Knowing that at the instant shown bar AB has a constant angular velocity of 4 rad/s clockwise, determine the angular acceleration ( a ) of bar BD , ( b ) of bar DE .

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Answer 1

Textbook Question

Chapter 15.4, Problem 15.132P

Knowing that at the instant shown bar AB has a constant angular velocity of 4 rad/s clockwise, determine the angular acceleration (a) of bar BD, (b) of bar DE.

Expert Solution

To determine

(a)

The angular acceleration of bar BD.

Answer to Problem 15.132P

The angular acceleration of bar BD is 4.55 rad/s2 in anticlockwise direction.

Explanation of Solution

Given information:

The constant angular velocity of bar AB is 4 rad/s in clockwise direction.

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 velocity of point B.

VB=ωAB×rBA ...... (III)

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) ...... (IV)

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 ...... (V)

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 ...... (VI)

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 ...... (VII)

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 ...... (VIII)

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

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

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

VB=|ijk00−4 rad/s−175 mm00|=[{0−(−4 rad/s)×0}i0−{0−(−4 rad/s)×(−175 mm)}j−{0−(0×−175 mm)}k]=0i−(−700 rad⋅mm/s)j=(700 mm/s)j

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

VD=(700 mm/s)j+{(ωDB)k×−(200 mm)j} ...... (X)

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

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

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

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

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

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

Compare the term of j unit vector on both sides of Equation (XII).

(700 mm/s)=−ωDE(275 mm)ωDE=−(700 mm/s)(275 mm)ωDE=2.55 rad/s

Compare the term of i on both sides of Equation (XII) and substitute 2.55 rad/s for ωDE.

ωDB(200 mm)=−ωDE(75 mm)ωDB(200 mm)=−(2.55 rad/s)(75 mm)ωDB=−(2.55 rad/s)×(75 mm)(200 mm)ωDB=−0.956 rad/s

Substitute 0 for aAB, −(175 mm)i for rBA and (4 rad/s) for ω′AB in Equation (VI).

aB=0×−(175 mm)i−{(4 rad/s)}2×{−(175 mm)i}=(16 rad2/s2)×(175 mm)i=(2800 mm/s2)i

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

−(200 mm)j for rDB and −0.956 rad/s for ωDB in Equation (VII).

aD=[{(2800 mm/s2)i}+{(aDBk)×−(200 mm)j}−{(−0.956 rad/s)2×−(200 mm)j}]={(2800 mm/s2)i}+{(aDBk)×−(200 mm)j}+(182.78 mm/s2)j ...... (XIII)

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

aD={(2800 mm/s2)i}+|ijk00aDB0−200 mm0|+(182.78 mm/s2)j=(2800 mm/s2)i+[{0−(−200 mm×aDB)}i−{0−0}j−{0−0}k]+(182.78 mm/s2)j=(2800 mm/s2)i+(200 mm×aDB)i+(182.78 mm/s2)j=(2800 mm/s2+aDB200 mm)i+(182.78 mm/s2)j

Substitute (2800 mm/s2+aDB200 mm)i+(182.78 mm/s2)j for aD, aDEk for aDE

−(275 mm)i+(75 mm)j for rDE and 2.55 rad/s for ωDE in Equation (VIII).

[(2800 mm/s2+aDB200 mm)i+(182.78 mm/s2)j]=[(aDEk)×{−(275 mm)i+(75 mm)j}−(2.55 rad/s)2{−(275 mm)i+(75 mm)j}][(2800 mm/s2+aDB200 mm)i+(182.78 mm/s2)j]=[(aDEk)×{−(275 mm)i+(75 mm)j}+(1788.19 mm/s2)i−(191.25 mm/s2)j] ...... (XIV)

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

[(2800 mm/s2+aDB200 mm)i+(182.78 mm/s2)j]=[|ijk00aDE−275 mm75 mm0|+(1788.19 mm/s2)i−(191.25 mm/s2)j]

[(2800 mm/s2+aDB200 mm)i+(182.78 mm/s2)j]=[{0−aDE×75 mm}i−{0−(aDE×−275 mm)}j−{0−0}k+(1788.19 mm/s2)i−(191.25 mm/s2)j]

[(2800 mm/s2+aDB200 mm)i+(182.78 mm/s2)j]=[−aDE(75 mm)i+(−aDE275 mm)j+(1788.19 mm/s2)i−(191.25 mm/s2)j]

[(2800 mm/s2+aDB200 mm)i+(182.78 mm/s2)j]=[{−aDE(75 mm)+(1788.19 mm/s2)}i+{(−aDE275 mm)−(191.25 mm/s2)}j]. ...... (XV)

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

(182.78 mm/s2)={(−aDE275 mm)−(191.25 mm/s2)}(−aDE275 mm)=(182.78 mm/s2)+(191.25 mm/s2)aDE=−374.03 mm/s2275 mmaDE=−1.36 rad/s2

Compare the term of i on both sides of Equation (XV) and substitute −1.36 rad/s2 for aDE.

(2800 mm/s2+aDB200 mm)=−(−1.36 rad/s2)(75 mm)+(1788.19 mm/s2)(aDB200 mm)=102 mm/s2+1788.19 mm/s2−2800 mm/s2aDB=−909.81 mm/s2200 mmaDB=−4.55 rad/s2

Conclusion:

The angular acceleration of bar BD is 4.55 rad/s2 in anticlockwise direction.

Expert Solution

To determine

(b)

The angular acceleration of bar DE.

Answer to Problem 15.132P

Explanation of Solution

Calculation:

Compare the terms of j on both sides of Equation (XV).

(182.78 mm/s2)={(−aDE275 mm)−(191.25 mm/s2)}(−aDE275 mm)=(182.78 mm/s2)+(191.25 mm/s2)aDE=−374.03 mm/s2275 mmaDE=−1.36 rad/s2

Conclusion:

The angular acceleration of bar DE is 1.36 rad/s2 in anticlockwise direction.

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