N, calculated from (0.18 m)/(595 N) C. 6×10 +02m*N, calculated from (595 N) D. 1.1×10 +02m*N, calculated from (0.18 m)*(595 N) 5. For the same force and crank length, what is the torque when the angle between the force and the crank arm is 120 degrees? (See figure.) A. 0 m*N, calculated from 0 B. -54 m*N, calculated from (0.18 m)*(595 N)*cos((120 degrees)*pi/180) C. 93 m*N, calculated from (0.18 m)*(595 N)*sin((120 degrees)*pi/180) D. 5.2×10 +02m*N, calculated from (595 N)*sin((120 degrees)*pi/180) E. -0.00015 m*N, calculated from (0.18 m)/(595 N)*cos((120 degrees)*pi/180) 6. Suppose the bicyclist were apply the same downward force to both pedals at the same time. What would the total torque be? A. -54 m*N, calculated from (0.18 m)*(595 N)*cos((120 degrees)*pi/180) B. 0 m*N, calculated from 0 C. -0.00015 m*N, calculated from (0.18 m)/(595 N)*cos((120 degrees)*pi/180) D. 1.9×10 +02m*N, calculated from 2*(0.18 m)*(595 N)*sin((120 degrees)*pi/180) +02 E. 5.2×10 m*N, calculated from (595 N)*sin((120 degrees)*pi/180) page 2 of 9 7. A centrifuge whirls 5 samples of mass 0.091 kg at a distance 0.13 m from the axis of rotation. What is the total moment of inertia of the 5 samples? A. 0.00031 kg*m , calculated from (0.091 kg)*(0.13 m) /(5)2 2 2 B. 0.0038 kg*m , calculated from (5)*(0.091 kg)*(0.13 m) /2 C. 0.0077 kg*m , calculated from (5)*(0.091 kg)*(0.13 m) 2 2 D. 0.059 kg*m , calculated from (5)*(0.091 kg)*(0.13 m) 8. Suppose it takes 2.9 s for a centrifuge to come up to an angular speed of 135 radians/s when a torque 1.6 m*N is applied. What is the angular acceleration of the centrifuge? A. 0.34 radians/s, calculated from 1/(2.9 s) B. 1.8×10 +04 radians/s , calculated from (135 radians/s)2 2 C. 47 radians/s , calculated from (135 radians/s)/(2.9 s) D. 0.12 radians/s , calculated from 1/(2.9 s) 2 9. What is the TOTAL moment of inertia of the centrifuge plus samples? (It is greater than that of the samples due to the mass of the rotor in the centrifuge.) A. 2.2×10 +02 kg*m , calculated from (1.6 m*N)*(135 radians/s) B. 0.034 kg*m , calculated from (1.6 m*N)/((135 radians/s)/(2.9 s)) 2 C. 4.6 kg*m , calculated from (1.6 m*N)*(2.9 s) D. 6.3×10 +02 kg*m , calculated from (1.6 m*N)*(135 radians/s)*(2.9 s) 10. What torque is required to bring the centrifuge to rest from an angular speed of 135 radians/s* in 0.85 s? A. 5.5 m*N, calculated from (1.6 m*N)*(2.9 s)/(0.85 s) B. 1.5 m*N, calculated from (0.85 s)*(2.9 s)/(1.6 m*N) C. 0.18 m*N, calculated from (0.85 s)/((2.9 s)*(1.6 m*N)) D. 2.1 m*N, calculated from (2.9 s)/((0.85 s)*(1.6 m*N)) page 3 of 9 Drawing of a tetherball. 11. A 0.7 kg "tether ball" rotates on a string of length 2.3 m a distance 1.1 m from the pole. (See figure.) What is the moment of inertia of the tether ball? Ignore the mass of the string. 2 2 A. 0.85 kg*m , calculated from (0.7 kg)*(1.1 m) B. 3.7 kg*m , calculated from (0.7 kg)*(2.3 m) 2 C. 1.8 kg*m/s, calculated from (0.7 kg)*(2.6 m/s) D. 8.8 kg*m /s, calculated from (0.7 kg)*(2.3 m) *((2.6 m/s)/(1.1 m)) E. 0.77 kg*m, calculated from (0.7 kg)*(1.1 m) 12. What is the torque on the tether ball if air resistance exerts a force 1.8 N on it? A. 2 m*N, calculated from (1.1 m)*(1.8 N) B. 1.8 m*N, calculated from (1.8 N) C. 1.6 m*N, calculated from (1.8 N)/(1.1 m) D. 0.61 m*N, calculated from (1.1 m)/(1.8 N) page 4 of 9 13. What is the angular acceleration of the tether ball due to air resistance? [Reminder, units are important. Remember Mars Climate Orbiter!] 2 A. 2.6 m/s , calculated from (1.8 N)/(0.7 kg) B. 2.3 radians/s, calculated from (1.1 m)*(1.8 N)/((0.7 kg)*(1.1 m) ) 2 2 C. 2.3 radians/s , calculated from (1.1 m)*(1.8 N)/((0.7 kg)*(1.1 m) ) D. 2.3 rpm/s, calculated from (1.1 m)*(1.8 N)/((0.7 kg)*(1.1 m) ) 3 2 E. 2.3 radians/s , calculated from (1.1 m)*(1.8 N)/((0.7 kg)*(1.1 m) ) 14. An almost horizontal telephone wire pulls the top of a 5.3 m tall pole with a horizontal component of force 263 N A "guy wire" on the opposite side of the pole has an angle 48 deg from horizontal. (See figure.) What is the tension in the "guy wire" such that the pole experiences no torque about its base? A. 2.6×10 +02 N, calculated from (263 N) +02 B. 5.3×10 N, calculated from 2*(263 N) C. 3.9×10 +02 N, calculated from (263 N)/cos((48 deg)*pi/180) +02 D. 3.5×10 N, calculated from (263 N)/sin((48 deg)*pi/180) 15. Does the weight of the pole exert a torque about the base of the pole? A. No B. Yes page 5 of 9 Cartoon of a person climbing a wall using a rope. 16. A 66 kg climber stands perpendicular to a vertical wall, hanging on to a rope with makes an angle 62 deg from horizontal. (See figure.) The climber's center of gravity is 0.98 m from the wall, and his shoulders are 1.47 m from the wall. What is the tension in the rope? [Hint: use the feet of the climber as the center of rotation when calculating torque.] +02 2 A. 4.9×10 N, calculated from (0.98 m)*(66 kg)*(9.80 m/s )/((1.47 m)*sin((62 deg)*pi/180)) B. 9.7×10 +02 N, calculated from (66 kg)*(9.80 m/s )*(1.47 m)/(0.98 m) +02 2 C. 6.5×10 N, calculated from (66 kg)*(9.80 m/s ) D. 4.3×10 +02 N, calculated from (66 kg)*(9.80 m/s )*(0.98 m)/(1.47 m) 17. What force horizontal force does the climber feel in his legs? (I.e., what is the normal force from the wall on the climber?) A. 6.5×10 +02 N, calculated from (66 kg)*(9.80 m/s ) B. 4.3×10 +02 N, calculated from (66 kg)*(9.80 m/s )*(0.98 m)/(1.47 m) +02 2 C. 2.3×10 N, calculated from (0.98 m)*(66 kg)*(9.80 m/s )/((1.47 m)*tan((62 deg)*pi/180)) D. 9.7×10 +02 N, calculated from (66 kg)*(9.80 m/s )*(1.47 m)/(0.98 m) 18. What force of friction on the climber's feet is required? A. 4.3×10 +02 N, calculated from (66 kg)*(9.80 m/s )*(0.98 m)/(1.47 m) +02 2 2 B. 2.2×10 N, calculated from (66 kg)*(9.80 m/s ) - (0.98 m)*(66 kg)*(9.80 m/s )/(1.47 m) C. 6.5×1

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