Given information:
The angular velocity of bar AB is 4 rad/s and angular acceleration of bar Y AB is 1 2 rad/s2.
Write the expression of position vector of a point.
YY1 r=ai+bj+ck ...... (I)
Here, the coordinate in 3 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 (−20 in, −40 in, 0) and the position vector is rBA.
The coordinate of point D with respect to point B are (40 in, 0 , 0) and the position vector is rDB.
The coordinate of point D with respect to point E are (20 in, −25 in, 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, −20 in for a, −40 in for b and 0 for c in Equation (I).
rBA=(−20 in)i+(−40 in)j+(0)k=−(20 in)i−(40 in)j
Substitute rDB for r, 40 in for a, 0 for b and 0 for c in Equation (I).
rDB=(40 in)i+(0)j+(0)k=(40 in)i
Substitute rDE for r, −275 mm for a, 75 mm for b and 0 for c in Equation (I).
rDE=(20 in)i+(−25 in)j+(0)k=(20 in)i−(25 in)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 −(20 in)i−(40 in)j for rBA in Equation (IV).
VB={−(4 rad/s)k}×{−(20 in)i−(40 in)j} ...... (X)
Here, the Equation (X) is the cross product of two vector, now change it in determinant form.
VB=|ijk00−4 rad/s−20 in−40 in0|=[{0−(−40 in×(−4 rad/s))}i−{0−((−20 in)×−4 rad/s)}j−{0−0}k]=(−160 in⋅rad/s)i+(80 in⋅rad/s)j=−(160 in/s)i+(80 in/s)j
Substitute −(160 in/s)i+(80 in/s)j for VB, ωDBk for ωDB and (40 in)i for rDB in Equation (V).
VD={−(160 in/s)i+(80 in/s)j}+{ωDBk×(40 in)i}VD=−(160 in/s)i+(80 in/s)j+{ωDBk×(40in)i} ...... (XI)
Here, the Equation (XI) is the cross product of two vector, now change it in determinant form.
VD=−(160 in/s)i+(80 in/s)j+|ijk00ωDB40 in00|=−(160 in/s)i+(80 in/s)j+{0−0}i−{0−(40 in)ωDB}j−{0−0}k=−(160 in/s)i+(80 in/s)j+ωDB(40 in)j=−(160 in/s)i+{(80 in/s)+ωDB(40 in)}j
Substitute −(160 in/s)i+{(80 in/s)+ωDB(40 in)}j for VD, (20 in)i−(25 in)j for rDE and ωDEk for ωDE in Equation (VI).
−(160 in/s)i+{(80 in/s)+ωDB(40 in)}j={ωDEk}×{(20 in)i−(25 in)j} ...... (XII)
Here, the Equation (XII) is the cross product of two vector, now change it in determinant form.
[−(160 in/s)i+{(80 in/s)+ωDB(40 in)}j]=|ijk00ωDE20 in−25 in0|[−(160 in/s)i+{(80 in/s)+ωDB(40 in)}j]=[{0−(−25 in)×ωDE}i−{0−(ωDE20 in)}j−{0−0}k][−(160 in/s)i+{(80 in/s)+ωDB(40 in)}j]=ωDE(25 in)i+ωDE{20 in}j−0k[−(160 in/s)i+{(80 in/s)+ωDB(40 in)}j]=ωDE(25 in)i+ωDE{20 in}j ...... (XIII)
Compare the term of i on both sides of Equation (XIII).
−(160 in/s)=ωDE(25 in)ωDE=−(160 in/s)(25 in)ωDE=−6.4 rad/s
Compare the term of j on both sides of Equation (XIII) and substitute −6.4 rad/s for ωDE.
{(80 in/s)+ωDB(40 in)}=(−6.4 rad/s){20 in}ωDB(40 in)=−128 in/s−80 in/sωDB=−208 in/s40 inωDB=−5.2 rad/s
Substitute −(2 rad/s2)k for aAB, −(20 in)i−(40 in)j for rBA and (4 rad/s) for ω′AB in Equation (VI).
aB={−(2 rad/s2)k×((−20 in)i−(40 in)j)}−(4 rad/s)2{−(20 in)i−(40 in)j}={−(2 rad/s2)k×((−20 in)i−(40 in)j)}+(320 in/s2)i+(640 in/s2)j ...... (XIV)
Here, the Equation (XIV) is the cross product of two vector, now change it in determinant form.
aB=|ijk00−2 rad/s2−20 in−40 in0|+(320 in/s2)i+(640 in/s2)j=[{0−(−2 rad/s2)×(−40 in)}i−{0−(−2 rad/s2)×(−20 in)}j−{0−0}k+(320 in/s2)i+(640 in/s2)j]=(−80 in⋅rad/s2)i−(−40 in⋅rad/s2)j+(320 in/s2)i+(640 in/s2)j=(240 in/s2)i+(680 in/s2)j
Substitute (240 in/s2)i+(680 in/s2)j for aB, (aDB)k for aDB
(40 in)i for rDB and −5.2 rad/s for ωDB in Equation (VIII).
aD=[{(240 in/s2)i+(680 in/s2)j}+{(aDB)k}×{(40 in)i}−(−5.2 rad/s)2×{(40 in)i}]=(240 in/s2)i+(680 in/s2)j+{(aDB)k×(40 in)i}−(1081.6 in/s2)i ...... (XV)
Here, the Equation (XV) is the cross product of two vector, now change it in determinant form.
aD=(240 in/s2)i+(680 in/s2)j+|ijk00aDB40 in00|−(1081.6 in/s2)i=[(240 in/s2)i+(680 in/s2)j+{0−0}i−{0−(aDB×40 in)}j−{0−0}k−(1081.6 in/s2)i]=(240 in/s2)i+(680 in/s2)j+(aDB×40 in)j−(1081.6 in/s2)i=−(841.6 in/s2)i+{(680 in/s2)+(aDB×40 in)}j
Substitute −(841.6 in/s2)i+{(680 in/s2)+(aDB×40 in)}j for aD, aDEk for aDE
(20 in)i−(25 in)j for rDE and −6.4 rad/s for ωDE in Equation (IX).
[−(841.6 in/s2)i+{(680 in/s2)+(aDB×40 in)}j]=[(aDEk)×{(20 in)i−(25 in)j}−(−6.4 rad/s)2×{(20 in)i−(25 in)j}][−(841.6 in/s2)i+{(680 in/s2)+(aDB×40 in)}j]=[(aDEk)×{(20 in)i−(25 in)j}−(819.2 in/s2)i+(1024in/s2)j] ...... (XVI)
Here, the Equation (XVI) is the cross product of two vector, now change it in determinant form
[−(841.6 in/s2)i+{(680 in/s2)+(aDB×40 in)}j]=|ijk00aDE20 in−25 in0|−{(819.2 in/s2)i+(1024in/s2)j}
[−(841.6 in/s2)i+{(680 in/s2)+(aDB×40 in)}j]=[{0−(−25 in)×aDE}i−{0−(20 in)×aDE}j−{0−0}k−(819.2 in/s2)i+(1024in/s2)j]
[−(841.6 in/s2)i+{(680 in/s2)+(aDB×40 in)}j]=[aDE(25 in)i+aDE(20 in)j−(819.2 in/s2)i+(1024in/s2)j]
[−(841.6 in/s2)i+{(680 in/s2)+(aDB×40 in)}j]=[{aDE(25 in)−(819.2 in/s2)}i+{aDE(20 in)+(1024in/s2)}j] ....... (XVII)
Compare the term of i on both sides of Equation (XVII).
−(841.6 in/s2)=aDE(25 in)−(819.2 in/s2)aDE(25 in)=−(841.6 in/s2)+(819.2 in/s2)aDE=−22.4 in/s225 inaDE=−0.896 rad/s2
Compare the term of j on both sides of Equation (XVII) and substitute −0.896 rad/s2 for aDE.
(680 in/s2)+(aDB×40 in)=(−0.896 rad/s2)(20 in)+(1024in/s2)(aDB×40 in)=(−17.92 in/s2)+(1024in/s2)−(680 in/s2)aDB=326.08 in/s240 inaDB=8.152 rad/s2
Conclusion:
The angular acceleration of bar BD is 8.152 rad/s2 in clockwise direction.