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University Physics Volume 1

10.1 Rotational Variables

University Physics Volume 110.1 Rotational Variables
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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

Learning Objectives

By the end of this section, you will be able to:
  • Describe the physical meaning of rotational variables as applied to fixed-axis rotation
  • Explain how angular velocity is related to tangential speed
  • Calculate the instantaneous angular velocity given the angular position function
  • Find the angular velocity and angular acceleration in a rotating system
  • Calculate the average angular acceleration when the angular velocity is changing
  • Calculate the instantaneous angular acceleration given the angular velocity function

So far in this text, we have mainly studied translational motion, including the variables that describe it: displacement, velocity, and acceleration. Now we expand our description of motion to rotation—specifically, rotational motion about a fixed axis. We will find that rotational motion is described by a set of related variables similar to those we used in translational motion.

Angular Velocity

Uniform circular motion (discussed previously in Motion in Two and Three Dimensions) is motion in a circle at constant speed. Although this is the simplest case of rotational motion, it is very useful for many situations, and we use it here to introduce rotational variables.

In Figure 10.2, we show a particle moving in a circle. The coordinate system is fixed and serves as a frame of reference to define the particle’s position. Its position vector from the origin of the circle to the particle sweeps out the angle θθ, which increases in the counterclockwise direction as the particle moves along its circular path. The angle θθ is called the angular position of the particle. As the particle moves in its circular path, it also traces an arc length s.

Figure is a graph that shows a particle moving counterclockwise. Vector r from the origin of the co-ordinate system to the point s on the pass of a particle forms an angle theta with the X axis.
Figure 10.2 A particle follows a circular path. As it moves counterclockwise, it sweeps out a positive angle θθ with respect to the x-axis and traces out an arc length s.

The angle is related to the radius of the circle and the arc length by

θ=sr.θ=sr.
(10.1)

The angle θθ, the angular position of the particle along its path, has units of radians (rad). There are 2π2π radians in 360°.360°. Note that the radian measure is a ratio of length measurements, and therefore is a dimensionless quantity. As the particle moves along its circular path, its angular position changes and it undergoes angular displacements Δθ.Δθ.

We can assign vectors to the quantities in Equation 10.1. The angle θθ is a vector out of the page in Figure 10.2. The angular position vector rr and the arc length ss both lie in the plane of the page. These three vectors are related to each other by

s=θ×r..s=θ×r..
(10.2)

That is, the arc length is the cross product of the angle vector and the position vector, as shown in Figure 10.3.

Figure is an XYZ coordinate system that shows three vectors. Vector Theta points in the positive Z direction. Vector s is in the XY plane. Vector r is directed from the origin of the coordinate system to the beginning of the vector s.
Figure 10.3 The angle vector points along the z-axis and the position vector and arc length vector both lie in the xy-plane. We see that s=θ×rs=θ×r. All three vectors are perpendicular to each other.

The magnitude of the angular velocity, denoted by ωω, is the time rate of change of the angle θθ as the particle moves in its circular path. The instantaneous angular velocity is defined as the limit in which Δt0Δt0 in the average angular velocity ω=ΔθΔtω=ΔθΔt:

ω=limΔt0ΔθΔt=dθdt,ω=limΔt0ΔθΔt=dθdt,
(10.3)

where θθ is the angle of rotation (Figure 10.2). The units of angular velocity are radians per second (rad/s). Angular velocity can also be referred to as the rotation rate in radians per second. In many situations, we are given the rotation rate in revolutions/s or cycles/s. To find the angular velocity, we must multiply revolutions/s by 2π2π, since there are 2π2π radians in one complete revolution. Since the direction of a positive angle in a circle is counterclockwise, we take counterclockwise rotations as being positive and clockwise rotations as negative.

We can see how angular velocity is related to the tangential speed of the particle by differentiating Equation 10.1 with respect to time. We rewrite Equation 10.1 as

s=rθ.s=rθ.

Taking the derivative with respect to time and noting that the radius r is a constant, we have

dsdt=ddt(rθ)=θdrdt+rdθdt=rdθdtdsdt=ddt(rθ)=θdrdt+rdθdt=rdθdt

where θdrdt=0θdrdt=0. Here dsdtdsdt is just the tangential speed vtvt of the particle in Figure 10.2. Thus, by using Equation 10.3, we arrive at

vt=rω.vt=rω.
(10.4)

That is, the tangential speed of the particle is its angular velocity times the radius of the circle. From Equation 10.4, we see that the tangential speed of the particle increases with its distance from the axis of rotation for a constant angular velocity. This effect is shown in Figure 10.4. Two particles are placed at different radii on a rotating disk with a constant angular velocity. As the disk rotates, the tangential speed increases linearly with the radius from the axis of rotation. In Figure 10.4, we see that v1=r1ω1v1=r1ω1 and v2=r2ω2v2=r2ω2. But the disk has a constant angular velocity, so ω1=ω2ω1=ω2. This means v1r1=v2r2v1r1=v2r2 or v2=(r2r1)v1v2=(r2r1)v1. Thus, since r2>r1r2>r1, v2>v1v2>v1.

Figure shows two particles on a rotating disk. Particle 1 is at the distance r1 from the axis of rotation and moved with the speed v1. Particle 2 is at the distance r2 from the axis of roation and moves with the speed v2.
Figure 10.4 Two particles on a rotating disk have different tangential speeds, depending on their distance to the axis of rotation.

Up until now, we have discussed the magnitude of the angular velocity ω=dθ/dt,ω=dθ/dt, which is a scalar quantity—the change in angular position with respect to time. The vector ωω is the vector associated with the angular velocity and points along the axis of rotation. This is useful because when a rigid body is rotating, we want to know both the axis of rotation and the direction that the body is rotating about the axis, clockwise or counterclockwise. The angular velocity ωω gives us this information. The angular velocity ωω has a direction determined by what is called the right-hand rule. The right-hand rule is such that if the fingers of your right hand wrap counterclockwise from the x-axis (the direction in which θθ increases) toward the y-axis, your thumb points in the direction of the positive z-axis (Figure 10.5). An angular velocity ωω that points along the positive z-axis therefore corresponds to a counterclockwise rotation, whereas an angular velocity ωω that points along the negative z-axis corresponds to a clockwise rotation.

Figure is a graph that shows the XYZ coordinate system with the counterclockwise rotation in the XY plane. The angular velocity points in the positive Z-direction.
Figure 10.5 For counterclockwise rotation in the coordinate system shown, the angular velocity points in the positive z-direction by the right-hand-rule.

Similar to Equation 10.2, one can state a cross product relation to the vector of the tangential velocity as stated in Equation 10.4. Therefore, we have

v=ω×r.v=ω×r.
(10.5)

That is, the tangential velocity is the cross product of the angular velocity and the position vector, as shown in Figure 10.6. From part (a) of this figure, we see that with the angular velocity in the positive z-direction, the rotation in the xy-plane is counterclockwise. In part (b), the angular velocity is in the negative z-direction, giving a clockwise rotation in the xy-plane.

Figure A is an XYZ coordinate system that shows three vectors. Vector Omega points in the positive Z direction. Vector v is in the XY plane. Vector r is directed from the origin of the coordinate system to the beginning of the vector v. Figure B is an XYZ coordinate system that shows three vectors. Vector Omega points in the negative Z direction. Vector v is in the XY plane. Vector r is directed from the origin of the coordinate system to the beginning of the vector v.
Figure 10.6 The vectors shown are the angular velocity, position, and tangential velocity. (a) The angular velocity points in the positive z-direction, giving a counterclockwise rotation in the xy-plane. (b) The angular velocity points in the negative z-direction, giving a clockwise rotation.

Example 10.1

Rotation of a Flywheel A flywheel rotates such that it sweeps out an angle at the rate of θ=ωt=(45.0rad/s)tθ=ωt=(45.0rad/s)t radians. The wheel rotates counterclockwise when viewed in the plane of the page. (a) What is the angular velocity of the flywheel? (b) What direction is the angular velocity? (c) How many radians does the flywheel rotate through in 30 s? (d) What is the tangential speed of a point on the flywheel 10 cm from the axis of rotation?

Strategy The functional form of the angular position of the flywheel is given in the problem as θ(t)=ωtθ(t)=ωt, so by taking the derivative with respect to time, we can find the angular velocity. We use the right-hand rule to find the angular velocity. To find the angular displacement of the flywheel during 30 s, we seek the angular displacement ΔθΔθ, where the change in angular position is between 0 and 30 s. To find the tangential speed of a point at a distance from the axis of rotation, we multiply its distance times the angular velocity of the flywheel.

Solution

  1. ω=dθdt=45rad/sω=dθdt=45rad/s. We see that the angular velocity is a constant.
  2. By the right-hand rule, we curl the fingers in the direction of rotation, which is counterclockwise in the plane of the page, and the thumb points in the direction of the angular velocity, which is out of the page.
  3. Δθ=θ(30s)θ(0s)=45.0(30.0s)45.0(0s)=1350.0radΔθ=θ(30s)θ(0s)=45.0(30.0s)45.0(0s)=1350.0rad.
  4. vt=rω=(0.1m)(45.0rad/s)=4.5m/svt=rω=(0.1m)(45.0rad/s)=4.5m/s.

Significance In 30 s, the flywheel has rotated through quite a number of revolutions, about 215 if we divide the angular displacement by 2π2π. A massive flywheel can be used to store energy in this way, if the losses due to friction are minimal. Recent research has considered superconducting bearings on which the flywheel rests, with zero energy loss due to friction.

Angular Acceleration

We have just discussed angular velocity for uniform circular motion, but not all motion is uniform. Envision an ice skater spinning with his arms outstretched—when he pulls his arms inward, his angular velocity increases. Or think about a computer’s hard disk slowing to a halt as the angular velocity decreases. We will explore these situations later, but we can already see a need to define an angular acceleration for describing situations where ωω changes. The faster the change in ωω, the greater the angular acceleration. We define the instantaneous angular acceleration αα as the derivative of angular velocity with respect to time:

α=limΔt0ΔωΔt=dωdt=d2θdt2,α=limΔt0ΔωΔt=dωdt=d2θdt2,
(10.6)

where we have taken the limit of the average angular acceleration, α=ΔωΔtα=ΔωΔt as Δt0Δt0.

The units of angular acceleration are (rad/s)/s, or rad/s2rad/s2.

In the same way as we defined the vector associated with angular velocity ωω, we can define αα, the vector associated with angular acceleration (Figure 10.7). If the angular velocity is along the positive z-axis, as in Figure 10.5, and dωdtdωdt is positive, then the angular acceleration αα is positive and points along the +z-+z- axis. Similarly, if the angular velocity ωω is along the positive z-axis and dωdtdωdt is negative, then the angular acceleration is negative and points along the +z-+z- axis.

Figure A shows rotation in the counterclockwise direction. The angular acceleration is in the same direction as the angular velocity. Text under the figure states “Rotation rate counterclockwise and increasing. Figure B shows rotation in the clockwise direction. The angular acceleration is in the direction opposite to the angular velocity. Text under the figure states “Rotation rate clockwise and decreasing.
Figure 10.7 The rotation is counterclockwise in both (a) and (b) with the angular velocity in the same direction. (a) The angular acceleration is in the same direction as the angular velocity, which increases the rotation rate. (b) The angular acceleration is in the opposite direction to the angular velocity, which decreases the rotation rate.

We can express the tangential acceleration vector as a cross product of the angular acceleration and the position vector. This expression can be found by taking the time derivative of v=ω×rv=ω×r and is left as an exercise:

a=α×r.a=α×r.
(10.7)

The vector relationships for the angular acceleration and tangential acceleration are shown in Figure 10.8.

Figure A is an XYZ coordinate system that shows three vectors. Vector Alpha points in the positive Z direction. Vector a is in the XY plane. Vector r is directed from the origin of the coordinate system to the beginning of the vector a. Figure B is an XYZ coordinate system that shows three vectors. Vector Alpha points in the negative Z direction. Vector a is in the XY plane. Vector r is directed from the origin of the coordinate system to the beginning of the vector a.
Figure 10.8 (a) The angular acceleration is the positive z-direction and produces a tangential acceleration in a counterclockwise sense. (b) The angular acceleration is in the negative z-direction and produces a tangential acceleration in the clockwise sense.

We can relate the tangential acceleration of a point on a rotating body at a distance from the axis of rotation in the same way that we related the tangential speed to the angular velocity. If we differentiate Equation 10.4 with respect to time, noting that the radius r is constant, we obtain

at=rα.at=rα.
(10.8)

Thus, the tangential acceleration atat is the radius times the angular acceleration. Equation 10.4 and Equation 10.8 are important for the discussion of rolling motion (see Angular Momentum).

Let’s apply these ideas to the analysis of a few simple fixed-axis rotation scenarios. Before doing so, we present a problem-solving strategy that can be applied to rotational kinematics: the description of rotational motion.

Problem-Solving Strategy: Rotational Kinematics

  1. Examine the situation to determine that rotational kinematics (rotational motion) is involved.
  2. Identify exactly what needs to be determined in the problem (identify the unknowns). A sketch of the situation is useful.
  3. Make a complete list of what is given or can be inferred from the problem as stated (identify the knowns).
  4. Solve the appropriate equation or equations for the quantity to be determined (the unknown). It can be useful to think in terms of a translational analog, because by now you are familiar with the equations of translational motion.
  5. Substitute the known values along with their units into the appropriate equation and obtain numerical solutions complete with units. Be sure to use units of radians for angles.
  6. Check your answer to see if it is reasonable: Does your answer make sense?

Now let’s apply this problem-solving strategy to a few specific examples.

Example 10.2

A Spinning Bicycle Wheel A bicycle mechanic mounts a bicycle on the repair stand and starts the rear wheel spinning from rest to a final angular velocity of 250 rpm in 5.00 s. (a) Calculate the average angular acceleration in rad/s2rad/s2. (b) If she now hits the brakes, causing an angular acceleration of −87.3 rad/s2rad/s2, how long does it take the wheel to stop?

Strategy The average angular acceleration can be found directly from its definition α=ΔωΔtα=ΔωΔt because the final angular velocity and time are given. We see that Δω=ωfinalωinitial=250rev/minΔω=ωfinalωinitial=250rev/min and ΔtΔt is 5.00 s. For part (b), we know the angular acceleration and the initial angular velocity. We can find the stopping time by using the definition of average angular acceleration and solving for ΔtΔt, yielding

Δt=Δωα.Δt=Δωα.

Solution

  1. Entering known information into the definition of angular acceleration, we get
    α=ΔωΔt=250rpm5.00s.α=ΔωΔt=250rpm5.00s.

    Because ΔωΔω is in revolutions per minute (rpm) and we want the standard units of rad/s2rad/s2 for angular acceleration, we need to convert from rpm to rad/s:
    Δω=250revmin·2πradrev·1min60s=26.2rads.Δω=250revmin·2πradrev·1min60s=26.2rads.

    Entering this quantity into the expression for αα, we get
    α=ΔωΔt=26.2rad/s5.00s=5.24rad/s2.α=ΔωΔt=26.2rad/s5.00s=5.24rad/s2.
  2. Here the angular velocity decreases from 26.2 rad/s (250 rpm) to zero, so that ΔωΔω is −26.2 rad/s, and αα is given to be –87.3 rad/s2rad/s2. Thus,
    Δt=−26.2rad/s−87.3rad/s2=0.300s.Δt=−26.2rad/s−87.3rad/s2=0.300s.

Significance Note that the angular acceleration as the mechanic spins the wheel is small and positive; it takes 5 s to produce an appreciable angular velocity. When she hits the brake, the angular acceleration is large and negative. The angular velocity quickly goes to zero.

Check Your Understanding 10.1

The fan blades on a turbofan jet engine (shown below) accelerate from rest up to a rotation rate of 40.0 rev/s in 20 s. The increase in angular velocity of the fan is constant in time. (The GE90-110B1 turbofan engine mounted on a Boeing 777, as shown, is currently the largest turbofan engine in the world, capable of thrusts of 330–510 kN.)

(a) What is the average angular acceleration?

(b) What is the instantaneous angular acceleration at any time during the first 20 s?

Picture is a photo of an air turbine under the wing of an airplane.
Figure 10.9 (credit: “Bubinator”/ Wikimedia Commons)

Example 10.3

Wind Turbine A wind turbine (Figure 10.10) in a wind farm is being shut down for maintenance. It takes 30 s for the turbine to go from its operating angular velocity to a complete stop in which the angular velocity function is ω(t)=[(ts−1−30.0)2/100.0]rad/sω(t)=[(ts−1−30.0)2/100.0]rad/s. If the turbine is rotating counterclockwise looking into the page, (a) what are the directions of the angular velocity and acceleration vectors? (b) What is the average angular acceleration? (c) What is the instantaneous angular acceleration at t=0.0,15.0,30.0s?t=0.0,15.0,30.0s?

Figure is a drawing of a wind turbine that is rotating counterclockwise, as seen head on.
Figure 10.10 A wind turbine that is rotating counterclockwise, as seen head on.

Strategy

  1. We are given the rotational sense of the turbine, which is counterclockwise in the plane of the page. Using the right hand rule (Figure 10.5), we can establish the directions of the angular velocity and acceleration vectors.
  2. We calculate the initial and final angular velocities to get the average angular acceleration. We establish the sign of the angular acceleration from the results in (a).
  3. We are given the functional form of the angular velocity, so we can find the functional form of the angular acceleration function by taking its derivative with respect to time.

Solution

  1. Since the turbine is rotating counterclockwise, angular velocity ωω points out of the page. But since the angular velocity is decreasing, the angular acceleration αα points into the page, in the opposite sense to the angular velocity.
  2. The initial angular velocity of the turbine, setting t=0,isω=9.0rad/st=0,isω=9.0rad/s. The final angular velocity is zero, so the average angular acceleration is
    α=ΔωΔt=ωω0tt0=09.0rad/s30.00s=−0.3rad/s2.α=ΔωΔt=ωω0tt0=09.0rad/s30.00s=−0.3rad/s2.
  3. Taking the derivative of the angular velocity with respect to time gives α=dωdt=(t30.0)/50.0rad/s2α=dωdt=(t30.0)/50.0rad/s2
    α(0.0s)=−0.6rad/s2,α(15.0s)=−0.3rad/s2,andα(30.0s)=0rad/s.α(0.0s)=−0.6rad/s2,α(15.0s)=−0.3rad/s2,andα(30.0s)=0rad/s.

Significance We found from the calculations in (a) and (b) that the angular acceleration αα and the average angular acceleration αα are negative. The turbine has an angular acceleration in the opposite sense to its angular velocity.

We now have a basic vocabulary for discussing fixed-axis rotational kinematics and relationships between rotational variables. We discuss more definitions and connections in the next section.

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