Given information:
Angular velocity of disk in z-direction is 60 rad/s, radius of the disk is 3 in ,length of BC member is 4 in.
The Figure-(1) shows a schematic diagram.
Figure (1)
Write the equation for the mass of the disk.
m=Wg .........(1)
Here, weight of disk is W and acceleration due to gravity is g.
Write the expression for the angular momentum about point A.
HA=I¯xωx⋅i+I¯yωy⋅j+I¯zωz⋅k .........(2)
Here, mass moment of inertia about the x-axis is I¯x, mass moment of inertia about the y-axis is I¯y, mass moment of inertia about the z-axis is I¯z, angular velocity of disk about x axis is ωx, angular velocity of disk about y axis is ωy .and angular velocity of disk about z axis is ωz.
Write the expression for the angular velocity of disk in x direction.
ωx=0 .........(3)
Substitute 0 for ωx, ω2 for ωy, and ω1 for ωz .in Equation (2).
HA=I¯x×0×⋅i+I¯yω2⋅j+I¯zω1⋅kHA=I¯yω2⋅j+I¯zω1⋅k .........(4)
Here, ω2 is the angular velocity in y-direction and ω1 is the angular velocity in z-direction.
Write the expression for angular velocity in vector form.
Ω=ω2j ..........(5)
Write the expression for rate of angular velocity of the reference frame Axyz.
(H˙A)xyz=I¯yω˙2⋅j+I¯zω˙1⋅k .........(6)
Write the expression for rate of total angular velocity.
H˙A=(H˙)Ax'y'z'+Ω×HA .........(7)
Substitute (I¯yω2⋅j+I¯zω1⋅k) for HA, I¯yω˙2⋅j+I¯zω˙1⋅k for (H˙A)xyz and ω2j for Ω in Equation (7).
H˙A=I¯yω˙2⋅j+I¯zω˙1⋅k+ω2j×(I¯yω2⋅j+I¯zω1⋅k) .........(8)
Write the expression for Matrix multiplication of the vector product for Equation (8).
H˙A=(I¯yω˙2⋅j+I¯zω˙1⋅k)+I¯zω1ω2i=I¯zω1ω2i+I¯yω˙2⋅j+I¯zω˙1⋅k .........(9)
Write the expression for the mass moment of inertia about the y-direction.
I¯y=14mr2 .........(10)
Here mass of the disk is m and radius of the disk is r.
Write the expression for the mass moment of inertia about the z- direction.
I¯z=12mr2 .........(11)
Substitute 14mr2 for I¯y and 12mr2 for I¯z in Equation (9).
H˙A=12mr2×ω1ω2i+14mr2×ω˙2⋅j+12mr2×ω˙1⋅k .........(12)
Write the expression for the velocity of mass centre of the disk.
v¯=ω2j×ci .........(13)
Here, velocity of mass centre is v¯ and distance between B and A is c.
Write the expression for the matrix multiplication of the vector product for Equation (13).
v¯=−cω2k .........(14)
Write the expression for the acceleration of the mass centre of the disk.
a¯=ω˙2j×ci+ω˙2j×v¯ .........(15)
Write the expression for the matrix multiplication of the vector product for Equation (15).
a¯=−cω˙2k−cω22i .........(16)
Write the expression for the the sum of the forces acting on the system.
F=Cxi+Czk+Dxi+Dzk .........(17)
Write the expression for the force in terms of mass and acceleration.
F=ma¯ .........(18)
Substitute Cxi+Czk+Dxi+Dzk for F in Equation (18).
Cxi+Czk+Dxi+Dzk=m a¯ .........(19)
Here, force at C in x-direction is Cx, force at C in z-direction is Cz. force at D in x-direction is Dx, and force at D in z-direction is Dz.
Substitute −cω˙2k−cω22i for a¯ in Equation (19)
Cxi+Czk+Dxi+Dzk=m ×(−cω˙2k−cω22i)Cxi+Czk+Dxi+Dzk=−mcω˙2k−mcω22i .........(20)
Compare the coefficients of the unit vector of i on both side of Equation (20).
Cx+Dx=−mcω22 .........(21)
Compare the coefficients of the unit vector of i on both side of Equation.
Cz+Dz=−mcω˙2 .........(22)
Write the expression for the rate of angular momentum about D.
H˙D=H˙A+rA/D×ma¯ .........(23)
Here, distance between A and D is rA/D.
Write the expression for rA/D in vector form.
rA/D=−ci−bj
Here, distance from the centre of disk to point B is c and distance of BD is b.
Substitute (12mr2×ω1ω2i+14mr2×ω˙2⋅j+12mr2×ω˙1⋅k) for H˙A, (−ci−bj) for rA/D and ω˙2j×ci+ω˙2j×v¯ for a¯ in Equation(23).
H˙D=(12mr2×ω1ω2i+14mr2×ω˙2⋅j+12mr2×ω˙1⋅k)+(−ci−bj)×m(ω˙2j×ci+ω˙2j×v¯) .........(24)
Write the expression for the matrix multiplication for vector product for equation (24).
H˙D=[(12mr2×ω1ω2i+14mr2×ω˙2⋅j+12mr2×ω˙1⋅k)−mbω˙2i+mc2ω˙2j+mbcω22k]H˙D=[m(12r2ω1ω2−bcω˙2)i+m(14r2+c2)ω˙2j+m(12r2ω˙1+bcω22)k] .........(25)
Write the expression for the moment about D
MD=M0j+2bj×(Cxi+Czk) .........(26)
Here, length of BC member is b and moment couple when system is at rest is M0.
Write the expression for the matrix multiplication for the vector product for equation (26).
MD=2bCzi+M0j−2bCxk .........(27)
Here 2b is the length of DC.
Write the given expression for couple when system is at rest.
M0=(0.25 ft⋅lb) .........(28)
The sum of the moment at D is equal to the rate of change of angular momentum at D.
MD=H˙D .........(29)
Substitute 2bCzi+M0j−2bCxk for MD in equation (25).
[2bCzi+M0j−2bCxk]=[m(12r2ω1ω2−bcω˙2)i+m(14r2+c2)ω˙2j+m(12r2ω˙1+bcω22)k] .........(30)
Compare the coefficients of the unit vector of j on both side of Equation (30).
M0=m(14r2+c2)ω˙2 .........(31)
Compare the coefficients of the unit vector of k on both side of Equation (30).
−2bCx=m(12r2ω˙1+bcω22)Cx=m−2b(12r2ω˙1+bcω22) .........(32).
Compare the coefficients of the unit vector of i on both side of Equation (30).
2bCz=m(12r2ω1ω2−bcω˙2)Cz=m2b(12r2ω1ω2−bcω˙2) .........(33)
Substitute m−2b(12r2ω˙1+bcω22) for Cx in Equation (21).
[m−2b(12r2ω˙1+bcω22)+Dx]=−mcω22Dx=m2b(12r2ω˙1+bcω22)−mcω22Dx=mcω222+m2b12r2ω˙1Dx=−(m2b)(−12r2ω˙1+bcω22) .........(34)
Substitute m2b(12r2ω1ω2−bcω˙2) for Cz in Equation (22).
[m2b(12r2ω1ω2−bcω˙2)+Dz]=−mcω˙2Dz=−mcω˙2−m2b(12r2ω1ω2−bcω˙2)Dz=−mcω˙22−(m2b)(12r2ω1ω2)Dz=−(m2b)(12r2ω1ω2+bcω˙2) .........(35)
Write the expression for the angular velocity in terms of time in y-direction.
ω2=(ω2)0+ω˙2t .........(36)
Here time is t.
Calculation:
Substitute 6 lb for W and 32.2 ft/s2 for g in Equation (1).
m=6 lb32.2 ft/s2=0.1863 lb⋅s2ft
Substitute values of 0.1863 lb⋅s2ft for m, (3 in) for r, (5 in) for c and 0.25 ft⋅lb for M0 in Equation (31).
0.25 ft⋅lb=0.1863 lb⋅s2ft(14(3 in)2+(5 in)2)ω˙20.25 ft⋅lb=0.1863 lb⋅s2ft×(14(3 in(1 ft12 in))2+(5 in(1 ft12 in))2)ω˙20.25 ft⋅lb=0.1863 lb⋅s2ft×((0.25 ft)2+(0.4166 ft)2)ω˙2ω˙2=7.089 rad/s2
Substitute 0 for (ω2)0, 7.089 rad/s2 for ω˙2 and 2 s for t in Equation (36).
ω2=0+7.089 rad/s2×2 s=14.179 rad/s
Substitute values of 0.1863 lb⋅s2ft for m, (3 in) for r, (4 in) for b, (5 in) for c, 0 for ω˙1, and 14.179 rad/s for ω2 in Equation (32).
Cx=[0.1863 lb⋅s2ft−2(4 in)×(12(3 in)2(0)+(4 in)(5 in)(14.179 rad/s)2)]Cx=[0.1863 lb⋅s2ft−2(4 in(1 ft12 in))×(12(3 in(1 ft12 in))2(0)+(4 in(1 ft12 in))(5 in(1 ft12 in))(14.179 rad/s)2)]=−0.2795(0+27.9183)=−7.81 lb
Substitute values of 0.1863 lb⋅s2ft for m, (3 in) for r, (4 in) for b, (5 in) for c, 60 rad/s for ω1, 14.179 rad/s for ω2 and 0 for ω˙2 in Equation (33).
Cz=[0.1863 lb⋅s2ft2(4 in)×(12(3 in)2(60 rad/s)(14.179 rad/s)−(4 in)(5 in)×0)]Cz=[0.1863 lb⋅s2ft2(4 in(1 ft12 in))×(12(3 in(1 ft12 in))2(60 rad/s)(14.179 rad/s)−(4 in(1 ft12 in))(5 in(1 ft12 in))×0)]=0.2795(2.7313+0)=7.43 lb
Hence, dynamic reaction at C is −(7.81 lb)i+(7.43 lb)k.
Substitute values of 0.1863 lb⋅s2ft for m, (3 in) for r, (4 in) for b, (5 in) for c, 0 for ω˙1, and 14.179 rad/s for ω2, Equation (34).
Dx=−(0.1863 lb⋅s2ft2(4 in))(−12(3 in)2×0+(4 in)(5 in)(14.179 rad/s)2)Dx=−(0.1863 lb⋅s2ft2(4 in(1 ft12 in)))(−12(3 in(1 ft12 in))2×0+(4 in(1 ft12 in))(5 in(1 ft12 in))(14.179 rad/s)2)Dx=−7.81 lb
Substitute values of 0.1863 lb⋅s2ft for m, (3 in) for r, (4 in) for b, (5 in) for c, 60 rad/s for ω1, 14.179 rad/s for ω2 and 0 for ω˙2 in Equation (35).
Dz=−(0.1863 lb⋅s2ft2(4 in))(12(3 in)2×(60 rad/s)×(14.179 rad/s)+(4 in)(5 in)×0)Dz=−(0.1863 lb⋅s2ft2(4 in(1 ft12 in)))(12(3 in(1 ft12 in))2×(60 rad/s)×(14.179 rad/s)+(4 in(1 ft12 in))(5 in(1 ft12 in))×0)Dz=−7.43 lb
Hence, dynamic reaction at D is −(7.81 lb)i+(7.43 lb)k
Conclusion:
The dynamic reactions at C is −(7.81 lb)i+(7.43 lb)k.
The dynamic reactions at D is −(7.81 lb)i+(7.43 lb)k.