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  1. Preface
  2. Unit 1. Mechanics
    1. 1 Units and Measurement
      1. Introduction
      2. 1.1 The Scope and Scale of Physics
      3. 1.2 Units and Standards
      4. 1.3 Unit Conversion
      5. 1.4 Dimensional Analysis
      6. 1.5 Estimates and Fermi Calculations
      7. 1.6 Significant Figures
      8. 1.7 Solving Problems in Physics
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    2. 2 Vectors
      1. Introduction
      2. 2.1 Scalars and Vectors
      3. 2.2 Coordinate Systems and Components of a Vector
      4. 2.3 Algebra of Vectors
      5. 2.4 Products of Vectors
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    3. 3 Motion Along a Straight Line
      1. Introduction
      2. 3.1 Position, Displacement, and Average Velocity
      3. 3.2 Instantaneous Velocity and Speed
      4. 3.3 Average and Instantaneous Acceleration
      5. 3.4 Motion with Constant Acceleration
      6. 3.5 Free Fall
      7. 3.6 Finding Velocity and Displacement from Acceleration
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    4. 4 Motion in Two and Three Dimensions
      1. Introduction
      2. 4.1 Displacement and Velocity Vectors
      3. 4.2 Acceleration Vector
      4. 4.3 Projectile Motion
      5. 4.4 Uniform Circular Motion
      6. 4.5 Relative Motion in One and Two Dimensions
      7. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    5. 5 Newton's Laws of Motion
      1. Introduction
      2. 5.1 Forces
      3. 5.2 Newton's First Law
      4. 5.3 Newton's Second Law
      5. 5.4 Mass and Weight
      6. 5.5 Newton’s Third Law
      7. 5.6 Common Forces
      8. 5.7 Drawing Free-Body Diagrams
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    6. 6 Applications of Newton's Laws
      1. Introduction
      2. 6.1 Solving Problems with Newton’s Laws
      3. 6.2 Friction
      4. 6.3 Centripetal Force
      5. 6.4 Drag Force and Terminal Speed
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    7. 7 Work and Kinetic Energy
      1. Introduction
      2. 7.1 Work
      3. 7.2 Kinetic Energy
      4. 7.3 Work-Energy Theorem
      5. 7.4 Power
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    8. 8 Potential Energy and Conservation of Energy
      1. Introduction
      2. 8.1 Potential Energy of a System
      3. 8.2 Conservative and Non-Conservative Forces
      4. 8.3 Conservation of Energy
      5. 8.4 Potential Energy Diagrams and Stability
      6. 8.5 Sources of Energy
      7. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
    9. 9 Linear Momentum and Collisions
      1. Introduction
      2. 9.1 Linear Momentum
      3. 9.2 Impulse and Collisions
      4. 9.3 Conservation of Linear Momentum
      5. 9.4 Types of Collisions
      6. 9.5 Collisions in Multiple Dimensions
      7. 9.6 Center of Mass
      8. 9.7 Rocket Propulsion
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    10. 10 Fixed-Axis Rotation
      1. Introduction
      2. 10.1 Rotational Variables
      3. 10.2 Rotation with Constant Angular Acceleration
      4. 10.3 Relating Angular and Translational Quantities
      5. 10.4 Moment of Inertia and Rotational Kinetic Energy
      6. 10.5 Calculating Moments of Inertia
      7. 10.6 Torque
      8. 10.7 Newton’s Second Law for Rotation
      9. 10.8 Work and Power for Rotational Motion
      10. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    11. 11 Angular Momentum
      1. Introduction
      2. 11.1 Rolling Motion
      3. 11.2 Angular Momentum
      4. 11.3 Conservation of Angular Momentum
      5. 11.4 Precession of a Gyroscope
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    12. 12 Static Equilibrium and Elasticity
      1. Introduction
      2. 12.1 Conditions for Static Equilibrium
      3. 12.2 Examples of Static Equilibrium
      4. 12.3 Stress, Strain, and Elastic Modulus
      5. 12.4 Elasticity and Plasticity
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    13. 13 Gravitation
      1. Introduction
      2. 13.1 Newton's Law of Universal Gravitation
      3. 13.2 Gravitation Near Earth's Surface
      4. 13.3 Gravitational Potential Energy and Total Energy
      5. 13.4 Satellite Orbits and Energy
      6. 13.5 Kepler's Laws of Planetary Motion
      7. 13.6 Tidal Forces
      8. 13.7 Einstein's Theory of Gravity
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    14. 14 Fluid Mechanics
      1. Introduction
      2. 14.1 Fluids, Density, and Pressure
      3. 14.2 Measuring Pressure
      4. 14.3 Pascal's Principle and Hydraulics
      5. 14.4 Archimedes’ Principle and Buoyancy
      6. 14.5 Fluid Dynamics
      7. 14.6 Bernoulli’s Equation
      8. 14.7 Viscosity and Turbulence
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
  3. Unit 2. Waves and Acoustics
    1. 15 Oscillations
      1. Introduction
      2. 15.1 Simple Harmonic Motion
      3. 15.2 Energy in Simple Harmonic Motion
      4. 15.3 Comparing Simple Harmonic Motion and Circular Motion
      5. 15.4 Pendulums
      6. 15.5 Damped Oscillations
      7. 15.6 Forced Oscillations
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    2. 16 Waves
      1. Introduction
      2. 16.1 Traveling Waves
      3. 16.2 Mathematics of Waves
      4. 16.3 Wave Speed on a Stretched String
      5. 16.4 Energy and Power of a Wave
      6. 16.5 Interference of Waves
      7. 16.6 Standing Waves and Resonance
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    3. 17 Sound
      1. Introduction
      2. 17.1 Sound Waves
      3. 17.2 Speed of Sound
      4. 17.3 Sound Intensity
      5. 17.4 Normal Modes of a Standing Sound Wave
      6. 17.5 Sources of Musical Sound
      7. 17.6 Beats
      8. 17.7 The Doppler Effect
      9. 17.8 Shock Waves
      10. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
  4. A | Units
  5. B | Conversion Factors
  6. C | Fundamental Constants
  7. D | Astronomical Data
  8. E | Mathematical Formulas
  9. F | Chemistry
  10. G | The Greek Alphabet
  11. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
    8. Chapter 8
    9. Chapter 9
    10. Chapter 10
    11. Chapter 11
    12. Chapter 12
    13. Chapter 13
    14. Chapter 14
    15. Chapter 15
    16. Chapter 16
    17. Chapter 17
  12. Index

Check Your Understanding

11.1

a. μStanθ1+(mr2/ICM)μStanθ1+(mr2/ICM); inserting the angle and noting that for a hollow cylinder ICM=mr2,ICM=mr2, we have μStan60°1+(mr2/mr2)=12tan60°=0.87;μStan60°1+(mr2/mr2)=12tan60°=0.87; we are given a value of 0.6 for the coefficient of static friction, which is less than 0.87, so the condition isn’t satisfied and the hollow cylinder will slip; b. The solid cylinder obeys the condition μS13tanθ=13tan60°=0.58.μS13tanθ=13tan60°=0.58. The value of 0.6 for μSμS satisfies this condition, so the solid cylinder will not slip.

11.2

From the figure, we see that the cross product of the radius vector with the momentum vector gives a vector directed out of the page. Inserting the radius and momentum into the expression for the angular momentum, we have
l=r×p=(0.4mi^)×(1.67×10−27kg(4.0×106m/s)j^)=2.7×10−21kg·m2/sk^l=r×p=(0.4mi^)×(1.67×10−27kg(4.0×106m/s)j^)=2.7×10−21kg·m2/sk^

11.3

Isphere=25mr2,Icylinder=12mr2Isphere=25mr2,Icylinder=12mr2; Taking the ratio of the angular momenta, we have:
LcylinderLsphere=Icylinderω0Isphereω0=12mr225mr2=54LcylinderLsphere=Icylinderω0Isphereω0=12mr225mr2=54. Thus, the cylinder has 25%25% more angular momentum. This is because the cylinder has more mass distributed farther from the axis of rotation.

11.4

Using conservation of angular momentum, we have
I(4.0rev/min)=1.25Iωf,ωf=1.01.25(4.0rev/min)=3.2rev/minI(4.0rev/min)=1.25Iωf,ωf=1.01.25(4.0rev/min)=3.2rev/min

11.5

The Moon’s gravity is 1/6 that of Earth’s. By examining Equation 11.12, we see that the top’s precession frequency is linearly proportional to the acceleration of gravity. All other quantities, mass, moment of inertia, and spin rate are the same on the Moon. Thus, the precession frequency on the Moon is
ωP(Moon)=16ωP(Earth)=16(5.0rad/s)=0.83rad/s.ωP(Moon)=16ωP(Earth)=16(5.0rad/s)=0.83rad/s.

Conceptual Questions

1.

No, the static friction force is zero.

3.

The wheel is more likely to slip on a steep incline since the coefficient of static friction must increase with the angle to keep rolling motion without slipping.

5.

The cylinder reaches a greater height. By Equation 11.4, its acceleration in the direction down the incline would be less.

7.

All points on the straight line will give zero angular momentum, because a vector crossed into a parallel vector is zero.

9.

The particle must be moving on a straight line that passes through the chosen origin.

11.

Without the small propeller, the body of the helicopter would rotate in the opposite sense to the large propeller in order to conserve angular momentum. The small propeller exerts a thrust at a distance R from the center of mass of the aircraft to prevent this from happening.

13.

The angular velocity increases because the moment of inertia is decreasing.

15.

More mass is concentrated near the rotational axis, which decreases the moment of inertia causing the star to increase its angular velocity.

17.

A torque is needed in the direction perpendicular to the angular momentum vector in order to change its direction. These forces on the space vehicle are external to the container in which the gyroscope is mounted and do not impart torques to the gyroscope’s rotating disk.

Problems

19.

vCM=Rωω=66.7rad/svCM=Rωω=66.7rad/s

21.

α=3.3rad/s2α=3.3rad/s2

23.

ICM=25mr2,aCM=3.5m/s2;x=15.75mICM=25mr2,aCM=3.5m/s2;x=15.75m

25.

positive is down the incline plane;
aCM=mgsinθm+(ICM/r2)ICM=r2[mgsin30aCMm]aCM=mgsinθm+(ICM/r2)ICM=r2[mgsin30aCMm],
xx0=v0t12aCMt2aCM=2.96m/s2,xx0=v0t12aCMt2aCM=2.96m/s2,
ICM=0.66mr2ICM=0.66mr2

27.

α=67.9rad/s2α=67.9rad/s2,
(aCM)x=1.5m/s2(aCM)x=1.5m/s2

29.

W=−1080.0JW=−1080.0J

31.

Mechanical energy at the bottom equals mechanical energy at the top;
12mv02+12(12mr2)(v0r)2=mghh=1g(12+14)v0212mv02+12(12mr2)(v0r)2=mghh=1g(12+14)v02,
h=7.7m,h=7.7m, so the distance up the incline is 22.5m22.5m.

33.

Use energy conservation
12mv02+12ICylω02=mghCyl12mv02+12ICylω02=mghCyl,
12mv02+12ISphω02=mghSph12mv02+12ISphω02=mghSph.
Subtracting the two equations, eliminating the initial translational energy, we have
12ICylω0212ISphω02=mg(hCylhSph)12ICylω0212ISphω02=mg(hCylhSph),
12mr2(v0r)21223mr2(v0r)2=mg(hCylhSph)12mr2(v0r)21223mr2(v0r)2=mg(hCylhSph),
12v021223v02=g(hCylhSph)12v021223v02=g(hCylhSph),
hCylhSph=1g(1213)v02=19.8m/s2(16)(5.0m/s)2=0.43mhCylhSph=1g(1213)v02=19.8m/s2(16)(5.0m/s)2=0.43m.
Thus, the hollow sphere, with the smaller moment of inertia, rolls up to a lower height of 1.00.43=0.57m.1.00.43=0.57m.

35.

The magnitude of the cross product of the radius to the bird and its momentum vector yields rpsinθrpsinθ, which gives rsinθrsinθ as the altitude of the bird h. The direction of the angular momentum is perpendicular to the radius and momentum vectors, which we choose arbitrarily as k^k^, which is in the plane of the ground:
L=r×p=hmvk^=(300.0m)(2.0kg)(20.0m/s)k^=12,000.0kg·m2/sk^L=r×p=hmvk^=(300.0m)(2.0kg)(20.0m/s)k^=12,000.0kg·m2/sk^

37.

a. l=45.0kg·m2/sk^l=45.0kg·m2/sk^;
b. τ=10.0N·mk^τ=10.0N·mk^

39.

a. l1=−0.4kg·m2/sk^l1=−0.4kg·m2/sk^, l2=l4=0l2=l4=0,
l3=1.35kg·m2/sk^l3=1.35kg·m2/sk^; b. L=0.95kg·m2/sk^L=0.95kg·m2/sk^

41.

a. L=1.0×1011kg·m2/sL=1.0×1011kg·m2/s; b. No, the angular momentum stays the same since the cross-product involves only the perpendicular distance from the plane to the ground no matter where it is along its path.

43.

a. v=gtj^,r=di^,l=mdgtk^v=gtj^,r=di^,l=mdgtk^;
b. F=mgj^,τ=dmgk^F=mgj^,τ=dmgk^; c. yes

45.

a. mgh=12m(rω)2+1225mr2ω2mgh=12m(rω)2+1225mr2ω2;
ω=51.2rad/sω=51.2rad/s;
L=16.4kg·m2/sL=16.4kg·m2/s;
b. ω=72.5rad/sω=72.5rad/s;
L=23.2kg·m2/sL=23.2kg·m2/s

47.

a. I=720.0kg·m2I=720.0kg·m2; α=4.20rad/s2α=4.20rad/s2;
ω(10s)=42.0rad/sω(10s)=42.0rad/s; L=3.02×104kg·m2/sL=3.02×104kg·m2/s;
ω(20s)=84.0rad/sω(20s)=84.0rad/s;
b. τ=3.03×103N·mτ=3.03×103N·m

49.

a. L=1.131×107kg·m2/sL=1.131×107kg·m2/s;
b. τ=3.77×104N·mτ=3.77×104N·m

51.

ω=28.6rad/sL=2.6kg·m2/sω=28.6rad/sL=2.6kg·m2/s

53.

Lf=25MS(3.5×103km)22πTfLf=25MS(3.5×103km)22πTf,
(7.0×105km)22π28days=(3.5×103km)22πTfTf=28days(3.5× 10 3km)2(7.0× 10 5km)2=7.0×104day=60.5s(7.0×105km)22π28days=(3.5×103km)22πTfTf=28days(3.5× 10 3km)2(7.0× 10 5km)2=7.0×104day=60.5s

55.

ff=2.1rev/sf0=0.5rev/sff=2.1rev/sf0=0.5rev/s

57.

rPmvP=rAmvAvP=18.3km/srPmvP=rAmvAvP=18.3km/s

59.

a. Idisk=5.0×10−4kg·m2Idisk=5.0×10−4kg·m2,
Ibug=2.0×10−4kg·m2Ibug=2.0×10−4kg·m2,
(Idisk+Ibug)ω1=Idiskω2,ω2=14.0rad/s(Idisk+Ibug)ω1=Idiskω2,ω2=14.0rad/s
b. ΔK=0.014JΔK=0.014J;
c. ω3=10.0rad/sω3=10.0rad/s back to the original value;
d. 12(Idisk+Ibug)ω32=0.035J12(Idisk+Ibug)ω32=0.035J back to the original value;
e. work of the bug crawling on the disk

61.

Li=400.0kg·m2/sLi=400.0kg·m2/s,
Lf=500.0kg·m2ωLf=500.0kg·m2ω,
ω=0.80rad/sω=0.80rad/s

63.

I0=340.48kg·m2I0=340.48kg·m2,
If=268.8kg·m2If=268.8kg·m2,
ωf=25.33rpmωf=25.33rpm

65.

a. L=280kg·m2/sL=280kg·m2/s,
If=89.6kg·m2If=89.6kg·m2,
ωf=3.125rad/sωf=3.125rad/s; b. Ki=437.5JKi=437.5J,
Kf=437.5JKf=437.5J

67.

Moment of inertia in the record spin: I0=0.5kg·m2I0=0.5kg·m2,
If=1.1kg·m2If=1.1kg·m2,
ωf=I0Ifω0ff=155.5rev/minωf=I0Ifω0ff=155.5rev/min

69.

Her spin rate in the air is: ff=2.0rev/sff=2.0rev/s;
She can do four flips in the air.

71.

Moment of inertia with all children aboard:
I0=2.4×105kg·m2I0=2.4×105kg·m2;
If=1.5×105kg·m2If=1.5×105kg·m2;
ff=0.3rev/sff=0.3rev/s

73.

I0=1.00×1010kg·m2I0=1.00×1010kg·m2,
If=9.94×109kg·m2If=9.94×109kg·m2,
ff=3.32rev/minff=3.32rev/min

75.

I=2.5×10−3kg·m2I=2.5×10−3kg·m2,
ωP=0.78rad/sωP=0.78rad/s

77.

a. LEarth=7.06×1033kg·m2/sLEarth=7.06×1033kg·m2/s,
ΔL=5.63×1033kg·m2/sΔL=5.63×1033kg·m2/s;
b. τ=1.4×1022N·mτ=1.4×1022N·m;
c. The two forces at the equator would have the same magnitude but different directions, one in the north direction and the other in the south direction on the opposite side of Earth. The angle between the forces and the lever arms to the center of Earth is 90°90°, so a given torque would have magnitude τ=FREsin90°=FREτ=FREsin90°=FRE. Both would provide a torque in the same direction:
τ=2FREF=1.3×1015Nτ=2FREF=1.3×1015N

Additional Problems

79.

aCM=310gaCM=310g,
v2=v02+2aCMxv2=(7.0m/s)22(310g)x,v2=v02+2aCMxv2=(7.0m/s)22(310g)x, v2=0x=8.34m;v2=0x=8.34m;
b. t=vv0aCM,t=vv0aCM, v=v0+aCMtt=2.38sv=v0+aCMtt=2.38s;
The hollow sphere has a larger moment of inertia, and therefore is harder to bring to a rest than the marble, or solid sphere. The distance travelled is larger and the time elapsed is longer.

81.

a. W=−500.0JW=−500.0J;
b. K+Ugrav=constantK+Ugrav=constant,
500J+0=0+(6.0kg)(9.8m/s2)h500J+0=0+(6.0kg)(9.8m/s2)h,
h=8.5m,d=17.0mh=8.5m,d=17.0m;
The moment of inertia is less for the hollow sphere, therefore less work is required to stop it. Likewise it rolls up the incline a shorter distance than the hoop.

83.

a. τ=34.0N·mτ=34.0N·m;
b. l=mr2ωω=3.6rad/sl=mr2ωω=3.6rad/s

85.

a. dM=3.85×108mdM=3.85×108m average distance to the Moon; orbital period 27.32d=2.36×106s27.32d=2.36×106s; speed of the Moon 2π3.85×108m2.36×106s=1.0×103m/s2π3.85×108m2.36×106s=1.0×103m/s; mass of the Moon 7.35×1022kg7.35×1022kg,
L=2.90×1034kgm2/sL=2.90×1034kgm2/s;
b. radius of the Moon 1.74×106m1.74×106m; the orbital period is the same as (a): ω=2.66×10−6rad/sω=2.66×10−6rad/s,
L=2.37×1029kg·m2/sL=2.37×1029kg·m2/s;
The orbital angular momentum is 1.22×1051.22×105 times larger than the rotational angular momentum for the Moon.

87.

I=0.135kg·m2I=0.135kg·m2,
α=4.19rad/s2,α=4.19rad/s2, ω=ω0+αtω=ω0+αt,
ω(5s)=21.0rad/s,ω(5s)=21.0rad/s, L=2.84kg·m2/sL=2.84kg·m2/s,
ω(10s)=41.9rad/s,ω(10s)=41.9rad/s, L=5.66kg·m/s2L=5.66kg·m/s2

89.

In the conservation of angular momentum equation, the rotation rate appears on both sides so we keep the (rev/min) notation as the angular velocity can be multiplied by a constant to get (rev/min):
Li=−0.04kg·m2(300.0rev/min), Lf=0.08kg·m2ffff=−150.0rev/min clockwiseLi=−0.04kg·m2(300.0rev/min), Lf=0.08kg·m2ffff=−150.0rev/min clockwise

91.

I0ω0=IfωfI0ω0=Ifωf,
I0=6120.0kg·m2I0=6120.0kg·m2,
If=1180.0kg·m2If=1180.0kg·m2,
ωf=31.1rev/minωf=31.1rev/min

93.

Li=1.00×107kg·m2/sLi=1.00×107kg·m2/s,
If=2.025×105kg·m2If=2.025×105kg·m2,
ωf=7.86rev/sωf=7.86rev/s

Challenge Problems

95.

Assume the roll accelerates forward with respect to the ground with an acceleration aa. Then it accelerates backwards relative to the truck with an acceleration (aa)(aa).

The forces on a cylinder on a horizontal surface are shown. The cylinder has radius R and moment of inertia one half m R squared and is centered on an x y coordinate system that has positive x to the right and positive y up. Force m g acts on the center of the cylinder and points down. Force N points up and acts at the contact point where the cylinder touches the surface. Force f sub s points to the right and acts at the contact point where the cylinder touches the surface.


Also, Rα=aaI=12mR2 Fx=fs=maRα=aaI=12mR2 Fx=fs=ma,
τ=fsR=Iα=IaaRfs=IR2(aa)=12m(aa) τ=fsR=Iα=IaaRfs=IR2(aa)=12m(aa),
Solving for aa: fs=12m(aa)fs=12m(aa); a=a3a=a3,
xx0=v0t+12at2;d=13at2;t=3da;xx0=v0t+12at2;d=13at2;t=3da;
therefore, s=1.5ds=1.5d

97.

a. The tension in the string provides the centripetal force such that Tsinθ=mrω2Tsinθ=mrω2. The component of the tension that is vertical opposes the gravitational force such that Tcosθ=mgTcosθ=mg. This gives T=5.7NT=5.7N. We solve for r=0.16mr=0.16m. This gives the length of the string as r=0.32mr=0.32m.
At ω=10.0rad/sω=10.0rad/s, there is a new angle, tension, and perpendicular radius to the rod. Dividing the two equations involving the tension to eliminate it, we have sinθcosθ=(0.32msinθ)ω2g1cosθ=0.32mω2gsinθcosθ=(0.32msinθ)ω2g1cosθ=0.32mω2g;
cosθ=0.31θ=72.2°cosθ=0.31θ=72.2°; b. linitial=0.08kg·m2/slinitial=0.08kg·m2/s,
lfinal=0.46kg·m2/slfinal=0.46kg·m2/s; c. No, the cosine of the angle is inversely proportional to the square of the angular velocity, therefore in order for θ90°,ωθ90°,ω. The rod would have to spin infinitely fast.

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