EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
9th Edition
ISBN: 8220100454899
Author: Jewett
Publisher: Cengage Learning US
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Chapter 10, Problem 10.88CP

As a gasoline engine operates, a flywheel turning with the crankshaft stores energy after each fuel explosion, providing the energy required to compress the next charge of fuel and air. For the engine of a certain lawn tractor, suppose a flywheel must be no more than 18.0 cm in diameter. Its thickness, measured along its axis of rotation, must be no larger than 8.00 cm. The flywheel must release energy 60.0 J when its angular speed drops from 800 rev/min to 600 rev/min. Design a sturdy steel (density 7.85 × 103 kg/m3) flywheel to meet these requirements with the smallest mass you can reasonably attain. Specify the shape and mass of the flywheel.

Expert Solution & Answer
Check Mark
To determine

The design of sturdy steel flywheel with the small mass and specify it shape.

Answer to Problem 10.88CP

The shape of the flywheel is hollow cylinder of inner radius 0.08m and outer radius 0.09m with the mass 7.34kg.

Explanation of Solution

The diameter of the flywheel must be no more than 18.0cm and the length is not larger than 8.00cm. The energy release by the flywheel is 60.0J and drops in angular speed is 800rev/min to 600rev/min. The density of the steel is 7.85×103kg/m3.

From the law of energy conservation,

    KrfKri=Wout

Here, Kri is the initial rotational energy of the flywheel, Kri is the initial rotational energy of the flywheel and Wout is the output work produce by the flywheel.

Write the expression for the initial rotational energy of the flywheel is,

    Kri=12Iωi2

Here, I is the moment of inertia of the flywheel and ωi is the initial angular speed of the flywheel.

Write the expression for the final rotational energy of the flywheel is,

    Krf=12Iωf2

Here, ωi is the final angular speed of the flywheel.

Substitute 12Iωi2 for Kri and 12Iωf2 for Krf in equation (1) and rearrange the equation for I.

    12Iωf212Iωi2=W12I(ωf2ωf2)=WI=2W(ωf2ωf2)

Substitute 60.0J for W, 600rev/min for ωi and 800rev/min for ωf in above equation to find I.

    I=2×60.0J((800rev/min×(1min60s)(2πrad1rev))2(600rev/min×(1min60s)(2πrad1rev))2)=120J×1kgm2/s21J3070.54s2=0.03908kgm2

Formula to calculate the outer radius of the hollow cylinder is,

  Router=D2

Here, D is the diameter of the cylinder.

Substitute 18.0cm for D in above equation to find Router.

  Router=18.0cm2=9cm×102m1cm=0.09m

For large energy storage by the moment of inertia of flywheel must be large but the mass should be small as much as possible for design to fulfill this requirement the mass should place as far away from the axis as possible to increase the moment of inertia.

Let choose a hollow cylinder to make the flywheel of 18.0cm in diameter and 8.00cm long. To support this rim, place a disk across its center. Assume the disk is 2.00cm thick will be enough to support the hollow cylinder securely.

Formula to calculate the moment of inertia of the flywheel is,

    I=Idisk+Ihollowcylinder        (1)

Here, I is the total moment of inertia of the flywheel, Idisk is the moment of inertia of the sturdy disk and Ihollowcylinder is the moment of inertia of the hollow cylinder.

Write the expression for the moment of inertia of the disk is,

    Idisk=12MdiskRdisk2

Here, Mdisk is the mass of the disk and Rdisk is the radius of the disk.

Assume the radius of the disk is equal to the outer radius of the hollow cylinder.

Write the expression for the moment of inertia of the hollow cylinder is,

    Ihollowcylinder=12Mwall(Router2+Rinner2)

Here, Mwall is the mass of the hollow cylinder wall, Router is the outer radius of the hollow cylinder and Rinner is the inner radius of the hollow cylinder.

Substitute 12MdiskRdisk2 for Idisk and 12Mwall(Router2+Rinner2) for Ihollowcylinder in equation (1).

    I=12MdiskRdisk2+12Mwall(Router2+Rinner2)        (2)

Write the expression for the mass of the hollow cylinder wall is,

    Mwall=ρπ(Router2Rinner2)L

Here, ρ is the density of the disk and L is the length of the cylinder wall.

Write the expression for the mass of the disk is,

    Mdisk=ρ(πRdisk2tdisk)

Here, tdisk is the thickness of the disk.

Substitute ρπ(Router2Rinner2)L for Mwall and ρ(πRdisktdisk) for Mdisk in equation (2).

    I=12(ρ(πRdisk2tdisk))Rdisk2+12(ρπ(Router2Rinner2)L)(Router2+Rinner2)=ρπ2[Rdisk4tdisk+(Router4Rinner4)L]

Substitute 0.03908kgm2 for I, 2.00cm for tdisk, 0.09m for Router and Rdisk , 8.00cm for L and 7.85×103kg/m3 for ρ to find Rinner.

    (7.85×103kg/m3)π2[(0.09m)4(2.00cm×102m1cm)+((0.09m)4Rinner4)(8.00cm×102m1cm)]=0.03908kgm2[1.31×106m5(6.5×105m4Rinner4)(0.08m)]=0.03908kgm212330.75kg/m3(6.5×105m4Rinner4)(0.08m)=(1.31×106m53.17×106m5)6.5×105m4Rinner4=1.86×106m50.08m

Solve the equation further,

    6.5×105m4Rinner4=2.325×105m4Rinner4=6.5×105m42.325×105m4Rinner4=4.175×105m4Rinner=0.080m

Formula to calculate the mass of the flywheel is,

    M=Mdisk+Mwall

Substitute ρ(πRdisk2tdisk) for Mdisk and ρπ(Router2Rinner2)L for Mwall to find M.

    M=ρ(πRdisk2tdisk)+ρπ(Router2Rinner2)L=ρπ[Rdisk2tdisk+(Router2Rinner2)L]

Conclusion:

Substitute 7.85×103kg/m3 for ρ, 2.00cm for tdisk, 0.09m for Router and Rdisk , 8.00cm for L and 0.080m for Rinner to find M.

    M=(7.85×103kg/m3)π[(0.09m)2(2.00cm×102m1cm)+((0.09m)2(0.08m)2)(8.00cm×102m1cm)]=2.466×104kg/m3×2.98×104m3=7.34kg

Therefore, shape of the flywheel is hollow cylinder of inner radius 0.08m and outer radius 0.09m with the mass 7.34kg.

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Chapter 10 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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