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Chapter 12, Problem 57PQ

(a)

To determine

The values of b and ω0.

(a)

Expert Solution
Check Mark

Answer to Problem 57PQ

The values of b and ω0 are 0.277s1 and 4.00rad/s respectively.

Explanation of Solution

Write the equation of angular speed of the wheel.

  dθdt=ω0ebt                                                                                            (I)

Here, dθ/dt is the rate of change of angular displacement, ω0 is the initial angular velocity of the wheel, b is the constant, and t is the time interval.

Write the expression for angular velocity.

  ω=dθdt                                                                                                  (II)

Here, ω is the angular speed of the wheel.

Compare equation (I) and (II).

  ω=ω0ebt                                                                                               (III)

Conclusion:

Substitute 0s for t and 4.00rad/s for ω in the equation (III) to find ω0.

  4.00rad/s=ω0eb(0s)4.00rad/s=ω0e04.00rad/s=ω0

Substitute 5.00s for t, 4.00rad/s for ω0, and 1.00rad/s for ω in the equation (III) to find b.

  1.00rad/s=(4.00rad/s)eb(5.00s)eb(5.00s)=1.00rad/s4.00rad/seb(5.00s)=0.25

Take natural logarithms on both the side of the equation.

  ln(eb(5.00s))=ln(0.25)b(5.00s)=1.386b(5.00s)=1.386b=0.277s1

Therefore, the values of b and ω0 are 0.277s1 and 4.00rad/s.

(b)

To determine

The magnitude of the angular acceleration of the disk at t=5.00s.

(b)

Expert Solution
Check Mark

Answer to Problem 57PQ

The magnitude of the angular acceleration of the disk at t=5.00s is 0.277rad/s2.

Explanation of Solution

Write the expression for angular acceleration.

  α=dωdt                                                                                          (IV)

Here, α is the angular acceleration and dω/dt is the rate of change of angular velocity.

Substitute equation (III) in the equation (IV).

  α=d(ω0ebt)dt=ω0debtdt=ω0(b)ebt                                                                                    (V)

Conclusion:

Substitute 5.00s for t, 4.00rad/s for ω0, and 0.277s1 for b in the equation (V) to find α.

  α=(4.00rad/s)(0.277s1)e(0.277s1)(5.00s)=(1.108rad/s2)e1.385=(1.108rad/s2)(0.250)=0.277rad/s2

Write the magnitude of the angular acceleration.

  |α|=0.277rad/s2

Therefore, the magnitude of the angular acceleration of the disk at t=5.00s is 0.277rad/s2.

(c)

To determine

The revolution of the disk during the time interval t=0 to t=5.00s.

(c)

Expert Solution
Check Mark

Answer to Problem 57PQ

The revolution of the disk during the time interval t=0 to t=5.00s is 1.72rev.

Explanation of Solution

Rewrite the expression (II) for angular displacement.

  Δθ=ωdt                                                                                    (VI)

Here, Δθ is the change in angular displacement.

Integrating the equation (VI) using the limits.

  Δθ=0tωdt

Substitute equation (III) in the above equation.

  Δθ=0tω0ebtdt=ω0b[ebte0]=ω0b[ebt1]=ω0b[1ebt]                                                                        (VII)

Conclusion:

Substitute 5.00s for t, 4.00rad/s for ω0, and 0.277s1 for b in the equation (VII) to find Δθ.

  Δθ=4.00rad/s0.277s1[1e(0.277s1)(5.00s)]=14.4rad[1e1.385]=14.4rad(0.75)=10.8rad

Convert the radians to revolutions.

  Δθ=10.8rad(1rev2πrad)=1.72rev

Therefore, the revolution of the disk during the time interval t=0 to t=5.00s is 1.72rev.

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Chapter 12 Solutions

Webassign Printed Access Card For Katz's Physics For Scientists And Engineers: Foundations And Connections, 1st Edition, Single-term

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