VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS
VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS
12th Edition
ISBN: 9781260265521
Author: BEER
Publisher: MCG
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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
Check Mark

Answer to Problem 19.50P

Distance d=227mm

Explanation of Solution

Given information:

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 19.2, Problem 19.50P , additional homework tip  1

Mass of collar mC=1Κg

Mass of rod mR=3Κg

Length of rod L=750mm

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

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, 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

WRL2sinθWCdsinθ=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, sinθθ 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+d23L2×2d2d×d=0L2+d23Ld2d2=0L23Ldd2=0(0.75)23(0.75)dd2=00.56252.25dd2=0or,d2+2.25d0.5625=0

By solving the above equation we get,

d=0.2270or2.477

Conclusion:

The distance d=227mm.

To determine

(b)

The period of oscillation.

Expert Solution
Check Mark

Answer to Problem 19.50P

Period of vibration, τn=1.3512s

Explanation of Solution

Given information:

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 19.2, Problem 19.50P , additional homework tip  3

Mass of collar mC=1Κg

Mass of rod mR=3Κg

Length of rod L=750mm

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

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, 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

WRL2sinθWCdsinθ=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, sinθθ 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+d23L2×2d2d×d=0L2+d23Ld2d2=0L23Ldd2=0(0.75)23(0.75)dd2=00.56252.25dd2=0or,d2+2.25d0.5625=0

By solving the above equation we get,

d=0.2270or2.477

Thus, d=0.2270mord=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=4.6476rad/s

Then, Time period τn=2πωn

τn=2×3.144.6476τn=1.3512s

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Chapter 19 Solutions

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS

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