A cam of mass M is in the shape of a circular disk of diameter 2 R 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?

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Answer 1

Textbook Question

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

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.

$Icam=Idisk−Ismall disk$ (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).

$Icam=34MdiskR2−132MdiskR2=MdiskR2(34−132)=MdiskR2(2432−132)=2332MdiskR2$ (VII)

Write the equation for the mass of the cam.

$Mcam=Mdisk−Msmall disk$

Here, $Mcam$ is the mass of the cam.

Put equation (V) in the above equation.

$Mcam=Mdisk−14Mdisk=34Mdisk$ (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

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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Answer 2

Textbook Question

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

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.

$Icam=Idisk−Ismall disk$ (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).

$Icam=34MdiskR2−132MdiskR2=MdiskR2(34−132)=MdiskR2(2432−132)=2332MdiskR2$ (VII)

Write the equation for the mass of the cam.

$Mcam=Mdisk−Msmall disk$

Here, $Mcam$ is the mass of the cam.

Put equation (V) in the above equation.

$Mcam=Mdisk−14Mdisk=34Mdisk$ (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

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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Answer 3

Textbook Question

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

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.

$Icam=Idisk−Ismall disk$ (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).

$Icam=34MdiskR2−132MdiskR2=MdiskR2(34−132)=MdiskR2(2432−132)=2332MdiskR2$ (VII)

Write the equation for the mass of the cam.

$Mcam=Mdisk−Msmall disk$

Here, $Mcam$ is the mass of the cam.

Put equation (V) in the above equation.

$Mcam=Mdisk−14Mdisk=34Mdisk$ (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

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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Answer 4

Textbook Question

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

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.

$Icam=Idisk−Ismall disk$ (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).

$Icam=34MdiskR2−132MdiskR2=MdiskR2(34−132)=MdiskR2(2432−132)=2332MdiskR2$ (VII)

Write the equation for the mass of the cam.

$Mcam=Mdisk−Msmall disk$

Here, $Mcam$ is the mass of the cam.

Put equation (V) in the above equation.

$Mcam=Mdisk−14Mdisk=34Mdisk$ (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

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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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

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.

$Icam=Idisk−Ismall disk$ (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).

$Icam=34MdiskR2−132MdiskR2=MdiskR2(34−132)=MdiskR2(2432−132)=2332MdiskR2$ (VII)

Write the equation for the mass of the cam.

$Mcam=Mdisk−Msmall disk$

Here, $Mcam$ is the mass of the cam.

Put equation (V) in the above equation.

$Mcam=Mdisk−14Mdisk=34Mdisk$ (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

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 $2348McamR2ω2+116MshaftR2ω2$ .

Answer 8

(a)

Expert Solution

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.

$Icam=Idisk−Ismall disk$ (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).

$Icam=34MdiskR2−132MdiskR2=MdiskR2(34−132)=MdiskR2(2432−132)=2332MdiskR2$ (VII)

Write the equation for the mass of the cam.

$Mcam=Mdisk−Msmall disk$

Here, $Mcam$ is the mass of the cam.

Put equation (V) in the above equation.

$Mcam=Mdisk−14Mdisk=34Mdisk$ (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 $2324McamR2+12Mshaft(R2)2$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

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

Answer 13

Answer to Problem 78PQ

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

Answer 14

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.

$Icam=Idisk−Ismall disk$ (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).

$Icam=34MdiskR2−132MdiskR2=MdiskR2(34−132)=MdiskR2(2432−132)=2332MdiskR2$ (VII)

Write the equation for the mass of the cam.

$Mcam=Mdisk−Msmall disk$

Here, $Mcam$ is the mass of the cam.

Put equation (V) in the above equation.

$Mcam=Mdisk−14Mdisk=34Mdisk$ (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 $2324McamR2+12Mshaft(R2)2$ .

Answer 15

(b)

Expert Solution

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 $2348McamR2ω2+116MshaftR2ω2$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

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

Answer 20

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 $2348McamR2ω2+116MshaftR2ω2$ .

Answer 21

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 $2348McamR2ω2+116MshaftR2ω2$ .