<LCPO> VECTOR MECH,STAT+DYNAMICS
<LCPO> VECTOR MECH,STAT+DYNAMICS
12th Edition
ISBN: 9781265566296
Author: BEER
Publisher: MCG
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Chapter 17.2, Problem 17.88P
To determine

Find the speed of the rod relative to the tube when x=400mm.

Expert Solution & Answer
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Answer to Problem 17.88P

The speed of the rod relative to the tube when x=400mm is 2.51m/s_.

Explanation of Solution

Given information:

The mass (mR) of the rod AB is 4 kg.

The mass (mT) of the tube CD is 6 kg.

The angular velocity (ω) of the assembly is 5 rad/s.

Calculation:

Consider l is the length of the rod and the tube and O is the point of intersection of the tube and axle.

Write the equation of the centroidal mass moment of inertia (I¯T) of the tube.

I¯T=112mTl2

Write the equation of the centroidal mass moment of inertia (I¯R) of the rod.

I¯R=112mRl2

Write the equation of the tangential equation of the tube using kinematics.

(v¯θ)T=r¯Tω

Substitute l2 for r¯T.

(v¯θ)T=l2ω

Write the equation of the tangential equation of the rod using kinematics.

(v¯θ)R=r¯Rω

Substitute l2+x for r¯R.

(v¯θ)R=(l2+x)ω

Find the equation of angular momentum about point O.

HO=mTr¯T(v¯θ)T+I¯Tω+mRr¯R(v¯θ)R+I¯Rω

Substitute l2 for r¯T, l2ω for (v¯θ)T, 112mTl2 for I¯T, l2+x for r¯R, (l2+x)ω for (v¯θ)R, and 112mRl2 for I¯R.

HO=mTl2(l2ω)+112mTl2ω+mR(l2+x)(l2+x)ω+112mRl2ω=412mTl2ω+mR(l24+2l2x+x2+112l2)ω=13mTl2ω+mR(4l212+lx+x2)ω=[13mTl2+mR(13l2+lx+x2)]ω (1)

Find the equation of kinetic energy.

T=12mT(v¯θ)T2+12I¯Tω2+12mR(v¯θ)R2+12mRvr2+12I¯Rω2

Substitute l2ω for (v¯θ)T, 112mTl2 for I¯T, (l2+x)ω for (v¯θ)R, and 112mRl2 for I¯R.

T=12mT(l2ω)2+12(112mTl2)ω2+12mR(l2+x)2ω2+12mRvr2+12(112mRl2)ω2=12[mTl24ω2+112mTl2ω2+mR(l24+lx+x2)ω2+112mRl2ω2]+12mRvr2=12[13mTl2+mR(13l2+lx+x2)]ω2+12mRvr2 (2)

Substitute Equation (1) in Equation (2),

T=12HOω+12mRvr2 (3)

All the motion in the system is horizontal. Therefore, the potential energy is zero.

Consider the initial position. (x=0).

At the initial position, the initial angular velocity of the system is 5 rad/s and the radial velocity is zero.(ω=ω1=5rad/sandvr=0).

Find the angular momentum at the initial position using equation (1).

(HO)1=[13mTl2+mR(13l2+lx+x2)]ω1

Substitute 0 for x.

(HO)1={13mTl2+mR[13l2+l(0)+(0)2]}ω1=(13mTl2+13mRl2)ω1=13(mT+mR)l2ω1

Substitute 6 kg for mT, 4 kg for mR, 0.8 m for l, and 5 rad/s for ω1.

(HO)1=13(6+4)(0.8)2(5)=10.6667kgm2/s

Find the kinetic energy at initial position using equation (3).

T1=12(HO)1ω1+12mRvr2

Substitute 10.6667kgm2/s for HO, 5 rad/s for ω1, and 0 for vr.

T1=12(10.6667)(5)+12mR(0)2=26.6667J

Consider the final position. (x=0.4m).

Find the angular momentum at the final position using equation (1).

(HO)2=[13mTl2+mR(13l2+lx+x2)]ω1

Substitute 0.4 m for x.

(HO)2={13mTl2+mR[13l2+lx+x2]}ω2

Substitute 6 kg for mT, 0.8 m for l, 4 kg for mR, and 0.4 m for x.

(HO)2={13(6)(0.8)2+4[13(0.8)2+(0.8)(0.4)+(0.4)2]}ω2=4.0533ω2

Find the kinetic energy at final position using equation (3).

T2=12(HO)2ω2+12mRvr2

Substitute 4.0533ω2 for (HO)2 and 4 kg for mR.

T2=12(4.0533ω2)ω2+12(4)vr2=2.02665ω22+2vr2

Consider the conservation of angular momentum.

(HO)1=(HO)2

Substitute 10.6667kgm2/s for (HO)1 and 4.0533ω2 for (HO)2.

10.6667=4.0533ω2ω2=2.6316rad/s

Find the speed of the rod relative to the tube when x=400mm.

Consider the conservation of energy.

T1+V1=T2+V2

Substitute 26.6667J for T1, 0 for V1, 2.02665ω22+2vr2 for T2, and 0 for V2.

26.6667+0=2.02665ω22+2vr2+02.02665ω22+2vr2=26.6667

Substitute 2.6316rad/s for ω2.

2.02665(2.6316)2+2vr2=26.6667vr=2.51m/s

Thus, the speed of the rod relative to the tube when x=400mm is 2.51m/s_.

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Chapter 17 Solutions

<LCPO> VECTOR MECH,STAT+DYNAMICS

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