Knowing that at the instant shown bar AB has angular velocity of 4 rad/s and an angular acceleration of 2 rad/s 2 , both clockwise, determine the angular acceleration ( a ) of bar BD , ( b ) of bar DE by using the vector approach as is done in Sample Prob. 15.16.

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Textbook Question

Chapter 15.4, Problem 15.134P

Knowing that at the instant shown bar AB has angular velocity of 4 rad/s and an angular acceleration of 2 rad/s², both clockwise, determine the angular acceleration (a) of bar BD, (b) of bar DE by using the vector approach as is done in Sample Prob. 15.16.

Expert Solution

To determine

(a)

The angular acceleration of bar BD.

Answer to Problem 15.134P

Explanation of Solution

Given information:

The angular velocity of bar AB is 4 rad/s and angular acceleration of bar AB is 2 rad/s2 both are clockwise.

Write the expression of position vector of a point.

r=ai+bj+ck ...... (I)

Here, the coordinate in x direction is a, the coordinate in y direction is b, the coordinate in z direction is c, the unit vector in x direction is i, the unit vector in y direction is j and the unit vector in z direction or angular direction is k.

The coordinate of point B with respect to point A are (−175 mm, 0, 0) and the position vector is rBA.

The coordinate of point D with respect to point B are (0, −200 mm, 0) and the position vector is rDB.

The coordinate of point D with respect to point E are (−275 mm, 75 mm, 0) and the position vector is rDE.

Write the expression of angular velocity of AB in vector form.

ωAB=−ω′ABk ...... (II)

Here, the angular velocity of AB is ω′AB and the unit vector in z direction or angular direction is k.

Write the expression of angular acceleration of AB in vector form.

aAB=−a′ABk ...... (III)

Here, the angular acceleration of AB is a′AB and the unit vector in z direction or angular direction is k.

Write the expression of velocity of point B.

VB=ωAB×rBA ...... (IV)

Here, the angular velocity of AB in vector form is ωAB and the position vector is rBA.

Write the expression of velocity of point D when it is considered as a part of link BD.

VD=VB+(ωDB×rDB) ...... (V)

Here, the angular velocity of BD in vector form is ωDB and the position vector is rDB.

Write the expression of velocity of point D when it is considered as a part of a link DE.

VD=ωDErDE ...... (VI)

Here, the angular velocity of DE in vector form is ωDE and the position vector is rDE.

Write the expression of acceleration of point B in vector form.

aB=aBA×rBA−ω′AB2rBA ...... (VII)

Here, acceleration of point B is aBA, the angular velocity of AB in vector form is ω′AB and position vector is rBA.

Write the expression of acceleration of point D when it is considered as a part of the link BD.

aD=aB+aDB×rDB−ωDB2×rDB ...... (VIII)

Here, the acceleration of point B in vector form is aB, the angular acceleration of link BD is aBD, the position vector is rDE and angular velocity of DB in vector form is ωDB.

Write the expression of acceleration of point D when it is considered as a part of the link DE.

aD=aDE×rDE−ωDE2rDE ...... (IX)

Here, the angular acceleration of link DE is aDE position vector is rDE and, the angular velocity of DE in vector form is ωDE.

Calculation:

Substitute rBA for r, −175 mm for a, 0 for b and 0 for c in Equation (I).

rBA=(−175 mm)i+(0)j+(0)k=−(175 mm)i

Substitute rDB for r, 0 for a, −200 mm for b and 0 for c in Equation (I).

rDB=(0)i+(−200 mm)j+(0)k=(−200 mm)j

Substitute rDE for r, −275 mm for a, 75 mm for b and 0 for c in Equation (I).

rDE=(−275 mm)i+(75 mm)j+(0)k=−(275 mm)i+(75 mm)j

Substitute 4 rad/s for ω′AB in Equation (II).

ωAB=−(4 rad/s)k

Substitute 2 rad/s2 for a′AB in Equation (III).

aAB=−(2 rad/s2)k

Substitute −(4 rad/s)k for ωAB and −(175 mm)i for rBA in Equation (IV).

VB={−(4 rad/s)k}×{−(175 mm)i} ...... (X)

Here, the Equation (X) is the cross product of two vectors.

Change Equation (X) in determinant form.

VB=|ijk00−4 rad/s−175 mm00|=[{0−0}i−{0−((−4 rad/s)×−175 mm)}j−{0−0}k]=−(−700 mm⋅rad/s)j=(700 mm/s)j

Substitute (700 mm/s)j for VB, ωDBk for ωDB and (−200 mm)j for rDB in Equation (V).

VD={(700 mm/s)j}+{ωDBk×(−200 mm)j}VD=(700 mm/s)j+{ωDBk×(−200 mm)j} ...... (XI)

Here, the Equation (XI) is the cross product of two vectors.

Change Equation (XI) in determinant form.

VD=(700 mm/s)j+|ijk00ωDB0−200 mm0|=(700 mm/s)j+{0−(−200 mm)×ωDB}i−{0−0}j−{0−0}k=(700 mm/s)j+ωDB(200 mm)i

Substitute (700 mm/s)j+ωDB(200 mm)i for VD, −(275 mm)i+(75 mm)j for rDE and ωDEk for ωDE in Equation (VI).

(700 mm/s)j+ωDB(200 mm)i={ωDEk}×{−(275 mm)i+(75 mm)j} ...... (XII)

Here, the Equation (XII) is the cross -product of two vector, now change it in determinant.

[(700 mm/s)j+ωDB(200 mm)i]=|ijk00ωDE−275 mm75 mm0|[(700 mm/s)j+ωDB(200 mm)i]=[{0−(75 mm)×ωDE}i−{0−(−275 mm)×ωDE}j−{0−0}k][(700 mm/s)j+ωDB(200 mm)i]=−ωDE(75 mm)i−ωDE(275 mm)j ...... (XIII)

Compare the term of j on both sides of Equation (XIII).

(700 mm/s)=−ωDE(275 mm)ωDE=−(700 mm/s)(275 mm)ωDE=−2.545 rad/s

Compare the term of i on both sides of Equation (XIII) and substitute −2.545 rad/s for ωDE.

ωDB(200 mm)=−(−2.545 rad/s)(75 mm)ωDB=(2.545 rad/s)(75 mm)(200 mm)ωDB=190.875 mm⋅rad/s200 mmωDB=0.954 rad/s

Substitute −(2 rad/s2)k for aAB, −(175 mm)i for rBA and (4 rad/s) for ω′AB in Equation (VI).

aB={−(2 rad/s2)k×−(175 mm)i}−(4 rad/s)2{−(175 mm)i}={−(2 rad/s2)k×−(175 mm)i}+(2800 mm/s2)i ...... (XIV)

Here, the Equation (XIV) is the cross product of two vector, now change it in determinant form.

aB=|ijk00−2 rad/s2−175 mm00|+(2800 mm/s2)i=[{0−0}i−{0−(−2 rad/s2)×(−175 mm)}j−{0−0}k+(2800 mm/s2)i]=(350 mm/s2)j+(2800 mm/s2)i

Substitute (350 mm/s2)j+(2800 mm/s2)i for aB, (aDB)k for aDB

(−200 mm)j for rDB and 0.954 rad/s for ωDB in Equation (VIII).

aD=[(350 mm/s2)j+(2800 mm/s2)i+{(aDB)k}×(−200 mm)j−(0.954 rad/s)2×{(−200 mm)j}]=[(2800 mm/s2)i+(350 mm/s2)j+{(aDB)k×(−200 mm)j}+(182.02 mm/s2)j] ...... (XV)

Here, the Equation (XV) is the cross product of two vector, now change it in determinant form.

aD=(2800 mm/s2)i+(350 mm/s2)j+|ijk00aDB0−200 mm0|+(182.02 mm/s2)j=[(2800 mm/s2)i+(350 mm/s2)j+{0−aDB×(−200 mm)}i−{0−0}j−{0−0}k+(182.02 mm/s2)j]=(2800 mm/s2)i+(350 mm/s2)j+(aDB×200 mm)i+(182.02 mm/s2)j={(2800 mm/s2)+(aDB×200 mm)}i+{532.02 mm/s2}j

Substitute {(2800 mm/s2)+(aDB×200 mm)}i+{532.02 mm/s2}j for aD, aDEk for aDE

−(275 mm)i+(75 mm)j for rDE and −2.545 rad/s for ωDE in Equation (IX).

[{(2800 mm/s2)+(aDB×200 mm)}i+{532.02 mm/s2}j]=[(aDEk)×{−(275 mm)i+(75 mm)j}−(−2.545 rad/s)2×{−(275 mm)i+(75 mm)j}][{(2800 mm/s2)+(aDB×200 mm)}i+{532.02 mm/s2}j]=[(aDEk)×{−(275 mm)i+(75 mm)j}+(1781.18 mm/s2)i−(485.78 mm/s2)j] ...... (XVI)

Here, the Equation (XVI) is the cross product of two vector, now change it in determinant form

[{(2800 mm/s2)+(aDB×200 mm)}i+{532.02 mm/s2}j]=[|ijk00aDE−275 mm75 mm0|+{(1781.18 mm/s2)i−(485.78 mm/s2)j}]

[{(2800 mm/s2)+(aDB×200 mm)}i+{532.02 mm/s2}j]=[{0−aDE(75 mm)}i−{0−aDE×(−275 mm)}j−{0−0}k+(1781.18 mm/s2)i−(485.78 mm/s2)j]

[{(2800 mm/s2)+(aDB×200 mm)}i+{532.02 mm/s2}j]={−aDE(75 mm)i−aDE(275 mm)j+(1781.18 mm/s2)i−(485.78 mm/s2)j}

[{(2800 mm/s2)+(aDB×200 mm)}i+{532.02 mm/s2}j]=[{(1781.18 mm/s2)−aDE(75 mm)}i−{aDE(275 mm)+(485.78 mm/s2)}j] ....... (XVII)

Compare the term of j on both sides of Equation (XVII).

{532.02 mm/s2}=−{aDE(275 mm)+(485.78 mm/s2)}−(532.02 mm/s2)=aDE(275 mm)+(485.78 mm/s2)aDE(275 mm)=−(485.78 mm/s2)−(532.02 mm/s2)aDE=−1017.8 mm/s2275 mmaDE=−3.701 rad/s2

Compare the term of i on both sides of Equation (XVII) and substitute −3.701 rad/s2 for aDE.

(2800 mm/s2)+(aDB×200 mm)=(1781.18 mm/s2)−{−3.701 rad/s2}(75 mm)(2800 mm/s2)+(aDB×200 mm)=(1781.18 mm/s2)−(277.575 mm/s2)(aDB×200 mm)=(1503.605 mm/s2)−(2800 mm/s2)aDB=−1296.395 mm/s2200 mm

aDB=−6.482 rad/s2

Conclusion:

The angular acceleration of bar BD is 6.482 rad/s2 in anticlockwise direction.

Expert Solution

To determine

(b)

The angular acceleration of bar DE by using the vector approach.

Answer to Problem 15.134P

The angular acceleration of bar DE by using the vector approach is 3.701 rad/s2 in anticlockwise direction.

Explanation of Solution

Calculation:

Compare the term of j on both sides of Equation (XVII).

{532.02 mm/s2}=−{aDE(275 mm)+(485.78 mm/s2)}−(532.02 mm/s2)=aDE(275 mm)+(485.78 mm/s2)aDE(275 mm)=−(485.78 mm/s2)−(532.02 mm/s2)aDE=−1017.8 mm/s2275 mmaDE=−3.701 rad/s2

Conclusion:

The angular acceleration of bar DE by using the vector approach is 3.701 rad/s2 in anti-clockwise direction.

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