Vector Mechanics for Engineers: Dynamics
Vector Mechanics for Engineers: Dynamics
11th Edition
ISBN: 9780077687342
Author: Ferdinand P. Beer, E. Russell Johnston Jr., Phillip J. Cornwell, Brian Self
Publisher: McGraw-Hill Education
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Chapter 16.1, Problem 16.35P

Each of the gears A and B has a mass of 9 kg and has a radius of gyration of 200 mm; gear C has a mass of 3 kg and has a radius of gyration of 75 mm. if a couple M of constant magnitude 5 N-m is applied to gear C, determine (a) the angular acceleration of gear A, (b) the tangential force that gear C exerts on gear A.

Expert Solution
Check Mark
To determine

(a)

The angular acceleration of the gear A.

Answer to Problem 16.35P

The angular acceleration of the gear A is 15.143rad/s2.

Explanation of Solution

Given Information:

The mass of the gear A is 9kg, the radius of gyration of the gear A is 200mm, the mass of the gear B is 9kg, the radius of gyration of the gear B is 200mm, the mass of the gear C is 3kg, the radius of gyration of the gear C is 75mm, the magnitude of the couple M is 5Nm, the radius of gear A is 250mm, the radius of gear B is 250mm and the radius of gear C is 100mm.

Write the expression for the tangential acceleration of the gear teeth.

αt=αr ...... (I)

Here, the angular acceleration is α and the radius of the gear is r.

Write the expression for the tangential acceleration of the gear A.

αtA=rAαA ...... (II)

Here, the angular acceleration of gear A is αA and the radius of the gear A is rA.

Write the expression for the tangential acceleration of the gear B.

αtB=rBαB ...... (III)

Here, the angular acceleration of gear B is αB and the radius of the gear B is rB.

Write the expression for the tangential acceleration of the gear C.

αtC=rCαC ...... (IV)

Here, the angular acceleration of gear C is αC and the radius of the gear C is rC.

Draw the free body diagram for the gear A.

Vector Mechanics for Engineers: Dynamics, Chapter 16.1, Problem 16.35P , additional homework tip  1

Figure-(1)

Draw the kinetic diagram of the gear A.

Vector Mechanics for Engineers: Dynamics, Chapter 16.1, Problem 16.35P , additional homework tip  2

Figure-(2)

Write the expression for the moment of inertia gear A.

IA=mAkA2 ...... (V)

Here, the mass of the gear is mA and the radius of gyration of the gear A is kA.

Write the expression for the external moment at A using the Figure-(1).

MA=rAF ...... (VI)

Here, the tangential force is F.

Write the expression for the effective forces using the Figure-(2).

(MA)eff=IAαA ...... (VII)

Since, the system of external forces is equivalent to system of effective forces hence

(MA)eff=(MA).

Substitute IAαA for (MA) in Equation (VI).

IAαA=rAFF=IAαArA ...... (VIII)

Draw the free body diagram for the gear C.

Vector Mechanics for Engineers: Dynamics, Chapter 16.1, Problem 16.35P , additional homework tip  3

Figure-(3)

Draw the kinetic diagram for the gear C.

Vector Mechanics for Engineers: Dynamics, Chapter 16.1, Problem 16.35P , additional homework tip  4

Figure-(4)

Write the expression for the moment of inertia gear C.

IC=mCkC2 ...... (IX)

Here, the mass of the gear is C

mC and the radius of gyration of the gear C is kC.

Write the expression for the external moment at C using the Figure-(3).

MC=M2rCF ...... (X)

Here, the tangential force is F.

Write the expression for the effective forces using the Figure-(4).

(MC)eff=ICαC ...... (XI)

Since, the system of external forces is equivalent to system of effective forces hence

(MC)eff=(MC).

Substitute ICαC for (MC) in Equation (VI).

ICαC=M2rCFF=MICαC2rC ...... (XII)

Calculation:

Substitute 250mm for rA in Equation (II).

αtA=(250mm)αA=(250mm(1m103mm))αA=(0.25m)αA ...... (XIII)

Substitute 250mm for rB in Equation (III).

αtB=(250mm)αB=(250mm(1m103mm))αB=(0.25m)αB ...... (XIV)

Substitute 100mm for rB in Equation (IV).

αtC=(100mm)αC=(100mm(1m103mm))αC=(0.1m)αC ...... (XV)

The tangential acceleration of the gear A and B is same that is αA=αB.

Substitute (0.25m)αB for αtA in Equation (XIII).

(0.25m)αA=(0.25m)αBαA=αB

The tangential acceleration of the gear A and C is same that is αA=αC.

Substitute (0.1m)αC for αtA in Equation (XIII).

(0.25m)αA=(0.1m)αCαC=2.5αA

Substitute 9kg for mA and 200mm for kA in Equation (V).

IA=(9kg)(200mm)2=(9kg)(200mm(1m103mm))2=(9kg)(0.2m)2=0.36kgm2

Substitute 0.36kgm2 for IA, 250mm for rA in Equation (VIII).

F=(0.36kgm2)αA(250mm)=(0.36kgm2)αA(250mm(1m103mm))=(1.44kgm)αA

Substitute 3kg for mC and 75mm for kC in Equation (IX).

IA=(3kg)(75mm)2=(3kg)(75mm(1m103mm))2=(3kg)(0.075m)2=0.01687kgm2

Substitute (2.5)αA for αC, 5Nm for M, 0.01687kgm2 for IC, 100mm for rC in Equation (XII).

F=(5Nm)(0.01687kgm2)((2.5)αA)2(100mm)=(5Nm)(0.04217kgm2)(αA)2(100mm(1m103mm))=(5Nm)(0.04217kgm2)(αA)0.2m ...... (XVI)

Substitute (1.44kgm)αA for F in Equation (XVI).

(1.44kgm)αA=(5Nm)(0.04217kgm2)(αA)0.2m(0.288kgm2)αA=(5Nm)(0.04217kgm2)(αA)αA=(5N)(1kgm/s2N)0.3301kgmαA=15.143rad/s2

Conclusion:

The angular acceleration of the gear A is 15.143rad/s2.

Expert Solution
Check Mark
To determine

(b)

The tangential force exerted by gear C on gear A.

Answer to Problem 16.35P

The tangential force exerted by gear C on gear A is 21.80N.

Explanation of Solution

Write the expression for the tangential force exerted by gear C on gear A.

F=IAαArA ...... (XVII).

Calculation:

Substitute 15.143rad/s2 for αA, 0.36kgm2 for IA, 250mm for rA in Equation (XVII).

F=(0.36kgm2)(15.143rad/s2)(250mm)=(0.36kgm2)(15.143rad/s2)(250mm(1m103mm))=(1.44kgm)(15.143rad/s2)(1N1kgm/s2)=21.80N

Conclusion:

The tangential force exerted by gear C on gear A is 21.80N.

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Chapter 16 Solutions

Vector Mechanics for Engineers: Dynamics

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