The number of revolutions of gear C required for its angular velocity to increase from 100 to 450 rpm.

Question

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Answer 1

Question

Chapter 17.1, Problem 17.11P

(a)

To determine

The number of revolutions of gear C required for its angular velocity to increase from 100 to 450 rpm.

(a)

Expert Solution

Answer to Problem 17.11P

The number of revolutions of gear C required for its angular velocity to increase from 100 to 450 rpm is $6.35 rev_$ .

Explanation of Solution

Given information:

The mass $(mA)$ of gear A is 2.4 kg.

The mass $(mB)$ of gear B is 2.4 kg.

The mass $(mC)$ of gear C is 12 kg.

The radius of gyration $(kA)$ of gear A is 60 mm.

The radius of gyration $(kB)$ of gear B is 60 mm.

The radius of gyration $(kC)$ of gear C is 150 mm.

The radius $(rA)$ of gear A is 80 mm.

The radius $(rB)$ of gear B is 80 mm.

The radius $(rC)$ of gear C is 200 mm.

The initial angular velocity $(ωC)1$ of gear C is 100 rpm.

The final angular velocity $(ωC)2$ of gear C is 450 rpm.

The couple acting on gear C is $10 N⋅m$ .

Calculation:

Find the mass moment of inertia $(IA)$ of gear A using the equation:

$IA=mAkA2$

Substitute 2.4 kg for $mA$ and 60 mm for $kA$ .

$IA=(2.4)(60 mm×1 m1,000 mm)2=8.64×10−3 kg⋅m2$

Find the mass moment of inertia $(IB)$ of gear B using the equation:

$IB=mBkB2$

Substitute 2.4 kg for $mB$ and 60 mm for $kB$ .

$IB=(2.4)(60 mm×1 m1,000 mm)2=8.64×10−3 kg⋅m2$

Find the mass moment of inertia $(IC)$ of gear C using the equation:

$IC=mCkC2$

Substitute 12 kg for $mC$ and 150 mm for $kA$ .

$IC=(12)(150 mm×1 m1,000 mm)2=270×10−3 kg⋅m2$

Find the initial angular velocity $(ωA)1$ of gear A using the kinematics:

$rA(ωA)1=rC(ωC)1(ωA)1=rCrA(ωC)1$

Substitute 200 mm for $rC$ , 80 mm for $rA$ , and 100 rpm for $(ωC)1$ .

$(ωA)1=20080(100 revmin×2π rad1 rev×1 min60 s)=26.18 rad/s$

Find the initial angular velocity $(ωB)1$ of gear B using the kinematics:

$rB(ωB)1=rC(ωC)1(ωB)1=rCrB(ωC)1$

Substitute 200 mm for $rC$ , 80 mm for $rB$ , and 100 rpm for $(ωC)1$ .

$(ωB)1=20080(100 revmin×2π rad1 rev×1 min60 s)=26.18 rad/s$

Find the initial kinetic energy $(T1)$ of the system using the equation:

$T1=(T1)A+(T1)B+(T1)C=12IA(ωA)12+12IB(ωB)12+12IC(ωC)12$

Here, $(T1)A$ is the initial angular velocity of gear A, $(T1)B$ is the initial angular velocity of gear B, and $(T1)C$ is the initial angular velocity of gear C.

Substitute $8.64×10−3 kg⋅m2$ for $IA$ , $26.18 rad/s$ for $(ωA)1$ , $8.64×10−3 kg⋅m2$ for $IB$ , $26.18 rad/s$ for $(ωB)1$ , $270×10−3 kg⋅m2$ for $IC$ , and 100 rpm for $ωC$ .

$T1={12[(8.64×10−3)(26.18)2+12(8.64×10−3)(26.18)2+12(270×10−3)](100 revmin×2π rad1 rev×1 min60 s)2}=2.9609+2.9609+14.8044=20.726 J$

Find the final angular velocity $(ωA)2$ of gear A using the kinematics:

$rA(ωA)2=rC(ωC)2(ωA)2=rCrA(ωC)2$

Substitute 200 mm for $rC$ , 80 mm for $rA$ , and 450 rpm for $(ωC)2$ .

$(ωA)2=20080(450 revmin×2π rad1 rev×1 min60 s)=117.81 rad/s$

Find the final angular velocity $(ωB)2$ of gear B using the kinematics:

$rB(ωB)2=rC(ωC)2(ωB)2=rCrB(ωC)2$

Substitute 200 mm for $rC$ , 80 mm for $rB$ , and 450 rpm for $(ωC)2$ .

$(ωB)2=20080(450 revmin×2π rad1 rev×1 min60 s)=117.81 rad/s$

Find the initial kinetic energy $(T2)$ of the system using the equation:

$T2=(T2)A+(T2)B+(T2)C=12IA(ωA)22+12IB(ωB)22+12IC(ωC)22$

Here, $(T2)A$ is the final angular velocity of gear A, $(T2)B$ is the final angular velocity of gear B, and $(T2)C$ is the final angular velocity of gear C.

Substitute $8.64×10−3 kg⋅m2$ for $IA$ , $117.81 rad/s$ for $(ωA)1$ , $8.64×10−3 kg⋅m2$ for $IB$ , $117.81 rad/s$ for $(ωB)1$ , $270×10−3 kg⋅m2$ for $IC$ , and 450 rpm for $ωC$ .

$T2={[12(8.64×10−3)(117.81)2+12(8.64×10−3)(117.81)2+12(270×10−3)](450 revmin×2π rad1 rev×1 min60 s)2}=59.958+59.958+299.8=419.7 J$

Find the work done $(U1→2)$ due to couple acting on gear C using the equation:

$U1→2=MCθC$

Substitute $10 N⋅m$ for $MC$ .

$U1→2=10θC$

Find the number of revolutions of gear C $(θC)$ required for its angular velocity to increase from 100 to 450 rpm.

Apply Principle of work and energy for system.

$T1+U1→2=T2$

Substitute $20.726 J$ for $T1$ , 419.7 J for $T2$ , and $10θC$ for $U1→2$ .

$20.726+10θC=419.7θC=39.897 rad×1 rev2π radθC=6.35 rev$

Thus, the number of revolutions of gear C required for its angular velocity to increase from 100 to 450 rpm is $6.35 rev_$ .

(b)

To determine

The tangential force acting on gear A.

(b)

Expert Solution

Answer to Problem 17.11P

The tangential force acting on gear A is $7.14 N_$ .

Explanation of Solution

Calculation:

Find the rotation of gear A using the equation:

$rAθA=rCθCθA=rCrAθC$

Substitute 200 mm for $rC$ , $80 mm$ for $rA$ , and $39.897 rad$ for $θC$ .

$θA=20080(39.897)=99.743 rad$

Find the tangential force $(Ft)$ acting on gear A.

Apply Principle of work and energy for gear A.

$(T1)A+MAθA=(T2)A12IA(ωA)12+MAθA=12IA(ωA)22$

Substitute $8.64×10−3 kg⋅m2$ for $IA$ , $26.18 rad/s$ for $(ωA)1$ , 99.743 rad/s for $θA$ , and $117.81 rad/s$ for $(ωA)2$ .

$12(8.64×10−3)(26.18)2+MA(99.743)=12(8.64×10−3)(117.81)22.9609+99.743MA=59.958MA=0.57144 N⋅m$

Find the tangential force $(Ft)$ acting on gear A using the equation:

$Ft=MArA$

Substitute $0.57144 N⋅m$ for $MA$ and 80 mm for $rA$ .

$Ft=0.57144(80 mm×1 m1,000 mm)=7.14 N$

Thus, the tangential force acting on gear A is $7.14 N_$ .

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Answer 2

Question

Chapter 17.1, Problem 17.11P

(a)

To determine

The number of revolutions of gear C required for its angular velocity to increase from 100 to 450 rpm.

(a)

Expert Solution

Answer to Problem 17.11P

The number of revolutions of gear C required for its angular velocity to increase from 100 to 450 rpm is $6.35 rev_$ .

Explanation of Solution

Given information:

The mass $(mA)$ of gear A is 2.4 kg.

The mass $(mB)$ of gear B is 2.4 kg.

The mass $(mC)$ of gear C is 12 kg.

The radius of gyration $(kA)$ of gear A is 60 mm.

The radius of gyration $(kB)$ of gear B is 60 mm.

The radius of gyration $(kC)$ of gear C is 150 mm.

The radius $(rA)$ of gear A is 80 mm.

The radius $(rB)$ of gear B is 80 mm.

The radius $(rC)$ of gear C is 200 mm.

The initial angular velocity $(ωC)1$ of gear C is 100 rpm.

The final angular velocity $(ωC)2$ of gear C is 450 rpm.

The couple acting on gear C is $10 N⋅m$ .

Calculation:

Find the mass moment of inertia $(IA)$ of gear A using the equation:

$IA=mAkA2$

Substitute 2.4 kg for $mA$ and 60 mm for $kA$ .

$IA=(2.4)(60 mm×1 m1,000 mm)2=8.64×10−3 kg⋅m2$

Find the mass moment of inertia $(IB)$ of gear B using the equation:

$IB=mBkB2$

Substitute 2.4 kg for $mB$ and 60 mm for $kB$ .

$IB=(2.4)(60 mm×1 m1,000 mm)2=8.64×10−3 kg⋅m2$

Find the mass moment of inertia $(IC)$ of gear C using the equation:

$IC=mCkC2$

Substitute 12 kg for $mC$ and 150 mm for $kA$ .

$IC=(12)(150 mm×1 m1,000 mm)2=270×10−3 kg⋅m2$

Find the initial angular velocity $(ωA)1$ of gear A using the kinematics:

$rA(ωA)1=rC(ωC)1(ωA)1=rCrA(ωC)1$

Substitute 200 mm for $rC$ , 80 mm for $rA$ , and 100 rpm for $(ωC)1$ .

$(ωA)1=20080(100 revmin×2π rad1 rev×1 min60 s)=26.18 rad/s$

Find the initial angular velocity $(ωB)1$ of gear B using the kinematics:

$rB(ωB)1=rC(ωC)1(ωB)1=rCrB(ωC)1$

Substitute 200 mm for $rC$ , 80 mm for $rB$ , and 100 rpm for $(ωC)1$ .

$(ωB)1=20080(100 revmin×2π rad1 rev×1 min60 s)=26.18 rad/s$

Find the initial kinetic energy $(T1)$ of the system using the equation:

$T1=(T1)A+(T1)B+(T1)C=12IA(ωA)12+12IB(ωB)12+12IC(ωC)12$

Here, $(T1)A$ is the initial angular velocity of gear A, $(T1)B$ is the initial angular velocity of gear B, and $(T1)C$ is the initial angular velocity of gear C.

Substitute $8.64×10−3 kg⋅m2$ for $IA$ , $26.18 rad/s$ for $(ωA)1$ , $8.64×10−3 kg⋅m2$ for $IB$ , $26.18 rad/s$ for $(ωB)1$ , $270×10−3 kg⋅m2$ for $IC$ , and 100 rpm for $ωC$ .

$T1={12[(8.64×10−3)(26.18)2+12(8.64×10−3)(26.18)2+12(270×10−3)](100 revmin×2π rad1 rev×1 min60 s)2}=2.9609+2.9609+14.8044=20.726 J$

Find the final angular velocity $(ωA)2$ of gear A using the kinematics:

$rA(ωA)2=rC(ωC)2(ωA)2=rCrA(ωC)2$

Substitute 200 mm for $rC$ , 80 mm for $rA$ , and 450 rpm for $(ωC)2$ .

$(ωA)2=20080(450 revmin×2π rad1 rev×1 min60 s)=117.81 rad/s$

Find the final angular velocity $(ωB)2$ of gear B using the kinematics:

$rB(ωB)2=rC(ωC)2(ωB)2=rCrB(ωC)2$

Substitute 200 mm for $rC$ , 80 mm for $rB$ , and 450 rpm for $(ωC)2$ .

$(ωB)2=20080(450 revmin×2π rad1 rev×1 min60 s)=117.81 rad/s$

Find the initial kinetic energy $(T2)$ of the system using the equation:

$T2=(T2)A+(T2)B+(T2)C=12IA(ωA)22+12IB(ωB)22+12IC(ωC)22$

Here, $(T2)A$ is the final angular velocity of gear A, $(T2)B$ is the final angular velocity of gear B, and $(T2)C$ is the final angular velocity of gear C.

Substitute $8.64×10−3 kg⋅m2$ for $IA$ , $117.81 rad/s$ for $(ωA)1$ , $8.64×10−3 kg⋅m2$ for $IB$ , $117.81 rad/s$ for $(ωB)1$ , $270×10−3 kg⋅m2$ for $IC$ , and 450 rpm for $ωC$ .

$T2={[12(8.64×10−3)(117.81)2+12(8.64×10−3)(117.81)2+12(270×10−3)](450 revmin×2π rad1 rev×1 min60 s)2}=59.958+59.958+299.8=419.7 J$

Find the work done $(U1→2)$ due to couple acting on gear C using the equation:

$U1→2=MCθC$

Substitute $10 N⋅m$ for $MC$ .

$U1→2=10θC$

Find the number of revolutions of gear C $(θC)$ required for its angular velocity to increase from 100 to 450 rpm.

Apply Principle of work and energy for system.

$T1+U1→2=T2$

Substitute $20.726 J$ for $T1$ , 419.7 J for $T2$ , and $10θC$ for $U1→2$ .

$20.726+10θC=419.7θC=39.897 rad×1 rev2π radθC=6.35 rev$

Thus, the number of revolutions of gear C required for its angular velocity to increase from 100 to 450 rpm is $6.35 rev_$ .

(b)

To determine

The tangential force acting on gear A.

(b)

Expert Solution

Answer to Problem 17.11P

The tangential force acting on gear A is $7.14 N_$ .

Explanation of Solution

Calculation:

Find the rotation of gear A using the equation:

$rAθA=rCθCθA=rCrAθC$

Substitute 200 mm for $rC$ , $80 mm$ for $rA$ , and $39.897 rad$ for $θC$ .

$θA=20080(39.897)=99.743 rad$

Find the tangential force $(Ft)$ acting on gear A.

Apply Principle of work and energy for gear A.

$(T1)A+MAθA=(T2)A12IA(ωA)12+MAθA=12IA(ωA)22$

Substitute $8.64×10−3 kg⋅m2$ for $IA$ , $26.18 rad/s$ for $(ωA)1$ , 99.743 rad/s for $θA$ , and $117.81 rad/s$ for $(ωA)2$ .

$12(8.64×10−3)(26.18)2+MA(99.743)=12(8.64×10−3)(117.81)22.9609+99.743MA=59.958MA=0.57144 N⋅m$

Find the tangential force $(Ft)$ acting on gear A using the equation:

$Ft=MArA$

Substitute $0.57144 N⋅m$ for $MA$ and 80 mm for $rA$ .

$Ft=0.57144(80 mm×1 m1,000 mm)=7.14 N$

Thus, the tangential force acting on gear A is $7.14 N_$ .

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Answer 3

Question

Chapter 17.1, Problem 17.11P

(a)

To determine

The number of revolutions of gear C required for its angular velocity to increase from 100 to 450 rpm.

(a)

Expert Solution

Answer to Problem 17.11P

The number of revolutions of gear C required for its angular velocity to increase from 100 to 450 rpm is $6.35 rev_$ .

Explanation of Solution

Given information:

The mass $(mA)$ of gear A is 2.4 kg.

The mass $(mB)$ of gear B is 2.4 kg.

The mass $(mC)$ of gear C is 12 kg.

The radius of gyration $(kA)$ of gear A is 60 mm.

The radius of gyration $(kB)$ of gear B is 60 mm.

The radius of gyration $(kC)$ of gear C is 150 mm.

The radius $(rA)$ of gear A is 80 mm.

The radius $(rB)$ of gear B is 80 mm.

The radius $(rC)$ of gear C is 200 mm.

The initial angular velocity $(ωC)1$ of gear C is 100 rpm.

The final angular velocity $(ωC)2$ of gear C is 450 rpm.

The couple acting on gear C is $10 N⋅m$ .

Calculation:

Find the mass moment of inertia $(IA)$ of gear A using the equation:

$IA=mAkA2$

Substitute 2.4 kg for $mA$ and 60 mm for $kA$ .

$IA=(2.4)(60 mm×1 m1,000 mm)2=8.64×10−3 kg⋅m2$

Find the mass moment of inertia $(IB)$ of gear B using the equation:

$IB=mBkB2$

Substitute 2.4 kg for $mB$ and 60 mm for $kB$ .

$IB=(2.4)(60 mm×1 m1,000 mm)2=8.64×10−3 kg⋅m2$

Find the mass moment of inertia $(IC)$ of gear C using the equation:

$IC=mCkC2$

Substitute 12 kg for $mC$ and 150 mm for $kA$ .

$IC=(12)(150 mm×1 m1,000 mm)2=270×10−3 kg⋅m2$

Find the initial angular velocity $(ωA)1$ of gear A using the kinematics:

$rA(ωA)1=rC(ωC)1(ωA)1=rCrA(ωC)1$

Substitute 200 mm for $rC$ , 80 mm for $rA$ , and 100 rpm for $(ωC)1$ .

$(ωA)1=20080(100 revmin×2π rad1 rev×1 min60 s)=26.18 rad/s$

Find the initial angular velocity $(ωB)1$ of gear B using the kinematics:

$rB(ωB)1=rC(ωC)1(ωB)1=rCrB(ωC)1$

Substitute 200 mm for $rC$ , 80 mm for $rB$ , and 100 rpm for $(ωC)1$ .

$(ωB)1=20080(100 revmin×2π rad1 rev×1 min60 s)=26.18 rad/s$

Find the initial kinetic energy $(T1)$ of the system using the equation:

$T1=(T1)A+(T1)B+(T1)C=12IA(ωA)12+12IB(ωB)12+12IC(ωC)12$

Here, $(T1)A$ is the initial angular velocity of gear A, $(T1)B$ is the initial angular velocity of gear B, and $(T1)C$ is the initial angular velocity of gear C.

Substitute $8.64×10−3 kg⋅m2$ for $IA$ , $26.18 rad/s$ for $(ωA)1$ , $8.64×10−3 kg⋅m2$ for $IB$ , $26.18 rad/s$ for $(ωB)1$ , $270×10−3 kg⋅m2$ for $IC$ , and 100 rpm for $ωC$ .

$T1={12[(8.64×10−3)(26.18)2+12(8.64×10−3)(26.18)2+12(270×10−3)](100 revmin×2π rad1 rev×1 min60 s)2}=2.9609+2.9609+14.8044=20.726 J$

Find the final angular velocity $(ωA)2$ of gear A using the kinematics:

$rA(ωA)2=rC(ωC)2(ωA)2=rCrA(ωC)2$

Substitute 200 mm for $rC$ , 80 mm for $rA$ , and 450 rpm for $(ωC)2$ .

$(ωA)2=20080(450 revmin×2π rad1 rev×1 min60 s)=117.81 rad/s$

Find the final angular velocity $(ωB)2$ of gear B using the kinematics:

$rB(ωB)2=rC(ωC)2(ωB)2=rCrB(ωC)2$

Substitute 200 mm for $rC$ , 80 mm for $rB$ , and 450 rpm for $(ωC)2$ .

$(ωB)2=20080(450 revmin×2π rad1 rev×1 min60 s)=117.81 rad/s$

Find the initial kinetic energy $(T2)$ of the system using the equation:

$T2=(T2)A+(T2)B+(T2)C=12IA(ωA)22+12IB(ωB)22+12IC(ωC)22$

Here, $(T2)A$ is the final angular velocity of gear A, $(T2)B$ is the final angular velocity of gear B, and $(T2)C$ is the final angular velocity of gear C.

Substitute $8.64×10−3 kg⋅m2$ for $IA$ , $117.81 rad/s$ for $(ωA)1$ , $8.64×10−3 kg⋅m2$ for $IB$ , $117.81 rad/s$ for $(ωB)1$ , $270×10−3 kg⋅m2$ for $IC$ , and 450 rpm for $ωC$ .

$T2={[12(8.64×10−3)(117.81)2+12(8.64×10−3)(117.81)2+12(270×10−3)](450 revmin×2π rad1 rev×1 min60 s)2}=59.958+59.958+299.8=419.7 J$

Find the work done $(U1→2)$ due to couple acting on gear C using the equation:

$U1→2=MCθC$

Substitute $10 N⋅m$ for $MC$ .

$U1→2=10θC$

Find the number of revolutions of gear C $(θC)$ required for its angular velocity to increase from 100 to 450 rpm.

Apply Principle of work and energy for system.

$T1+U1→2=T2$

Substitute $20.726 J$ for $T1$ , 419.7 J for $T2$ , and $10θC$ for $U1→2$ .

$20.726+10θC=419.7θC=39.897 rad×1 rev2π radθC=6.35 rev$

Thus, the number of revolutions of gear C required for its angular velocity to increase from 100 to 450 rpm is $6.35 rev_$ .

(b)

To determine

The tangential force acting on gear A.

(b)

Expert Solution

Answer to Problem 17.11P

The tangential force acting on gear A is $7.14 N_$ .

Explanation of Solution

Calculation:

Find the rotation of gear A using the equation:

$rAθA=rCθCθA=rCrAθC$

Substitute 200 mm for $rC$ , $80 mm$ for $rA$ , and $39.897 rad$ for $θC$ .

$θA=20080(39.897)=99.743 rad$

Find the tangential force $(Ft)$ acting on gear A.

Apply Principle of work and energy for gear A.

$(T1)A+MAθA=(T2)A12IA(ωA)12+MAθA=12IA(ωA)22$

Substitute $8.64×10−3 kg⋅m2$ for $IA$ , $26.18 rad/s$ for $(ωA)1$ , 99.743 rad/s for $θA$ , and $117.81 rad/s$ for $(ωA)2$ .

$12(8.64×10−3)(26.18)2+MA(99.743)=12(8.64×10−3)(117.81)22.9609+99.743MA=59.958MA=0.57144 N⋅m$

Find the tangential force $(Ft)$ acting on gear A using the equation:

$Ft=MArA$

Substitute $0.57144 N⋅m$ for $MA$ and 80 mm for $rA$ .

$Ft=0.57144(80 mm×1 m1,000 mm)=7.14 N$

Thus, the tangential force acting on gear A is $7.14 N_$ .

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Answer 4

Question

Chapter 17.1, Problem 17.11P

(a)

To determine

The number of revolutions of gear C required for its angular velocity to increase from 100 to 450 rpm.

(a)

Expert Solution

Answer to Problem 17.11P

The number of revolutions of gear C required for its angular velocity to increase from 100 to 450 rpm is $6.35 rev_$ .

Explanation of Solution

Given information:

The mass $(mA)$ of gear A is 2.4 kg.

The mass $(mB)$ of gear B is 2.4 kg.

The mass $(mC)$ of gear C is 12 kg.

The radius of gyration $(kA)$ of gear A is 60 mm.

The radius of gyration $(kB)$ of gear B is 60 mm.

The radius of gyration $(kC)$ of gear C is 150 mm.

The radius $(rA)$ of gear A is 80 mm.

The radius $(rB)$ of gear B is 80 mm.

The radius $(rC)$ of gear C is 200 mm.

The initial angular velocity $(ωC)1$ of gear C is 100 rpm.

The final angular velocity $(ωC)2$ of gear C is 450 rpm.

The couple acting on gear C is $10 N⋅m$ .

Calculation:

Find the mass moment of inertia $(IA)$ of gear A using the equation:

$IA=mAkA2$

Substitute 2.4 kg for $mA$ and 60 mm for $kA$ .

$IA=(2.4)(60 mm×1 m1,000 mm)2=8.64×10−3 kg⋅m2$

Find the mass moment of inertia $(IB)$ of gear B using the equation:

$IB=mBkB2$

Substitute 2.4 kg for $mB$ and 60 mm for $kB$ .

$IB=(2.4)(60 mm×1 m1,000 mm)2=8.64×10−3 kg⋅m2$

Find the mass moment of inertia $(IC)$ of gear C using the equation:

$IC=mCkC2$

Substitute 12 kg for $mC$ and 150 mm for $kA$ .

$IC=(12)(150 mm×1 m1,000 mm)2=270×10−3 kg⋅m2$

Find the initial angular velocity $(ωA)1$ of gear A using the kinematics:

$rA(ωA)1=rC(ωC)1(ωA)1=rCrA(ωC)1$

Substitute 200 mm for $rC$ , 80 mm for $rA$ , and 100 rpm for $(ωC)1$ .

$(ωA)1=20080(100 revmin×2π rad1 rev×1 min60 s)=26.18 rad/s$

Find the initial angular velocity $(ωB)1$ of gear B using the kinematics:

$rB(ωB)1=rC(ωC)1(ωB)1=rCrB(ωC)1$

Substitute 200 mm for $rC$ , 80 mm for $rB$ , and 100 rpm for $(ωC)1$ .

$(ωB)1=20080(100 revmin×2π rad1 rev×1 min60 s)=26.18 rad/s$

Find the initial kinetic energy $(T1)$ of the system using the equation:

$T1=(T1)A+(T1)B+(T1)C=12IA(ωA)12+12IB(ωB)12+12IC(ωC)12$

Here, $(T1)A$ is the initial angular velocity of gear A, $(T1)B$ is the initial angular velocity of gear B, and $(T1)C$ is the initial angular velocity of gear C.

Substitute $8.64×10−3 kg⋅m2$ for $IA$ , $26.18 rad/s$ for $(ωA)1$ , $8.64×10−3 kg⋅m2$ for $IB$ , $26.18 rad/s$ for $(ωB)1$ , $270×10−3 kg⋅m2$ for $IC$ , and 100 rpm for $ωC$ .

$T1={12[(8.64×10−3)(26.18)2+12(8.64×10−3)(26.18)2+12(270×10−3)](100 revmin×2π rad1 rev×1 min60 s)2}=2.9609+2.9609+14.8044=20.726 J$

Find the final angular velocity $(ωA)2$ of gear A using the kinematics:

$rA(ωA)2=rC(ωC)2(ωA)2=rCrA(ωC)2$

Substitute 200 mm for $rC$ , 80 mm for $rA$ , and 450 rpm for $(ωC)2$ .

$(ωA)2=20080(450 revmin×2π rad1 rev×1 min60 s)=117.81 rad/s$

Find the final angular velocity $(ωB)2$ of gear B using the kinematics:

$rB(ωB)2=rC(ωC)2(ωB)2=rCrB(ωC)2$

Substitute 200 mm for $rC$ , 80 mm for $rB$ , and 450 rpm for $(ωC)2$ .

$(ωB)2=20080(450 revmin×2π rad1 rev×1 min60 s)=117.81 rad/s$

Find the initial kinetic energy $(T2)$ of the system using the equation:

$T2=(T2)A+(T2)B+(T2)C=12IA(ωA)22+12IB(ωB)22+12IC(ωC)22$

Here, $(T2)A$ is the final angular velocity of gear A, $(T2)B$ is the final angular velocity of gear B, and $(T2)C$ is the final angular velocity of gear C.

Substitute $8.64×10−3 kg⋅m2$ for $IA$ , $117.81 rad/s$ for $(ωA)1$ , $8.64×10−3 kg⋅m2$ for $IB$ , $117.81 rad/s$ for $(ωB)1$ , $270×10−3 kg⋅m2$ for $IC$ , and 450 rpm for $ωC$ .

$T2={[12(8.64×10−3)(117.81)2+12(8.64×10−3)(117.81)2+12(270×10−3)](450 revmin×2π rad1 rev×1 min60 s)2}=59.958+59.958+299.8=419.7 J$

Find the work done $(U1→2)$ due to couple acting on gear C using the equation:

$U1→2=MCθC$

Substitute $10 N⋅m$ for $MC$ .

$U1→2=10θC$

Find the number of revolutions of gear C $(θC)$ required for its angular velocity to increase from 100 to 450 rpm.

Apply Principle of work and energy for system.

$T1+U1→2=T2$

Substitute $20.726 J$ for $T1$ , 419.7 J for $T2$ , and $10θC$ for $U1→2$ .

$20.726+10θC=419.7θC=39.897 rad×1 rev2π radθC=6.35 rev$

Thus, the number of revolutions of gear C required for its angular velocity to increase from 100 to 450 rpm is $6.35 rev_$ .

(b)

To determine

The tangential force acting on gear A.

(b)

Expert Solution

Answer to Problem 17.11P

The tangential force acting on gear A is $7.14 N_$ .

Explanation of Solution

Calculation:

Find the rotation of gear A using the equation:

$rAθA=rCθCθA=rCrAθC$

Substitute 200 mm for $rC$ , $80 mm$ for $rA$ , and $39.897 rad$ for $θC$ .

$θA=20080(39.897)=99.743 rad$

Find the tangential force $(Ft)$ acting on gear A.

Apply Principle of work and energy for gear A.

$(T1)A+MAθA=(T2)A12IA(ωA)12+MAθA=12IA(ωA)22$

Substitute $8.64×10−3 kg⋅m2$ for $IA$ , $26.18 rad/s$ for $(ωA)1$ , 99.743 rad/s for $θA$ , and $117.81 rad/s$ for $(ωA)2$ .

$12(8.64×10−3)(26.18)2+MA(99.743)=12(8.64×10−3)(117.81)22.9609+99.743MA=59.958MA=0.57144 N⋅m$

Find the tangential force $(Ft)$ acting on gear A using the equation:

$Ft=MArA$

Substitute $0.57144 N⋅m$ for $MA$ and 80 mm for $rA$ .

$Ft=0.57144(80 mm×1 m1,000 mm)=7.14 N$

Thus, the tangential force acting on gear A is $7.14 N_$ .

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Answer 5

Chapter 17.1, Problem 17.11P

(a)

To determine

The number of revolutions of gear C required for its angular velocity to increase from 100 to 450 rpm.

(a)

Expert Solution

Answer to Problem 17.11P

The number of revolutions of gear C required for its angular velocity to increase from 100 to 450 rpm is $6.35 rev_$ .

Explanation of Solution

Given information:

The mass $(mA)$ of gear A is 2.4 kg.

The mass $(mB)$ of gear B is 2.4 kg.

The mass $(mC)$ of gear C is 12 kg.

The radius of gyration $(kA)$ of gear A is 60 mm.

The radius of gyration $(kB)$ of gear B is 60 mm.

The radius of gyration $(kC)$ of gear C is 150 mm.

The radius $(rA)$ of gear A is 80 mm.

The radius $(rB)$ of gear B is 80 mm.

The radius $(rC)$ of gear C is 200 mm.

The initial angular velocity $(ωC)1$ of gear C is 100 rpm.

The final angular velocity $(ωC)2$ of gear C is 450 rpm.

The couple acting on gear C is $10 N⋅m$ .

Calculation:

Find the mass moment of inertia $(IA)$ of gear A using the equation:

$IA=mAkA2$

Substitute 2.4 kg for $mA$ and 60 mm for $kA$ .

$IA=(2.4)(60 mm×1 m1,000 mm)2=8.64×10−3 kg⋅m2$

Find the mass moment of inertia $(IB)$ of gear B using the equation:

$IB=mBkB2$

Substitute 2.4 kg for $mB$ and 60 mm for $kB$ .

$IB=(2.4)(60 mm×1 m1,000 mm)2=8.64×10−3 kg⋅m2$

Find the mass moment of inertia $(IC)$ of gear C using the equation:

$IC=mCkC2$

Substitute 12 kg for $mC$ and 150 mm for $kA$ .

$IC=(12)(150 mm×1 m1,000 mm)2=270×10−3 kg⋅m2$

Find the initial angular velocity $(ωA)1$ of gear A using the kinematics:

$rA(ωA)1=rC(ωC)1(ωA)1=rCrA(ωC)1$

Substitute 200 mm for $rC$ , 80 mm for $rA$ , and 100 rpm for $(ωC)1$ .

$(ωA)1=20080(100 revmin×2π rad1 rev×1 min60 s)=26.18 rad/s$

Find the initial angular velocity $(ωB)1$ of gear B using the kinematics:

$rB(ωB)1=rC(ωC)1(ωB)1=rCrB(ωC)1$

Substitute 200 mm for $rC$ , 80 mm for $rB$ , and 100 rpm for $(ωC)1$ .

$(ωB)1=20080(100 revmin×2π rad1 rev×1 min60 s)=26.18 rad/s$

Find the initial kinetic energy $(T1)$ of the system using the equation:

$T1=(T1)A+(T1)B+(T1)C=12IA(ωA)12+12IB(ωB)12+12IC(ωC)12$

Here, $(T1)A$ is the initial angular velocity of gear A, $(T1)B$ is the initial angular velocity of gear B, and $(T1)C$ is the initial angular velocity of gear C.

Substitute $8.64×10−3 kg⋅m2$ for $IA$ , $26.18 rad/s$ for $(ωA)1$ , $8.64×10−3 kg⋅m2$ for $IB$ , $26.18 rad/s$ for $(ωB)1$ , $270×10−3 kg⋅m2$ for $IC$ , and 100 rpm for $ωC$ .

$T1={12[(8.64×10−3)(26.18)2+12(8.64×10−3)(26.18)2+12(270×10−3)](100 revmin×2π rad1 rev×1 min60 s)2}=2.9609+2.9609+14.8044=20.726 J$

Find the final angular velocity $(ωA)2$ of gear A using the kinematics:

$rA(ωA)2=rC(ωC)2(ωA)2=rCrA(ωC)2$

Substitute 200 mm for $rC$ , 80 mm for $rA$ , and 450 rpm for $(ωC)2$ .

$(ωA)2=20080(450 revmin×2π rad1 rev×1 min60 s)=117.81 rad/s$

Find the final angular velocity $(ωB)2$ of gear B using the kinematics:

$rB(ωB)2=rC(ωC)2(ωB)2=rCrB(ωC)2$

Substitute 200 mm for $rC$ , 80 mm for $rB$ , and 450 rpm for $(ωC)2$ .

$(ωB)2=20080(450 revmin×2π rad1 rev×1 min60 s)=117.81 rad/s$

Find the initial kinetic energy $(T2)$ of the system using the equation:

$T2=(T2)A+(T2)B+(T2)C=12IA(ωA)22+12IB(ωB)22+12IC(ωC)22$

Here, $(T2)A$ is the final angular velocity of gear A, $(T2)B$ is the final angular velocity of gear B, and $(T2)C$ is the final angular velocity of gear C.

Substitute $8.64×10−3 kg⋅m2$ for $IA$ , $117.81 rad/s$ for $(ωA)1$ , $8.64×10−3 kg⋅m2$ for $IB$ , $117.81 rad/s$ for $(ωB)1$ , $270×10−3 kg⋅m2$ for $IC$ , and 450 rpm for $ωC$ .

$T2={[12(8.64×10−3)(117.81)2+12(8.64×10−3)(117.81)2+12(270×10−3)](450 revmin×2π rad1 rev×1 min60 s)2}=59.958+59.958+299.8=419.7 J$

Find the work done $(U1→2)$ due to couple acting on gear C using the equation:

$U1→2=MCθC$

Substitute $10 N⋅m$ for $MC$ .

$U1→2=10θC$

Find the number of revolutions of gear C $(θC)$ required for its angular velocity to increase from 100 to 450 rpm.

Apply Principle of work and energy for system.

$T1+U1→2=T2$

Substitute $20.726 J$ for $T1$ , 419.7 J for $T2$ , and $10θC$ for $U1→2$ .

$20.726+10θC=419.7θC=39.897 rad×1 rev2π radθC=6.35 rev$

Thus, the number of revolutions of gear C required for its angular velocity to increase from 100 to 450 rpm is $6.35 rev_$ .

(b)

To determine

The tangential force acting on gear A.

(b)

Expert Solution

Answer to Problem 17.11P

The tangential force acting on gear A is $7.14 N_$ .

Explanation of Solution

Calculation:

Find the rotation of gear A using the equation:

$rAθA=rCθCθA=rCrAθC$

Substitute 200 mm for $rC$ , $80 mm$ for $rA$ , and $39.897 rad$ for $θC$ .

$θA=20080(39.897)=99.743 rad$

Find the tangential force $(Ft)$ acting on gear A.

Apply Principle of work and energy for gear A.

$(T1)A+MAθA=(T2)A12IA(ωA)12+MAθA=12IA(ωA)22$

Substitute $8.64×10−3 kg⋅m2$ for $IA$ , $26.18 rad/s$ for $(ωA)1$ , 99.743 rad/s for $θA$ , and $117.81 rad/s$ for $(ωA)2$ .

$12(8.64×10−3)(26.18)2+MA(99.743)=12(8.64×10−3)(117.81)22.9609+99.743MA=59.958MA=0.57144 N⋅m$

Find the tangential force $(Ft)$ acting on gear A using the equation:

$Ft=MArA$

Substitute $0.57144 N⋅m$ for $MA$ and 80 mm for $rA$ .

$Ft=0.57144(80 mm×1 m1,000 mm)=7.14 N$

Thus, the tangential force acting on gear A is $7.14 N_$ .

Answer 6

Chapter 17.1, Problem 17.11P

(a)

To determine

The number of revolutions of gear C required for its angular velocity to increase from 100 to 450 rpm.

(a)

Expert Solution

Answer to Problem 17.11P

The number of revolutions of gear C required for its angular velocity to increase from 100 to 450 rpm is $6.35 rev_$ .

Explanation of Solution

Given information:

The mass $(mA)$ of gear A is 2.4 kg.

The mass $(mB)$ of gear B is 2.4 kg.

The mass $(mC)$ of gear C is 12 kg.

The radius of gyration $(kA)$ of gear A is 60 mm.

The radius of gyration $(kB)$ of gear B is 60 mm.

The radius of gyration $(kC)$ of gear C is 150 mm.

The radius $(rA)$ of gear A is 80 mm.

The radius $(rB)$ of gear B is 80 mm.

The radius $(rC)$ of gear C is 200 mm.

The initial angular velocity $(ωC)1$ of gear C is 100 rpm.

The final angular velocity $(ωC)2$ of gear C is 450 rpm.

The couple acting on gear C is $10 N⋅m$ .

Calculation:

Find the mass moment of inertia $(IA)$ of gear A using the equation:

$IA=mAkA2$

Substitute 2.4 kg for $mA$ and 60 mm for $kA$ .

$IA=(2.4)(60 mm×1 m1,000 mm)2=8.64×10−3 kg⋅m2$

Find the mass moment of inertia $(IB)$ of gear B using the equation:

$IB=mBkB2$

Substitute 2.4 kg for $mB$ and 60 mm for $kB$ .

$IB=(2.4)(60 mm×1 m1,000 mm)2=8.64×10−3 kg⋅m2$

Find the mass moment of inertia $(IC)$ of gear C using the equation:

$IC=mCkC2$

Substitute 12 kg for $mC$ and 150 mm for $kA$ .

$IC=(12)(150 mm×1 m1,000 mm)2=270×10−3 kg⋅m2$

Find the initial angular velocity $(ωA)1$ of gear A using the kinematics:

$rA(ωA)1=rC(ωC)1(ωA)1=rCrA(ωC)1$

Substitute 200 mm for $rC$ , 80 mm for $rA$ , and 100 rpm for $(ωC)1$ .

$(ωA)1=20080(100 revmin×2π rad1 rev×1 min60 s)=26.18 rad/s$

Find the initial angular velocity $(ωB)1$ of gear B using the kinematics:

$rB(ωB)1=rC(ωC)1(ωB)1=rCrB(ωC)1$

Substitute 200 mm for $rC$ , 80 mm for $rB$ , and 100 rpm for $(ωC)1$ .

$(ωB)1=20080(100 revmin×2π rad1 rev×1 min60 s)=26.18 rad/s$

Find the initial kinetic energy $(T1)$ of the system using the equation:

$T1=(T1)A+(T1)B+(T1)C=12IA(ωA)12+12IB(ωB)12+12IC(ωC)12$

Here, $(T1)A$ is the initial angular velocity of gear A, $(T1)B$ is the initial angular velocity of gear B, and $(T1)C$ is the initial angular velocity of gear C.

Substitute $8.64×10−3 kg⋅m2$ for $IA$ , $26.18 rad/s$ for $(ωA)1$ , $8.64×10−3 kg⋅m2$ for $IB$ , $26.18 rad/s$ for $(ωB)1$ , $270×10−3 kg⋅m2$ for $IC$ , and 100 rpm for $ωC$ .

$T1={12[(8.64×10−3)(26.18)2+12(8.64×10−3)(26.18)2+12(270×10−3)](100 revmin×2π rad1 rev×1 min60 s)2}=2.9609+2.9609+14.8044=20.726 J$

Find the final angular velocity $(ωA)2$ of gear A using the kinematics:

$rA(ωA)2=rC(ωC)2(ωA)2=rCrA(ωC)2$

Substitute 200 mm for $rC$ , 80 mm for $rA$ , and 450 rpm for $(ωC)2$ .

$(ωA)2=20080(450 revmin×2π rad1 rev×1 min60 s)=117.81 rad/s$

Find the final angular velocity $(ωB)2$ of gear B using the kinematics:

$rB(ωB)2=rC(ωC)2(ωB)2=rCrB(ωC)2$

Substitute 200 mm for $rC$ , 80 mm for $rB$ , and 450 rpm for $(ωC)2$ .

$(ωB)2=20080(450 revmin×2π rad1 rev×1 min60 s)=117.81 rad/s$

Find the initial kinetic energy $(T2)$ of the system using the equation:

$T2=(T2)A+(T2)B+(T2)C=12IA(ωA)22+12IB(ωB)22+12IC(ωC)22$

Here, $(T2)A$ is the final angular velocity of gear A, $(T2)B$ is the final angular velocity of gear B, and $(T2)C$ is the final angular velocity of gear C.

Substitute $8.64×10−3 kg⋅m2$ for $IA$ , $117.81 rad/s$ for $(ωA)1$ , $8.64×10−3 kg⋅m2$ for $IB$ , $117.81 rad/s$ for $(ωB)1$ , $270×10−3 kg⋅m2$ for $IC$ , and 450 rpm for $ωC$ .

$T2={[12(8.64×10−3)(117.81)2+12(8.64×10−3)(117.81)2+12(270×10−3)](450 revmin×2π rad1 rev×1 min60 s)2}=59.958+59.958+299.8=419.7 J$

Find the work done $(U1→2)$ due to couple acting on gear C using the equation:

$U1→2=MCθC$

Substitute $10 N⋅m$ for $MC$ .

$U1→2=10θC$

Find the number of revolutions of gear C $(θC)$ required for its angular velocity to increase from 100 to 450 rpm.

Apply Principle of work and energy for system.

$T1+U1→2=T2$

Substitute $20.726 J$ for $T1$ , 419.7 J for $T2$ , and $10θC$ for $U1→2$ .

$20.726+10θC=419.7θC=39.897 rad×1 rev2π radθC=6.35 rev$

Thus, the number of revolutions of gear C required for its angular velocity to increase from 100 to 450 rpm is $6.35 rev_$ .

Answer 7

Chapter 17.1, Problem 17.11P

(a)

To determine

The number of revolutions of gear C required for its angular velocity to increase from 100 to 450 rpm.

Answer 8

(a)

Expert Solution

Answer to Problem 17.11P

The number of revolutions of gear C required for its angular velocity to increase from 100 to 450 rpm is $6.35 rev_$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given information:

The mass $(mA)$ of gear A is 2.4 kg.

The mass $(mB)$ of gear B is 2.4 kg.

The mass $(mC)$ of gear C is 12 kg.

The radius of gyration $(kA)$ of gear A is 60 mm.

The radius of gyration $(kB)$ of gear B is 60 mm.

The radius of gyration $(kC)$ of gear C is 150 mm.

The radius $(rA)$ of gear A is 80 mm.

The radius $(rB)$ of gear B is 80 mm.

The radius $(rC)$ of gear C is 200 mm.

The initial angular velocity $(ωC)1$ of gear C is 100 rpm.

The final angular velocity $(ωC)2$ of gear C is 450 rpm.

The couple acting on gear C is $10 N⋅m$ .

Calculation:

Find the mass moment of inertia $(IA)$ of gear A using the equation:

$IA=mAkA2$

Substitute 2.4 kg for $mA$ and 60 mm for $kA$ .

$IA=(2.4)(60 mm×1 m1,000 mm)2=8.64×10−3 kg⋅m2$

Find the mass moment of inertia $(IB)$ of gear B using the equation:

$IB=mBkB2$

Substitute 2.4 kg for $mB$ and 60 mm for $kB$ .

$IB=(2.4)(60 mm×1 m1,000 mm)2=8.64×10−3 kg⋅m2$

Find the mass moment of inertia $(IC)$ of gear C using the equation:

$IC=mCkC2$

Substitute 12 kg for $mC$ and 150 mm for $kA$ .

$IC=(12)(150 mm×1 m1,000 mm)2=270×10−3 kg⋅m2$

Find the initial angular velocity $(ωA)1$ of gear A using the kinematics:

$rA(ωA)1=rC(ωC)1(ωA)1=rCrA(ωC)1$

Substitute 200 mm for $rC$ , 80 mm for $rA$ , and 100 rpm for $(ωC)1$ .

$(ωA)1=20080(100 revmin×2π rad1 rev×1 min60 s)=26.18 rad/s$

Find the initial angular velocity $(ωB)1$ of gear B using the kinematics:

$rB(ωB)1=rC(ωC)1(ωB)1=rCrB(ωC)1$

Substitute 200 mm for $rC$ , 80 mm for $rB$ , and 100 rpm for $(ωC)1$ .

$(ωB)1=20080(100 revmin×2π rad1 rev×1 min60 s)=26.18 rad/s$

Find the initial kinetic energy $(T1)$ of the system using the equation:

$T1=(T1)A+(T1)B+(T1)C=12IA(ωA)12+12IB(ωB)12+12IC(ωC)12$

Here, $(T1)A$ is the initial angular velocity of gear A, $(T1)B$ is the initial angular velocity of gear B, and $(T1)C$ is the initial angular velocity of gear C.

Substitute $8.64×10−3 kg⋅m2$ for $IA$ , $26.18 rad/s$ for $(ωA)1$ , $8.64×10−3 kg⋅m2$ for $IB$ , $26.18 rad/s$ for $(ωB)1$ , $270×10−3 kg⋅m2$ for $IC$ , and 100 rpm for $ωC$ .

$T1={12[(8.64×10−3)(26.18)2+12(8.64×10−3)(26.18)2+12(270×10−3)](100 revmin×2π rad1 rev×1 min60 s)2}=2.9609+2.9609+14.8044=20.726 J$

Find the final angular velocity $(ωA)2$ of gear A using the kinematics:

$rA(ωA)2=rC(ωC)2(ωA)2=rCrA(ωC)2$

Substitute 200 mm for $rC$ , 80 mm for $rA$ , and 450 rpm for $(ωC)2$ .

$(ωA)2=20080(450 revmin×2π rad1 rev×1 min60 s)=117.81 rad/s$

Find the final angular velocity $(ωB)2$ of gear B using the kinematics:

$rB(ωB)2=rC(ωC)2(ωB)2=rCrB(ωC)2$

Substitute 200 mm for $rC$ , 80 mm for $rB$ , and 450 rpm for $(ωC)2$ .

$(ωB)2=20080(450 revmin×2π rad1 rev×1 min60 s)=117.81 rad/s$

Find the initial kinetic energy $(T2)$ of the system using the equation:

$T2=(T2)A+(T2)B+(T2)C=12IA(ωA)22+12IB(ωB)22+12IC(ωC)22$

Here, $(T2)A$ is the final angular velocity of gear A, $(T2)B$ is the final angular velocity of gear B, and $(T2)C$ is the final angular velocity of gear C.

Substitute $8.64×10−3 kg⋅m2$ for $IA$ , $117.81 rad/s$ for $(ωA)1$ , $8.64×10−3 kg⋅m2$ for $IB$ , $117.81 rad/s$ for $(ωB)1$ , $270×10−3 kg⋅m2$ for $IC$ , and 450 rpm for $ωC$ .

$T2={[12(8.64×10−3)(117.81)2+12(8.64×10−3)(117.81)2+12(270×10−3)](450 revmin×2π rad1 rev×1 min60 s)2}=59.958+59.958+299.8=419.7 J$

Find the work done $(U1→2)$ due to couple acting on gear C using the equation:

$U1→2=MCθC$

Substitute $10 N⋅m$ for $MC$ .

$U1→2=10θC$

Find the number of revolutions of gear C $(θC)$ required for its angular velocity to increase from 100 to 450 rpm.

Apply Principle of work and energy for system.

$T1+U1→2=T2$

Substitute $20.726 J$ for $T1$ , 419.7 J for $T2$ , and $10θC$ for $U1→2$ .

$20.726+10θC=419.7θC=39.897 rad×1 rev2π radθC=6.35 rev$

Thus, the number of revolutions of gear C required for its angular velocity to increase from 100 to 450 rpm is $6.35 rev_$ .

Answer 13

(b)

To determine

The tangential force acting on gear A.

(b)

Expert Solution

Answer to Problem 17.11P

The tangential force acting on gear A is $7.14 N_$ .

Explanation of Solution

Calculation:

Find the rotation of gear A using the equation:

$rAθA=rCθCθA=rCrAθC$

Substitute 200 mm for $rC$ , $80 mm$ for $rA$ , and $39.897 rad$ for $θC$ .

$θA=20080(39.897)=99.743 rad$

Find the tangential force $(Ft)$ acting on gear A.

Apply Principle of work and energy for gear A.

$(T1)A+MAθA=(T2)A12IA(ωA)12+MAθA=12IA(ωA)22$

Substitute $8.64×10−3 kg⋅m2$ for $IA$ , $26.18 rad/s$ for $(ωA)1$ , 99.743 rad/s for $θA$ , and $117.81 rad/s$ for $(ωA)2$ .

$12(8.64×10−3)(26.18)2+MA(99.743)=12(8.64×10−3)(117.81)22.9609+99.743MA=59.958MA=0.57144 N⋅m$

Find the tangential force $(Ft)$ acting on gear A using the equation:

$Ft=MArA$

Substitute $0.57144 N⋅m$ for $MA$ and 80 mm for $rA$ .

$Ft=0.57144(80 mm×1 m1,000 mm)=7.14 N$

Thus, the tangential force acting on gear A is $7.14 N_$ .

Answer 14

(b)

To determine

The tangential force acting on gear A.

Answer 15

(b)

Expert Solution

Answer to Problem 17.11P

The tangential force acting on gear A is $7.14 N_$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation:

Find the rotation of gear A using the equation:

$rAθA=rCθCθA=rCrAθC$

Substitute 200 mm for $rC$ , $80 mm$ for $rA$ , and $39.897 rad$ for $θC$ .

$θA=20080(39.897)=99.743 rad$

Find the tangential force $(Ft)$ acting on gear A.

Apply Principle of work and energy for gear A.

$(T1)A+MAθA=(T2)A12IA(ωA)12+MAθA=12IA(ωA)22$

Substitute $8.64×10−3 kg⋅m2$ for $IA$ , $26.18 rad/s$ for $(ωA)1$ , 99.743 rad/s for $θA$ , and $117.81 rad/s$ for $(ωA)2$ .

$12(8.64×10−3)(26.18)2+MA(99.743)=12(8.64×10−3)(117.81)22.9609+99.743MA=59.958MA=0.57144 N⋅m$

Find the tangential force $(Ft)$ acting on gear A using the equation:

$Ft=MArA$

Substitute $0.57144 N⋅m$ for $MA$ and 80 mm for $rA$ .

$Ft=0.57144(80 mm×1 m1,000 mm)=7.14 N$

Thus, the tangential force acting on gear A is $7.14 N_$ .

Answer 20

Ch. 17.1 - A round object of mass m and radius r is released...Ch. 17.1 - Prob. 17.2CQ Ch. 17.1 - Prob. 17.3CQ Ch. 17.1 - Prob. 17.4CQ Ch. 17.1 - Slender bar A is rigidly connected to a massless...Ch. 17.1 - A 200-kg flywheel is at rest when a constant 300...Ch. 17.1 - The rotor of an electric motor has an angular...Ch. 17.1 - Prob. 17.3P Ch. 17.1 - Two disks of the same material are attached to a...Ch. 17.1 - Prob. 17.5P

The number of revolutions of gear C required for its angular velocity to increase from 100 to 450 rpm.

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The number of revolutions of gear C required for its angular velocity to increase from 100 to 450 rpm.

Answer to Problem 17.11P

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The tangential force acting on gear A.

Answer to Problem 17.11P

Explanation of Solution

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