VECTOR MECHANICS FOR ENGINEERS W/CON >B
VECTOR MECHANICS FOR ENGINEERS W/CON >B
12th Edition
ISBN: 9781260804638
Author: BEER
Publisher: MCG
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Chapter 17.2, Problem 17.83P

(a)

To determine

Find the angular velocity of the assembly after the tube has moved to end E.

(a)

Expert Solution
Check Mark

Answer to Problem 17.83P

The angular velocity of the assembly after the tube has moved to end E is 2.54rad/s_.

Explanation of Solution

Given information:

The mass (m) of the tube AB is 1.6 kg.

The initial angular velocity (ω1) of the system is 5 rad/s.

The moment of inertia of the rod and bracket (I¯DCE) about the vertical axis of rotation is.0.30kgm2

The centroidal moment of inertia of the tube about a vertical axis is 0.0025kgm2.

Calculation:

Find the radius of the tube AB (r(G/A)1) at initial position:

r(G/A)1=r(G/C)1=12(125)=62.5mm

Find the radius of the tube AB (r(G/A)2) at final position:

r(G/A)2=r(G/C)2=50062.5=437.5mm

Write the equation of the velocity of the tube (vG)θ at the center G.

(vG)θ=v¯θ=rG/Cω

Consider the principal of impulse and momentum.

Syst Momenta1+ Syst Ext Imp1y2 = Syst Momenta2

Sketch the impulse and momentum diagram of the given system from top view as shown in Figure (1).

VECTOR MECHANICS FOR ENGINEERS W/CON >B, Chapter 17.2, Problem 17.83P

Refer Figure (1).

Find the angular velocity of the assembly after the tube has moved to end E.

Take moment about C.

I¯ABω1+I¯DCEω1+mAB(v¯θ)1(rG/C)1+0=I¯ABω2+I¯DCEω2+mAB(v¯θ)2(rG/C)2I¯ABω1+I¯DCEω1+mAB(v¯θ)1(rG/C)1=I¯ABω2+I¯DCEω2+mAB(v¯θ)2(rG/C)2

Here, (v¯θ)1 is the velocity of the tube at the center G at initial position and (v¯θ)2 is the velocity of the tube at the center G at final position.

Substitute (rG/C)1ω1 for (v¯θ)1 and (rG/C)2ω2 for (v¯θ)2.

I¯ABω1+I¯DCEω1+mAB((rG/C)1ω1)(rG/C)1=I¯ABω2+I¯DCEω2+mAB((rG/C)2ω2)(rG/C)2I¯ABω1+I¯DCEω1+mAB(rG/C)12ω1=I¯ABω2+I¯DCEω2+mAB(rG/C)22ω2(I¯AB+I¯DCE+mAB(rG/C)12)ω1=(I¯AB+I¯DCE+mAB(rG/C)22)ω2

Substitute 0.0025kgm2 for I¯AB, 0.30kgm2 for I¯DCE, 1.6 kg for mAB, 5rad/s for ω1, 62.5mm for r(G/C)1, and 437.5mm for r(G/C)2.

[0.0025+0.30+1.6(62.5mm×1m1,000mm)2](5)=[0.0025+0.30+1.6(437.5mm×1m1,000mm)2]ω2(0.30875)(5)=0.60875ω20.60875ω2=1.54375ω2=1.543750.60875ω2=2.54rad/s

Thus, the angular velocity of the assembly after the tube has moved to end E is 2.54rad/s_.

(b)

To determine

Find the energy lost during the plastic impact at E.

(b)

Expert Solution
Check Mark

Answer to Problem 17.83P

The energy lost during the plastic impact at E.is 1.902J_.

Explanation of Solution

Calculation:

Write the equation of the kinetic energy (T1) of the system at initial position.

T1=12I¯ABω12+12I¯DCEω12+12mAB(v¯θ)1

Substitute (rG/C)1ω1 for (v¯θ)1.

T1=12I¯ABω12+12I¯DCEω12+12mAB((rG/C)1ω1)2=12I¯ABω12+12I¯DCEω12+12mAB(rG/C)12ω12=12(I¯AB+I¯DCE+mAB(rG/C)12)ω12

Substitute 0.0025kgm2 for I¯AB, 0.30kgm2 for I¯DCE, 1.6 kg for mAB, 5rad/s for ω1, and 62.5mm for r(G/C)1.

T1=12[0.0025+0.30+1.6(62.5mm×1m1,000mm)2](5)2=3.859375J

Write the equation of the kinetic energy (T2) of the system at final position.

T2=12I¯ABω22+12I¯DCEω22+12mAB(v¯θ)2

Substitute (rG/C)2ω2 for (v¯θ)2.

T2=12I¯ABω22+12I¯DCEω22+12mAB((rG/C)2ω2)2=12I¯ABω22+12I¯DCEω22+12mAB(rG/C)22ω22=12(I¯AB+I¯DCE+mAB(rG/C)22)ω22

Substitute 0.0025kgm2 for I¯AB, 0.30kgm2 for I¯DCE, 1.6 kg for mAB, 2.54rad/s for ω2, and 437.5mm for r(G/C)2.

T2=12[0.0025+0.30+1.6(437.5mm×1m1,000mm)2](2.54)2=1.9573J

Find the energy lost during plastic impact using the equation:

Energy lost=T1T2

Substitute 3.8593J for T1 and 1.9573J for T2.

Energy lost=3.85931.9573=1.902J

Thus, the speed of the ball at time (t1).is 16.95ft/s_.

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Chapter 17 Solutions

VECTOR MECHANICS FOR ENGINEERS W/CON >B

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