The position vector ( r ¯ ) , velocity vector ( r ¯ ˙ ) , acceleration vector ( r ¯ ¨ ) , kinetic energy ( T ) , angular momentum vector ( H O ) , and the time rate of angular momentum vector ( H ˙ O ) .

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Answer 1

Question

Chapter 4.8, Problem 93RP

To determine

The position vector $(r¯)$ , velocity vector $(r¯˙)$ , acceleration vector $(r¯¨)$ , kinetic energy $(T)$ , angular momentum vector $(HO)$ , and the time rate of angular momentum vector $(H˙O)$ .

Expert Solution & Answer

Answer to Problem 93RP

The position vector $(r¯)$ of the system of particles is $d7(i+4j+6k)$ .

The velocity vector $(r¯˙)$ of the system of particles is $v7(4i+2j+6k)$ .

The acceleration vector $(r¯¨)$ of the system of particles is $(F7m)k$ .

The kinetic energy $(T)$ of the system of particles is $13mv2$ .

The angular momentum vector $(HO)$ of the system about the origin is

$mvd[12i+6j+2k]$ .

The time rate of angular momentum vector $(H˙O)$ of the system of particles is $−(Fd)j$ .

Explanation of Solution

Draw the free body diagram of the system f particles is shown in Figure (1).

Engineering Mechanics: Dynamics, Chapter 4.8, Problem 93RP

Write the expression to determine the position vector $(r¯)$ of the system of particles.

$r¯=∑miri∑mi=mArA+mBrB+mCrCmA+mB+mC$ (I)

Here, mass of the particles is $mi$ , position of the articles is $ri$ , masses of the particles in positions are $mA$ , $mB$ , and $mC$ , distance between the particles are $rA$ , $rB$ , and $rC$ .

Write the expression to determine the velocity vector $(r¯˙)$ of the system of particles.

$r¯˙=∑mivi∑mi=mAvA+mBvB+mCvCmA+mB+mC$ (II)

Here, velocity acting on the particles is $vi$ .and velocity of the particles are $vA$ , $vB$ , and $vC$ .

Write the expression to determine the acceleration vector $(r¯¨)$ of the system of particles.

$r¯¨=∑Fi∑mi=FA+FB+FCmA+mB+mC$ (III)

Here, the external forces acting on the particles is $Fi$ , and forces of the particles are $FA$ , $FB$ , and $FC$ .

Write the expression to determine the kinetic energy $(T)$ of the system of particles.

$T=∑12mivi=12(mAvA2+mBvB2+mCvC2)$ (IV)

Write the expression to determine the angular momentum vector $(HO)$ of the system about the origin.

$HO=∑(ri×mivi)=(rA×mAvA+rB×mBvB+rC×mCvC)$ (V)

Write the expression to determine time rate of angular momentum vector $(H˙O)$ of the system of particles.

$H˙O=∑MO=∑ri×Fi=(rA×FA)+(rB×FB)+(rC×Fc)$ (VI)

Conclusion:

Susbtitute $4m$ for $mA$ , $2m$ for $mB$ , $m$ for $mC$ , $(1.5d)i$ for $rA$ , $(2d)j$ for $rB$ , and $(d)i$ for $rC$ in Equation (I).

$r¯=[4m×(1.5d)k]+[2m×((2d)j)]+[m×(d)i]4m+2m+m=m[6dk+4dj+di]7m=di+4dj+6dk7=d7(i+4j+6k)$

Thus, position vector $(r¯)$ of the system of particles is $d7(i+4j+6k)$ .

Susbtitute $m$ for $mA$ , $4m$ for $mB$ , $2m$ for $mC$ , $vi$ for $vA$ , $3vk$ for $vB$ , and $2vj$ for $vC$ in Equation (II).

$r¯˙=[4m×(v)i]+[2m×(3v)k]+[m×(2v)j]4m+2m+m=m[(4v)i+(6v)k+(2v)j]7m=(4v)i+(6v)k+(2v)j7=v7(4i+2j+6k)$

Thus, the velocity vector $(r¯˙)$ of the system of particles is $v7(4i+2j+6k)$ .

Susbtitute $4m$ for $mA$ , $2m$ for $mB$ , $m$ for $mC$ , $0$ for $FA$ , $0$ for $FB$ , and $Fk$ for $FC$ in Equation (III).

$r¯¨=(F)k4m+2m+m=(F7m)k$

Thus, the acceleration vector $(r¯¨)$ of the system of particles is $(F7m)k$ .

Susbtitute $4m$ for $mA$ , $2m$ for $mB$ , $m$ for $mC$ , $vi$ for $vA$ , $3vk$ for $vB$ , and $2vj$ for $vC$ in Equation (IV).

$T=12[(4m×v2)+(2m×(3v)2)+(m×(2v)2)]=12(4mv2+18mv2+4mv2)=12(26mv2)=13mv2$

Thus, the kinetic energy $(T)$ of the system of particles is $13mv2$ .

Susbtitute $4m$ for $mA$ , $2m$ for $mB$ , $m$ for $mC$ , $vi$ for $vA$ , $3vk$ for $vB$ , $2vj$ for $vC$ , $(1.5d)i$ for $rA$ , $(2d)j$ for $rB$ , and $(d)i$ for $rC$ in Equation (V).

$HO=[(1.5d)k+(2d)j+(d)i]×[(4m×vi)+(2m×(3v)k)+(m×2v)j]=[(1.5d)k+(2d)j+(d)i]×m[(4vi)+(6v)k+(2v)j]=6mvdj+12mvdi+2mvdk=mvd[12i+6j+2k]$

Thus, the angular momentum vector $(HO)$ of the system about the origin is $mvd[12i+6j+2k]$

Susbtitute $4m$ for $mA$ , $2m$ for $mB$ , $m$ for $mC$ , $0$ for $FA$ , $0$ for $FB$ , and $Fk$ for $FC$ , $(1.5d)i$ for $rA$ , $(2d)j$ for $rB$ , and $(d)i$ for $rC$ in in Equation (VI).

$H˙O=((1.5d)k×0)+[2dj×0]+((d)i×Fk)=0+0+(−Fd)j=−(Fd)j$

Thus, the time rate of angular momentum vector $(H˙O)$ of the system of particles is $−(Fd)j$ .

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