Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 9, Problem 38P

(a)

To determine

To Find: The average angular acceleration during 4.5-s spin-up and again during 0.24-s spin- down.

(a)

Expert Solution
Check Mark

Answer to Problem 38P

  αavg1=30rad/s2

  αavg2=1.2×102rad/s2

Explanation of Solution

Given:

Initial angular speed of the fly wheel, ωi=0

Final angular speed of the fly wheel

  ωf=4.5rev/s=(4.5rev/s)×( 2πrad 1rev)=9πrad/s

Time taken to pull the rope, Δt=0.95s

Time taken during spin up, Δtup=4.5s

Time taken during spin down, Δtdown=0.24s

Formula used:

  αavg=ΔωΔtαavg=ωfωiΔt

  αavg = average acceleration

  ωf = final angular velocity

  ωi = initial angular velocity

Calculation:

Average angular acceleration for spin up is:

  αavg1=( 9π0)rad/s0.95sαavg1=29.76rad/s230rad/s2

Average angular acceleration for spin down is:

  αavg2=( 09π)rad/s0.24sαavg2=117.8rad/s21.2×102rad/s2

Conclusion:

Spin down acceleration =1.2×102rad/s2

Spin up acceleration =30rad/s2

(b)

To determine

To find: Maximum angular speed reached by fly wheel.

(b)

Expert Solution
Check Mark

Answer to Problem 38P

  ωmax=28rad/s

Explanation of Solution

Given:

Initial angular speed of the fly wheel, ωi=0

Final angular speed of the fly wheel

  ωf=4.5rev/s=(4.5rev/s)×( 2πrad 1rev)=9πrad/s

Formula used:

  ωmax=ωf

Where,

  ωmax = maximum angular velocity

  ωf = final angular velocity

Calculation:

Substitute the given value in the equation,

  ωmax=ωf=9πrad/s28rad/s

Conclusion:

  ωmax=28rad/s is maximum angular speed reached by fly wheel.

(c)

To determine

To find: The ratio of the number of revolutions made during spin-up to the number made during spin down.

(c)

Expert Solution
Check Mark

Answer to Problem 38P

  ΔθupΔθdown4

Explanation of Solution

Given:

Time taken to pull the rope, Δt=0.95s

Time taken during spin up, Δtup=4.5s

Time taken during spin down, Δtdown=0.24s

Angular speed at spin up =0

Angular speed at spin down, ωmax=28rad/s

Angular acceleration for spin up, αup=30rad/s2

Angular acceleration for spin up, αdown=1.2×102rad/s2

Formula used:

Number of revolutions can be obtained by:

  Δθ=ωt+12αt2

Where,

  θ is the number of revolutions.

  Δt is the time taken.

  ω is the angular speed

  α is angular acceleration

Calculation:

For spin-up revolution:

  Δθup=ωtup+12αuptup2Δθup=0+12×30×0.952Δθup=13.5373.....(1)

For spin down revolution:

  Δθdown=ωtdown+12αdowntdown2Δθdown=(28×0.24)+12×(1.2× 102)(0.24)2Δθdown=6.723.456Δθdown=3.264...(2)

Divide equation (1) by equation (2)

  Δθ upΔθ down=13.53733.264Δθ upΔθ down=4.147Δθ upΔθ down4

Conclusion:

Ratio of the number of revolutions made during spin-up to the number made during spin down is 4 .

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Chapter 9 Solutions

Physics for Scientists and Engineers

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