EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
1st Edition
ISBN: 9780100546714
Author: Katz
Publisher: YUZU
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Chapter 13, Problem 78PQ

A cam of mass M is in the shape of a circular disk of diameter 2R with an off-center circular hole of diameter R is mounted on a uniform cylindrical shaft whose diameter matches that of the hole (Fig. P1 3.78). a. What is the rotational inertia of the cam and shaft around the axis of the shaft? b. What is the rotational kinetic energy of the cam and shaft if the system rotates with angular speed ω around this axis?

Chapter 13, Problem 78PQ, A cam of mass M is in the shape of a circular disk of diameter 2R with an off-center circular hole

(a)

Expert Solution
Check Mark
To determine

The rotational inertia of the cam and shaft around the axis of the shaft.

Answer to Problem 78PQ

The rotational inertia of the cam and shaft around the axis of the shaft is 2324McamR2+12Mshaft(R2)2 .

Explanation of Solution

Write the equation for the rotational inertia of the cam and shaft around the axis of the shaft.

  Icam-shaft=Icam+Ishaft                                                                                                   (I)

Here, Icam-shaft is the rotational inertia of the cam and shaft around the axis of the shaft, Icam is the rotational inertia of the cam and Ishaft is the rotational inertia of the shaft.

Rotational inertia of the cam is the difference of the rotational inertia of the solid disk about an axis R/2 from the center of mass and the rotational inertia of the small disk of radius R/2 with a rotation axis through its center.

Write the equation for the rotational inertia of the cam.

  Icam=IdiskIsmall disk                                                                                                 (II)

Here, Idisk is the rotational inertia of the solid disk about an axis R/2 from the center of mass and Ismall disk is the rotational inertia of the small disk of radius R/2 with a rotation axis through its center.

The rotational inertia of the solid disk bout an axis R/2 from its center of mass can be found using the parallel axis theorem.

Write the equation for Idisk .

  Idisk=ICM+Mdisk(R2)2                                                                                        (III)

Here, ICM is the rotational inertia of the disk about its center of mass and R is the radius of the cam and Mdisk is the mass of the disk.

Write the equation for the rotational inertia of the disk about its center of mass.

  ICM=12MdiskR2

Put the above equation in equation (III).

  Idisk=12MdiskR2+14MdiskR2=34MdiskR2                                                                                 (IV)

With half the radius, the cut away small disk has one-quarter the face area, one-quarter the volume and one-quarter the mass of the original solid disk.

Write the expression for the ratio of the mass of the small disk to the mass of the original solid disk.

  Msmall diskMdisk=(R/2)2R2=R24R2=14

Here, Msmall disk is the mass of the small disk.

Rewrite the above equation for Msmall disk .

  Msmall disk=14Mdisk                                                                                               (V)

Write the equation for the rotational inertia of the small disk about an axis through its center of mass.

  Ismall disk=12Msmall disk(R2)2=18Msmall diskR2

Here, Ismall disk is the rotational inertia of the small disk about an axis through its center of mass and (R/2) is the radius of the small disk.

Put equation (V) in the above equation.

  Ismall disk==18(14Mdisk)R2=132MdiskR2                                                                                      (VI)

Put equations (IV) and (VI) in equation (II).

  Icam=34MdiskR2132MdiskR2=MdiskR2(34132)=MdiskR2(2432132)=2332MdiskR2                                                                                (VII)

Write the equation for the mass of the cam.

  Mcam=MdiskMsmall disk

Here, Mcam is the mass of the cam.

Put equation (V) in the above equation.

  Mcam=Mdisk14Mdisk=34Mdisk                                                                                         (VIII)

Multiply and divide the right hand side of equation (VII) with Mcam .

  Icam=2332MdiskR2McamMcam

Put equation (VIII) in the denominator of the above equation.

  Icam=2332MdiskR2Mcam34Mdisk=McamR2(2332)(43)=2324McamR2                                                                                       (IX)

Write the equation for the rotational inertia of the shaft.

  Ishaft=12Mshaft(R2)2                                                                                                (X)

Here, Mshaft is the mass of the shaft.

Conclusion:

Put equations (IX) and (X) in equation in equation (I).

  Icam-shaft=2324McamR2+12Mshaft(R2)2

Therefore, the rotational inertia of the cam and shaft around the axis of the shaft is 2324McamR2+12Mshaft(R2)2 .

(b)

Expert Solution
Check Mark
To determine

The rotational kinetic energy of the cam and shaft if the system rotates with angular speed ω around the axis.

Answer to Problem 78PQ

The rotational kinetic energy of the cam and shaft if the system rotates with angular speed ω around the axis is 2348McamR2ω2+116MshaftR2ω2 .

Explanation of Solution

Write the equation for the rotational kinetic energy of the cam and the shaft.

  Ktotal=Kcam+Kshaft                                                                                               (XI)

Here, Ktotal is the rotational kinetic energy of the system of cam and the shaft, Kcam is the rotational kinetic energy of the cam and Kshaft is the rotational kinetic energy of the shaft.

Write the equation for Kcam .

  Kcam=12Icamω2

Put equation (IX) in the above equation.

  Kcam=12(2324McamR2)ω2=2348McamR2ω2                                                                                     (XII)

Write the equation for Kshaft .

  Kshaft=12Ishaftω2

Put equation (X) in the above equation.

  Kshaft=12(12Mshaft(R2)2)ω2=116MshaftR2ω2                                                                                (XIII)

Conclusion:

Put equations (XII) and (XIII) in equation (XI).

  Ktotal=2348McamR2ω2+116MshaftR2ω2

Therefore, the rotational kinetic energy of the cam and shaft if the system rotates with angular speed ω around the axis is 2348McamR2ω2+116MshaftR2ω2 .

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Chapter 13 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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