(a) Then the distance ‘d’ to maximise the frequency of oscillation when a small initial displacement is given.

Question

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Answer 1

Question

Chapter 19.2, Problem 19.50P

To determine

(a)

Then the distance ‘d’ to maximise the frequency of oscillation when a small initial displacement is given.

Expert Solution

Answer to Problem 19.50P

Distance $d=227mm$

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 19.2, Problem 19.50P , additional homework tip 1

Mass of collar $mC=1 Κg$

Mass of rod $mR=3 Κg$

Length of rod $L=750 mm$

The forces corresponding to the rod and collar are shown in the free body diagram below:

Vector Mechanics For Engineers, Chapter 19.2, Problem 19.50P , additional homework tip 2

$α=θ¨ (a¯t)R=L2α=L2θ¨(at)C=dα=dθ¨$

Now, from the equation of motion taking moment about A,

$∑MA=∑(MA)eff$

$−WRL2sinθ−WCdsinθ=IRα+mRL2(a¯t)R+mCd(at)CIRθ¨+mRL2×L2θ¨+mCd(dθ¨)+mRgL2θ+mCgdθ=0(IR+mR(L2)2+mCd2)θ¨+(mRgL2+mCgd)θ=0$ Since, $sinθ≈θ$ and $W=mg$

And, moment of inertia of rod is $IR=mRL212$

$(mRL212+mRL24+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL2+3mRL212+mCd2)θ¨+(mRgL2+mCgd)θ=0$

$(4mRL212+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL23+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL23+mCd2)θ¨+(mRL2+mCd)gθ=0$

$θ¨+(L2+mCmRd)g(L23+mCmRd2)θ=0$

Compare the above equation with un-damped equation of vibration:

$Mθ¨+ωn2θ=0$

Natural frequency:

$ωn2=(L2+mCmRd)g(L23+mCmRd2)ωn2=(L2+13d)g(L23+13d2)ωn2=(3L+2d6)g(L2+d23)ωn2=(3L2+d6)g×(3L2+d2)ωn2=(3L2+d)g(L2+d2)$

To maximise the frequency, we need to take the derivative with respect to d and set it equal to zero.

$1gd(ωn2)d(d)=(L2+d2)(1)−(3L2+d)(2d)(L2+d2)2=0(L2+d2)(1)−(3L2+d)(2d)=0L2+d2−3L2×2d−2d×d=0L2+d2−3Ld−2d2=0L2−3Ld−d2=0(0.75)2−3(0.75)d−d2=00.5625−2.25d−d2=0 or,d2+2.25d−0.5625=0$

By solving the above equation we get,

$d=0.2270 or −2.477$

Conclusion:

The distance $d=227mm.$

To determine

(b)

The period of oscillation.

Expert Solution

Answer to Problem 19.50P

Period of vibration, $τn=1.3512 s$

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 19.2, Problem 19.50P , additional homework tip 3

Mass of collar $mC=1 Κg$

Mass of rod $mR=3 Κg$

Length of rod $L=750 mm$

The forces corresponding to the rod and collar are shown in the free body diagram below:

Vector Mechanics For Engineers, Chapter 19.2, Problem 19.50P , additional homework tip 4

$α=θ¨ (a¯t)R=L2α=L2θ¨(at)C=dα=dθ¨$

Now, from the equation of motion taking moment about A,

$∑MA=∑(MA)eff$

$−WRL2sinθ−WCdsinθ=IRα+mRL2(a¯t)R+mCd(at)CIRθ¨+mRL2×L2θ¨+mCd(dθ¨)+mRgL2θ+mCgdθ=0(IR+mR(L2)2+mCd2)θ¨+(mRgL2+mCgd)θ=0$ Since, $sinθ≈θ$ and $W=mg$

And, moment of inertia of rod is $IR=mRL212$

$(mRL212+mRL24+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL2+3mRL212+mCd2)θ¨+(mRgL2+mCgd)θ=0$

$(4mRL212+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL23+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL23+mCd2)θ¨+(mRL2+mCd)gθ=0$

$θ¨+(L2+mCmRd)g(L23+mCmRd2)θ=0$

Compare the above equation with un-damped equation of vibration;

$Mθ¨+ωn2θ=0$

Natural frequency:

$ωn2=(L2+mCmRd)g(L23+mCmRd2)ωn2=(L2+13d)g(L23+13d2)ωn2=(3L+2d6)g(L2+d23)ωn2=(3L2+d6)g×(3L2+d2)ωn2=(3L2+d)g(L2+d2)$

To maximise the frequency, we need to take the derivative with respect to d and set it equal to zero.

$1gd(ωn2)d(d)=(L2+d2)(1)−(3L2+d)(2d)(L2+d2)2=0(L2+d2)(1)−(3L2+d)(2d)=0L2+d2−3L2×2d−2d×d=0L2+d2−3Ld−2d2=0L2−3Ld−d2=0(0.75)2−3(0.75)d−d2=00.5625−2.25d−d2=0 or,d2+2.25d−0.5625=0$

By solving the above equation we get,

$d=0.2270 or −2.477$

Thus, $d=0.2270 m or d=227mm$

Now, natural frequency:

$ωn2=(3L2+d)g(L2+d2)ωn2=(3(0.75)2+0.2270)9.81((0.75)2+(0.2270)2)ωn2=(1.125+0.2270)9.81(0.5625+0.051529)ωn2=21.6001524ω2=4.6476 rad/s$

Then, Time period $τn=2πωn$

$τn=2×3.144.6476τn=1.3512 s$

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Answer 2

Question

Chapter 19.2, Problem 19.50P

To determine

(a)

Then the distance ‘d’ to maximise the frequency of oscillation when a small initial displacement is given.

Expert Solution

Answer to Problem 19.50P

Distance $d=227mm$

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 19.2, Problem 19.50P , additional homework tip 1

Mass of collar $mC=1 Κg$

Mass of rod $mR=3 Κg$

Length of rod $L=750 mm$

The forces corresponding to the rod and collar are shown in the free body diagram below:

Vector Mechanics For Engineers, Chapter 19.2, Problem 19.50P , additional homework tip 2

$α=θ¨ (a¯t)R=L2α=L2θ¨(at)C=dα=dθ¨$

Now, from the equation of motion taking moment about A,

$∑MA=∑(MA)eff$

$−WRL2sinθ−WCdsinθ=IRα+mRL2(a¯t)R+mCd(at)CIRθ¨+mRL2×L2θ¨+mCd(dθ¨)+mRgL2θ+mCgdθ=0(IR+mR(L2)2+mCd2)θ¨+(mRgL2+mCgd)θ=0$ Since, $sinθ≈θ$ and $W=mg$

And, moment of inertia of rod is $IR=mRL212$

$(mRL212+mRL24+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL2+3mRL212+mCd2)θ¨+(mRgL2+mCgd)θ=0$

$(4mRL212+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL23+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL23+mCd2)θ¨+(mRL2+mCd)gθ=0$

$θ¨+(L2+mCmRd)g(L23+mCmRd2)θ=0$

Compare the above equation with un-damped equation of vibration:

$Mθ¨+ωn2θ=0$

Natural frequency:

$ωn2=(L2+mCmRd)g(L23+mCmRd2)ωn2=(L2+13d)g(L23+13d2)ωn2=(3L+2d6)g(L2+d23)ωn2=(3L2+d6)g×(3L2+d2)ωn2=(3L2+d)g(L2+d2)$

To maximise the frequency, we need to take the derivative with respect to d and set it equal to zero.

$1gd(ωn2)d(d)=(L2+d2)(1)−(3L2+d)(2d)(L2+d2)2=0(L2+d2)(1)−(3L2+d)(2d)=0L2+d2−3L2×2d−2d×d=0L2+d2−3Ld−2d2=0L2−3Ld−d2=0(0.75)2−3(0.75)d−d2=00.5625−2.25d−d2=0 or,d2+2.25d−0.5625=0$

By solving the above equation we get,

$d=0.2270 or −2.477$

Conclusion:

The distance $d=227mm.$

To determine

(b)

The period of oscillation.

Expert Solution

Answer to Problem 19.50P

Period of vibration, $τn=1.3512 s$

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 19.2, Problem 19.50P , additional homework tip 3

Mass of collar $mC=1 Κg$

Mass of rod $mR=3 Κg$

Length of rod $L=750 mm$

The forces corresponding to the rod and collar are shown in the free body diagram below:

Vector Mechanics For Engineers, Chapter 19.2, Problem 19.50P , additional homework tip 4

$α=θ¨ (a¯t)R=L2α=L2θ¨(at)C=dα=dθ¨$

Now, from the equation of motion taking moment about A,

$∑MA=∑(MA)eff$

$−WRL2sinθ−WCdsinθ=IRα+mRL2(a¯t)R+mCd(at)CIRθ¨+mRL2×L2θ¨+mCd(dθ¨)+mRgL2θ+mCgdθ=0(IR+mR(L2)2+mCd2)θ¨+(mRgL2+mCgd)θ=0$ Since, $sinθ≈θ$ and $W=mg$

And, moment of inertia of rod is $IR=mRL212$

$(mRL212+mRL24+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL2+3mRL212+mCd2)θ¨+(mRgL2+mCgd)θ=0$

$(4mRL212+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL23+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL23+mCd2)θ¨+(mRL2+mCd)gθ=0$

$θ¨+(L2+mCmRd)g(L23+mCmRd2)θ=0$

Compare the above equation with un-damped equation of vibration;

$Mθ¨+ωn2θ=0$

Natural frequency:

$ωn2=(L2+mCmRd)g(L23+mCmRd2)ωn2=(L2+13d)g(L23+13d2)ωn2=(3L+2d6)g(L2+d23)ωn2=(3L2+d6)g×(3L2+d2)ωn2=(3L2+d)g(L2+d2)$

To maximise the frequency, we need to take the derivative with respect to d and set it equal to zero.

$1gd(ωn2)d(d)=(L2+d2)(1)−(3L2+d)(2d)(L2+d2)2=0(L2+d2)(1)−(3L2+d)(2d)=0L2+d2−3L2×2d−2d×d=0L2+d2−3Ld−2d2=0L2−3Ld−d2=0(0.75)2−3(0.75)d−d2=00.5625−2.25d−d2=0 or,d2+2.25d−0.5625=0$

By solving the above equation we get,

$d=0.2270 or −2.477$

Thus, $d=0.2270 m or d=227mm$

Now, natural frequency:

$ωn2=(3L2+d)g(L2+d2)ωn2=(3(0.75)2+0.2270)9.81((0.75)2+(0.2270)2)ωn2=(1.125+0.2270)9.81(0.5625+0.051529)ωn2=21.6001524ω2=4.6476 rad/s$

Then, Time period $τn=2πωn$

$τn=2×3.144.6476τn=1.3512 s$

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Answer 3

Question

Chapter 19.2, Problem 19.50P

To determine

(a)

Then the distance ‘d’ to maximise the frequency of oscillation when a small initial displacement is given.

Expert Solution

Answer to Problem 19.50P

Distance $d=227mm$

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 19.2, Problem 19.50P , additional homework tip 1

Mass of collar $mC=1 Κg$

Mass of rod $mR=3 Κg$

Length of rod $L=750 mm$

The forces corresponding to the rod and collar are shown in the free body diagram below:

Vector Mechanics For Engineers, Chapter 19.2, Problem 19.50P , additional homework tip 2

$α=θ¨ (a¯t)R=L2α=L2θ¨(at)C=dα=dθ¨$

Now, from the equation of motion taking moment about A,

$∑MA=∑(MA)eff$

$−WRL2sinθ−WCdsinθ=IRα+mRL2(a¯t)R+mCd(at)CIRθ¨+mRL2×L2θ¨+mCd(dθ¨)+mRgL2θ+mCgdθ=0(IR+mR(L2)2+mCd2)θ¨+(mRgL2+mCgd)θ=0$ Since, $sinθ≈θ$ and $W=mg$

And, moment of inertia of rod is $IR=mRL212$

$(mRL212+mRL24+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL2+3mRL212+mCd2)θ¨+(mRgL2+mCgd)θ=0$

$(4mRL212+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL23+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL23+mCd2)θ¨+(mRL2+mCd)gθ=0$

$θ¨+(L2+mCmRd)g(L23+mCmRd2)θ=0$

Compare the above equation with un-damped equation of vibration:

$Mθ¨+ωn2θ=0$

Natural frequency:

$ωn2=(L2+mCmRd)g(L23+mCmRd2)ωn2=(L2+13d)g(L23+13d2)ωn2=(3L+2d6)g(L2+d23)ωn2=(3L2+d6)g×(3L2+d2)ωn2=(3L2+d)g(L2+d2)$

To maximise the frequency, we need to take the derivative with respect to d and set it equal to zero.

$1gd(ωn2)d(d)=(L2+d2)(1)−(3L2+d)(2d)(L2+d2)2=0(L2+d2)(1)−(3L2+d)(2d)=0L2+d2−3L2×2d−2d×d=0L2+d2−3Ld−2d2=0L2−3Ld−d2=0(0.75)2−3(0.75)d−d2=00.5625−2.25d−d2=0 or,d2+2.25d−0.5625=0$

By solving the above equation we get,

$d=0.2270 or −2.477$

Conclusion:

The distance $d=227mm.$

To determine

(b)

The period of oscillation.

Expert Solution

Answer to Problem 19.50P

Period of vibration, $τn=1.3512 s$

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 19.2, Problem 19.50P , additional homework tip 3

Mass of collar $mC=1 Κg$

Mass of rod $mR=3 Κg$

Length of rod $L=750 mm$

The forces corresponding to the rod and collar are shown in the free body diagram below:

Vector Mechanics For Engineers, Chapter 19.2, Problem 19.50P , additional homework tip 4

$α=θ¨ (a¯t)R=L2α=L2θ¨(at)C=dα=dθ¨$

Now, from the equation of motion taking moment about A,

$∑MA=∑(MA)eff$

$−WRL2sinθ−WCdsinθ=IRα+mRL2(a¯t)R+mCd(at)CIRθ¨+mRL2×L2θ¨+mCd(dθ¨)+mRgL2θ+mCgdθ=0(IR+mR(L2)2+mCd2)θ¨+(mRgL2+mCgd)θ=0$ Since, $sinθ≈θ$ and $W=mg$

And, moment of inertia of rod is $IR=mRL212$

$(mRL212+mRL24+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL2+3mRL212+mCd2)θ¨+(mRgL2+mCgd)θ=0$

$(4mRL212+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL23+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL23+mCd2)θ¨+(mRL2+mCd)gθ=0$

$θ¨+(L2+mCmRd)g(L23+mCmRd2)θ=0$

Compare the above equation with un-damped equation of vibration;

$Mθ¨+ωn2θ=0$

Natural frequency:

$ωn2=(L2+mCmRd)g(L23+mCmRd2)ωn2=(L2+13d)g(L23+13d2)ωn2=(3L+2d6)g(L2+d23)ωn2=(3L2+d6)g×(3L2+d2)ωn2=(3L2+d)g(L2+d2)$

To maximise the frequency, we need to take the derivative with respect to d and set it equal to zero.

$1gd(ωn2)d(d)=(L2+d2)(1)−(3L2+d)(2d)(L2+d2)2=0(L2+d2)(1)−(3L2+d)(2d)=0L2+d2−3L2×2d−2d×d=0L2+d2−3Ld−2d2=0L2−3Ld−d2=0(0.75)2−3(0.75)d−d2=00.5625−2.25d−d2=0 or,d2+2.25d−0.5625=0$

By solving the above equation we get,

$d=0.2270 or −2.477$

Thus, $d=0.2270 m or d=227mm$

Now, natural frequency:

$ωn2=(3L2+d)g(L2+d2)ωn2=(3(0.75)2+0.2270)9.81((0.75)2+(0.2270)2)ωn2=(1.125+0.2270)9.81(0.5625+0.051529)ωn2=21.6001524ω2=4.6476 rad/s$

Then, Time period $τn=2πωn$

$τn=2×3.144.6476τn=1.3512 s$

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Answer 4

Question

Chapter 19.2, Problem 19.50P

To determine

(a)

Then the distance ‘d’ to maximise the frequency of oscillation when a small initial displacement is given.

Expert Solution

Answer to Problem 19.50P

Distance $d=227mm$

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 19.2, Problem 19.50P , additional homework tip 1

Mass of collar $mC=1 Κg$

Mass of rod $mR=3 Κg$

Length of rod $L=750 mm$

The forces corresponding to the rod and collar are shown in the free body diagram below:

Vector Mechanics For Engineers, Chapter 19.2, Problem 19.50P , additional homework tip 2

$α=θ¨ (a¯t)R=L2α=L2θ¨(at)C=dα=dθ¨$

Now, from the equation of motion taking moment about A,

$∑MA=∑(MA)eff$

$−WRL2sinθ−WCdsinθ=IRα+mRL2(a¯t)R+mCd(at)CIRθ¨+mRL2×L2θ¨+mCd(dθ¨)+mRgL2θ+mCgdθ=0(IR+mR(L2)2+mCd2)θ¨+(mRgL2+mCgd)θ=0$ Since, $sinθ≈θ$ and $W=mg$

And, moment of inertia of rod is $IR=mRL212$

$(mRL212+mRL24+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL2+3mRL212+mCd2)θ¨+(mRgL2+mCgd)θ=0$

$(4mRL212+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL23+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL23+mCd2)θ¨+(mRL2+mCd)gθ=0$

$θ¨+(L2+mCmRd)g(L23+mCmRd2)θ=0$

Compare the above equation with un-damped equation of vibration:

$Mθ¨+ωn2θ=0$

Natural frequency:

$ωn2=(L2+mCmRd)g(L23+mCmRd2)ωn2=(L2+13d)g(L23+13d2)ωn2=(3L+2d6)g(L2+d23)ωn2=(3L2+d6)g×(3L2+d2)ωn2=(3L2+d)g(L2+d2)$

To maximise the frequency, we need to take the derivative with respect to d and set it equal to zero.

$1gd(ωn2)d(d)=(L2+d2)(1)−(3L2+d)(2d)(L2+d2)2=0(L2+d2)(1)−(3L2+d)(2d)=0L2+d2−3L2×2d−2d×d=0L2+d2−3Ld−2d2=0L2−3Ld−d2=0(0.75)2−3(0.75)d−d2=00.5625−2.25d−d2=0 or,d2+2.25d−0.5625=0$

By solving the above equation we get,

$d=0.2270 or −2.477$

Conclusion:

The distance $d=227mm.$

To determine

(b)

The period of oscillation.

Expert Solution

Answer to Problem 19.50P

Period of vibration, $τn=1.3512 s$

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 19.2, Problem 19.50P , additional homework tip 3

Mass of collar $mC=1 Κg$

Mass of rod $mR=3 Κg$

Length of rod $L=750 mm$

The forces corresponding to the rod and collar are shown in the free body diagram below:

Vector Mechanics For Engineers, Chapter 19.2, Problem 19.50P , additional homework tip 4

$α=θ¨ (a¯t)R=L2α=L2θ¨(at)C=dα=dθ¨$

Now, from the equation of motion taking moment about A,

$∑MA=∑(MA)eff$

$−WRL2sinθ−WCdsinθ=IRα+mRL2(a¯t)R+mCd(at)CIRθ¨+mRL2×L2θ¨+mCd(dθ¨)+mRgL2θ+mCgdθ=0(IR+mR(L2)2+mCd2)θ¨+(mRgL2+mCgd)θ=0$ Since, $sinθ≈θ$ and $W=mg$

And, moment of inertia of rod is $IR=mRL212$

$(mRL212+mRL24+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL2+3mRL212+mCd2)θ¨+(mRgL2+mCgd)θ=0$

$(4mRL212+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL23+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL23+mCd2)θ¨+(mRL2+mCd)gθ=0$

$θ¨+(L2+mCmRd)g(L23+mCmRd2)θ=0$

Compare the above equation with un-damped equation of vibration;

$Mθ¨+ωn2θ=0$

Natural frequency:

$ωn2=(L2+mCmRd)g(L23+mCmRd2)ωn2=(L2+13d)g(L23+13d2)ωn2=(3L+2d6)g(L2+d23)ωn2=(3L2+d6)g×(3L2+d2)ωn2=(3L2+d)g(L2+d2)$

To maximise the frequency, we need to take the derivative with respect to d and set it equal to zero.

$1gd(ωn2)d(d)=(L2+d2)(1)−(3L2+d)(2d)(L2+d2)2=0(L2+d2)(1)−(3L2+d)(2d)=0L2+d2−3L2×2d−2d×d=0L2+d2−3Ld−2d2=0L2−3Ld−d2=0(0.75)2−3(0.75)d−d2=00.5625−2.25d−d2=0 or,d2+2.25d−0.5625=0$

By solving the above equation we get,

$d=0.2270 or −2.477$

Thus, $d=0.2270 m or d=227mm$

Now, natural frequency:

$ωn2=(3L2+d)g(L2+d2)ωn2=(3(0.75)2+0.2270)9.81((0.75)2+(0.2270)2)ωn2=(1.125+0.2270)9.81(0.5625+0.051529)ωn2=21.6001524ω2=4.6476 rad/s$

Then, Time period $τn=2πωn$

$τn=2×3.144.6476τn=1.3512 s$

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Answer 5

Chapter 19.2, Problem 19.50P

To determine

(a)

Then the distance ‘d’ to maximise the frequency of oscillation when a small initial displacement is given.

Expert Solution

Answer to Problem 19.50P

Distance $d=227mm$

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 19.2, Problem 19.50P , additional homework tip 1

Mass of collar $mC=1 Κg$

Mass of rod $mR=3 Κg$

Length of rod $L=750 mm$

The forces corresponding to the rod and collar are shown in the free body diagram below:

Vector Mechanics For Engineers, Chapter 19.2, Problem 19.50P , additional homework tip 2

$α=θ¨ (a¯t)R=L2α=L2θ¨(at)C=dα=dθ¨$

Now, from the equation of motion taking moment about A,

$∑MA=∑(MA)eff$

$−WRL2sinθ−WCdsinθ=IRα+mRL2(a¯t)R+mCd(at)CIRθ¨+mRL2×L2θ¨+mCd(dθ¨)+mRgL2θ+mCgdθ=0(IR+mR(L2)2+mCd2)θ¨+(mRgL2+mCgd)θ=0$ Since, $sinθ≈θ$ and $W=mg$

And, moment of inertia of rod is $IR=mRL212$

$(mRL212+mRL24+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL2+3mRL212+mCd2)θ¨+(mRgL2+mCgd)θ=0$

$(4mRL212+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL23+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL23+mCd2)θ¨+(mRL2+mCd)gθ=0$

$θ¨+(L2+mCmRd)g(L23+mCmRd2)θ=0$

Compare the above equation with un-damped equation of vibration:

$Mθ¨+ωn2θ=0$

Natural frequency:

$ωn2=(L2+mCmRd)g(L23+mCmRd2)ωn2=(L2+13d)g(L23+13d2)ωn2=(3L+2d6)g(L2+d23)ωn2=(3L2+d6)g×(3L2+d2)ωn2=(3L2+d)g(L2+d2)$

To maximise the frequency, we need to take the derivative with respect to d and set it equal to zero.

$1gd(ωn2)d(d)=(L2+d2)(1)−(3L2+d)(2d)(L2+d2)2=0(L2+d2)(1)−(3L2+d)(2d)=0L2+d2−3L2×2d−2d×d=0L2+d2−3Ld−2d2=0L2−3Ld−d2=0(0.75)2−3(0.75)d−d2=00.5625−2.25d−d2=0 or,d2+2.25d−0.5625=0$

By solving the above equation we get,

$d=0.2270 or −2.477$

Conclusion:

The distance $d=227mm.$

To determine

(b)

The period of oscillation.

Expert Solution

Answer to Problem 19.50P

Period of vibration, $τn=1.3512 s$

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 19.2, Problem 19.50P , additional homework tip 3

Mass of collar $mC=1 Κg$

Mass of rod $mR=3 Κg$

Length of rod $L=750 mm$

The forces corresponding to the rod and collar are shown in the free body diagram below:

Vector Mechanics For Engineers, Chapter 19.2, Problem 19.50P , additional homework tip 4

$α=θ¨ (a¯t)R=L2α=L2θ¨(at)C=dα=dθ¨$

Now, from the equation of motion taking moment about A,

$∑MA=∑(MA)eff$

$−WRL2sinθ−WCdsinθ=IRα+mRL2(a¯t)R+mCd(at)CIRθ¨+mRL2×L2θ¨+mCd(dθ¨)+mRgL2θ+mCgdθ=0(IR+mR(L2)2+mCd2)θ¨+(mRgL2+mCgd)θ=0$ Since, $sinθ≈θ$ and $W=mg$

And, moment of inertia of rod is $IR=mRL212$

$(mRL212+mRL24+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL2+3mRL212+mCd2)θ¨+(mRgL2+mCgd)θ=0$

$(4mRL212+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL23+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL23+mCd2)θ¨+(mRL2+mCd)gθ=0$

$θ¨+(L2+mCmRd)g(L23+mCmRd2)θ=0$

Compare the above equation with un-damped equation of vibration;

$Mθ¨+ωn2θ=0$

Natural frequency:

$ωn2=(L2+mCmRd)g(L23+mCmRd2)ωn2=(L2+13d)g(L23+13d2)ωn2=(3L+2d6)g(L2+d23)ωn2=(3L2+d6)g×(3L2+d2)ωn2=(3L2+d)g(L2+d2)$

To maximise the frequency, we need to take the derivative with respect to d and set it equal to zero.

$1gd(ωn2)d(d)=(L2+d2)(1)−(3L2+d)(2d)(L2+d2)2=0(L2+d2)(1)−(3L2+d)(2d)=0L2+d2−3L2×2d−2d×d=0L2+d2−3Ld−2d2=0L2−3Ld−d2=0(0.75)2−3(0.75)d−d2=00.5625−2.25d−d2=0 or,d2+2.25d−0.5625=0$

By solving the above equation we get,

$d=0.2270 or −2.477$

Thus, $d=0.2270 m or d=227mm$

Now, natural frequency:

$ωn2=(3L2+d)g(L2+d2)ωn2=(3(0.75)2+0.2270)9.81((0.75)2+(0.2270)2)ωn2=(1.125+0.2270)9.81(0.5625+0.051529)ωn2=21.6001524ω2=4.6476 rad/s$

Then, Time period $τn=2πωn$

$τn=2×3.144.6476τn=1.3512 s$

Answer 6

Chapter 19.2, Problem 19.50P

To determine

(a)

Then the distance ‘d’ to maximise the frequency of oscillation when a small initial displacement is given.

Expert Solution

Answer to Problem 19.50P

Distance $d=227mm$

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 19.2, Problem 19.50P , additional homework tip 1

Mass of collar $mC=1 Κg$

Mass of rod $mR=3 Κg$

Length of rod $L=750 mm$

The forces corresponding to the rod and collar are shown in the free body diagram below:

Vector Mechanics For Engineers, Chapter 19.2, Problem 19.50P , additional homework tip 2

$α=θ¨ (a¯t)R=L2α=L2θ¨(at)C=dα=dθ¨$

Now, from the equation of motion taking moment about A,

$∑MA=∑(MA)eff$

$−WRL2sinθ−WCdsinθ=IRα+mRL2(a¯t)R+mCd(at)CIRθ¨+mRL2×L2θ¨+mCd(dθ¨)+mRgL2θ+mCgdθ=0(IR+mR(L2)2+mCd2)θ¨+(mRgL2+mCgd)θ=0$ Since, $sinθ≈θ$ and $W=mg$

And, moment of inertia of rod is $IR=mRL212$

$(mRL212+mRL24+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL2+3mRL212+mCd2)θ¨+(mRgL2+mCgd)θ=0$

$(4mRL212+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL23+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL23+mCd2)θ¨+(mRL2+mCd)gθ=0$

$θ¨+(L2+mCmRd)g(L23+mCmRd2)θ=0$

Compare the above equation with un-damped equation of vibration:

$Mθ¨+ωn2θ=0$

Natural frequency:

$ωn2=(L2+mCmRd)g(L23+mCmRd2)ωn2=(L2+13d)g(L23+13d2)ωn2=(3L+2d6)g(L2+d23)ωn2=(3L2+d6)g×(3L2+d2)ωn2=(3L2+d)g(L2+d2)$

To maximise the frequency, we need to take the derivative with respect to d and set it equal to zero.

$1gd(ωn2)d(d)=(L2+d2)(1)−(3L2+d)(2d)(L2+d2)2=0(L2+d2)(1)−(3L2+d)(2d)=0L2+d2−3L2×2d−2d×d=0L2+d2−3Ld−2d2=0L2−3Ld−d2=0(0.75)2−3(0.75)d−d2=00.5625−2.25d−d2=0 or,d2+2.25d−0.5625=0$

By solving the above equation we get,

$d=0.2270 or −2.477$

Conclusion:

The distance $d=227mm.$

Answer 7

Chapter 19.2, Problem 19.50P

To determine

(a)

Then the distance ‘d’ to maximise the frequency of oscillation when a small initial displacement is given.

Answer 8

Expert Solution

Answer to Problem 19.50P

Distance $d=227mm$

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 19.2, Problem 19.50P , additional homework tip 1

Mass of collar $mC=1 Κg$

Mass of rod $mR=3 Κg$

Length of rod $L=750 mm$

The forces corresponding to the rod and collar are shown in the free body diagram below:

Vector Mechanics For Engineers, Chapter 19.2, Problem 19.50P , additional homework tip 2

$α=θ¨ (a¯t)R=L2α=L2θ¨(at)C=dα=dθ¨$

Now, from the equation of motion taking moment about A,

$∑MA=∑(MA)eff$

$−WRL2sinθ−WCdsinθ=IRα+mRL2(a¯t)R+mCd(at)CIRθ¨+mRL2×L2θ¨+mCd(dθ¨)+mRgL2θ+mCgdθ=0(IR+mR(L2)2+mCd2)θ¨+(mRgL2+mCgd)θ=0$ Since, $sinθ≈θ$ and $W=mg$

And, moment of inertia of rod is $IR=mRL212$

$(mRL212+mRL24+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL2+3mRL212+mCd2)θ¨+(mRgL2+mCgd)θ=0$

$(4mRL212+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL23+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL23+mCd2)θ¨+(mRL2+mCd)gθ=0$

$θ¨+(L2+mCmRd)g(L23+mCmRd2)θ=0$

Compare the above equation with un-damped equation of vibration:

$Mθ¨+ωn2θ=0$

Natural frequency:

$ωn2=(L2+mCmRd)g(L23+mCmRd2)ωn2=(L2+13d)g(L23+13d2)ωn2=(3L+2d6)g(L2+d23)ωn2=(3L2+d6)g×(3L2+d2)ωn2=(3L2+d)g(L2+d2)$

To maximise the frequency, we need to take the derivative with respect to d and set it equal to zero.

$1gd(ωn2)d(d)=(L2+d2)(1)−(3L2+d)(2d)(L2+d2)2=0(L2+d2)(1)−(3L2+d)(2d)=0L2+d2−3L2×2d−2d×d=0L2+d2−3Ld−2d2=0L2−3Ld−d2=0(0.75)2−3(0.75)d−d2=00.5625−2.25d−d2=0 or,d2+2.25d−0.5625=0$

By solving the above equation we get,

$d=0.2270 or −2.477$

Conclusion:

The distance $d=227mm.$

Answer 13

To determine

(b)

The period of oscillation.

Expert Solution

Answer to Problem 19.50P

Period of vibration, $τn=1.3512 s$

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 19.2, Problem 19.50P , additional homework tip 3

Mass of collar $mC=1 Κg$

Mass of rod $mR=3 Κg$

Length of rod $L=750 mm$

The forces corresponding to the rod and collar are shown in the free body diagram below:

Vector Mechanics For Engineers, Chapter 19.2, Problem 19.50P , additional homework tip 4

$α=θ¨ (a¯t)R=L2α=L2θ¨(at)C=dα=dθ¨$

Now, from the equation of motion taking moment about A,

$∑MA=∑(MA)eff$

$−WRL2sinθ−WCdsinθ=IRα+mRL2(a¯t)R+mCd(at)CIRθ¨+mRL2×L2θ¨+mCd(dθ¨)+mRgL2θ+mCgdθ=0(IR+mR(L2)2+mCd2)θ¨+(mRgL2+mCgd)θ=0$ Since, $sinθ≈θ$ and $W=mg$

And, moment of inertia of rod is $IR=mRL212$

$(mRL212+mRL24+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL2+3mRL212+mCd2)θ¨+(mRgL2+mCgd)θ=0$

$(4mRL212+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL23+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL23+mCd2)θ¨+(mRL2+mCd)gθ=0$

$θ¨+(L2+mCmRd)g(L23+mCmRd2)θ=0$

Compare the above equation with un-damped equation of vibration;

$Mθ¨+ωn2θ=0$

Natural frequency:

$ωn2=(L2+mCmRd)g(L23+mCmRd2)ωn2=(L2+13d)g(L23+13d2)ωn2=(3L+2d6)g(L2+d23)ωn2=(3L2+d6)g×(3L2+d2)ωn2=(3L2+d)g(L2+d2)$

To maximise the frequency, we need to take the derivative with respect to d and set it equal to zero.

$1gd(ωn2)d(d)=(L2+d2)(1)−(3L2+d)(2d)(L2+d2)2=0(L2+d2)(1)−(3L2+d)(2d)=0L2+d2−3L2×2d−2d×d=0L2+d2−3Ld−2d2=0L2−3Ld−d2=0(0.75)2−3(0.75)d−d2=00.5625−2.25d−d2=0 or,d2+2.25d−0.5625=0$

By solving the above equation we get,

$d=0.2270 or −2.477$

Thus, $d=0.2270 m or d=227mm$

Now, natural frequency:

$ωn2=(3L2+d)g(L2+d2)ωn2=(3(0.75)2+0.2270)9.81((0.75)2+(0.2270)2)ωn2=(1.125+0.2270)9.81(0.5625+0.051529)ωn2=21.6001524ω2=4.6476 rad/s$

Then, Time period $τn=2πωn$

$τn=2×3.144.6476τn=1.3512 s$

Answer 14

To determine

(b)

The period of oscillation.

Answer 15

Expert Solution

Answer to Problem 19.50P

Period of vibration, $τn=1.3512 s$

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 19.2, Problem 19.50P , additional homework tip 3

Mass of collar $mC=1 Κg$

Mass of rod $mR=3 Κg$

Length of rod $L=750 mm$

The forces corresponding to the rod and collar are shown in the free body diagram below:

Vector Mechanics For Engineers, Chapter 19.2, Problem 19.50P , additional homework tip 4

$α=θ¨ (a¯t)R=L2α=L2θ¨(at)C=dα=dθ¨$

Now, from the equation of motion taking moment about A,

$∑MA=∑(MA)eff$

$−WRL2sinθ−WCdsinθ=IRα+mRL2(a¯t)R+mCd(at)CIRθ¨+mRL2×L2θ¨+mCd(dθ¨)+mRgL2θ+mCgdθ=0(IR+mR(L2)2+mCd2)θ¨+(mRgL2+mCgd)θ=0$ Since, $sinθ≈θ$ and $W=mg$

And, moment of inertia of rod is $IR=mRL212$

$(mRL212+mRL24+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL2+3mRL212+mCd2)θ¨+(mRgL2+mCgd)θ=0$

$(4mRL212+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL23+mCd2)θ¨+(mRgL2+mCgd)θ=0(mRL23+mCd2)θ¨+(mRL2+mCd)gθ=0$

$θ¨+(L2+mCmRd)g(L23+mCmRd2)θ=0$

Compare the above equation with un-damped equation of vibration;

$Mθ¨+ωn2θ=0$

Natural frequency:

$ωn2=(L2+mCmRd)g(L23+mCmRd2)ωn2=(L2+13d)g(L23+13d2)ωn2=(3L+2d6)g(L2+d23)ωn2=(3L2+d6)g×(3L2+d2)ωn2=(3L2+d)g(L2+d2)$

To maximise the frequency, we need to take the derivative with respect to d and set it equal to zero.

$1gd(ωn2)d(d)=(L2+d2)(1)−(3L2+d)(2d)(L2+d2)2=0(L2+d2)(1)−(3L2+d)(2d)=0L2+d2−3L2×2d−2d×d=0L2+d2−3Ld−2d2=0L2−3Ld−d2=0(0.75)2−3(0.75)d−d2=00.5625−2.25d−d2=0 or,d2+2.25d−0.5625=0$