VECTOR MECHANICS FOR ENGINEERS W/CON >B
VECTOR MECHANICS FOR ENGINEERS W/CON >B
12th Edition
ISBN: 9781260804638
Author: BEER
Publisher: MCG
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Chapter 18.2, Problem 18.104P

A 2.5-kg homogeneous disk of radius 80 mm rotates at the constant rate ω1 = 50 rad/s with respect to arm ABC, which is welded to a shaft DCE. Knowing that at the instant shown, shaft DCE has an angular velocity ω2 = (12 rad/s)k and an angular acceleration α2 = (8 rad/s2)k, determine (a) the couple that must be applied to shaft DCE to produce that acceleration, (b) the corresponding dynamic reactions at D and E.

Chapter 18.2, Problem 18.104P, A 2.5-kg homogeneous disk of radius 80 mm rotates at the constant rate 1 = 50 rad/s with respect to

Fig. P18.103 and P18.104

(a)

Expert Solution
Check Mark
To determine

The couple which must be applied to shaft DCE to produce that acceleration.

Answer to Problem 18.104P

The couple which must be applied to shaft DCE to produce that acceleration is (0.392Nm)k_.

Explanation of Solution

Given information:

The mass (m) of the disk is 2.5kg.

The radius (r) of the disk Ais 80 mm.

The angular velocity (ω2) of the shaft DCE is12 rad/s.

The angular acceleration (α2) of the shaft DCE is (8rad/s2)k.

Calculation:

The angular velocity (ωx) of disk A along the x-axis is zero.

Write the equation of angular velocity of disk A(ωy) along the y-axis:

ωy=ω1

Write the equation of angular velocity (ωz) of disk A along the z-axis:

ωz=ω2

Find the equation of angular velocity (ω) of disk.

ω=ωxi+ωyj+ωzk

Substitute 0 for ωx, ω1 for ωy, and ω2 for ωz.

ω=(0)i+(ω1)j+(ω2)k=ω1j+ω2k

Find the equation of angular momentum about A (HA) about A.

HA=I¯xωxi+I¯yωyj+I¯zωzk

Here, I¯x is the moment of inertia in the x direction, I¯y is the moment of inertia in the y direction, and I¯z is the moment of inertia in the z direction.

Substitute 0 for ωx, ω1 for ωy, and ω2 for ωz.

HA=I¯x(0)i+I¯yω1j+I¯zω2k=I¯yω1j+I¯zω2k (1)

Find the rate of change of angular momentum (H˙A)Axyz about the reference frame.

(H˙A)Axyz=I¯yω˙1j+I¯zω˙2k

Here, ω˙1 is the angular acceleration disk, and ω˙2 is the acceleration of shaft CBD and arm.

Write the equation of vector form of angular velocity (Ω) of the reference frame Axyz.

Ω=ω2k

Write the equation of the rate of change of angular momentum about A(H˙A).

(H˙A)=(H˙A)Axyz+Ω×HA

Substitute I¯yω˙1j+I¯zω˙2k for (H˙A)Axyz, ω2j for Ω and I¯yω˙1j+I¯zω˙2k for H˙A.

H˙A=(I¯yω˙1j+I¯zω˙2k)+ω2k×(I¯yω1j+I¯zω2k)=(I¯yω˙1j+I¯zω˙2k)I¯yω1ω2i+0=I¯yω1ω2i+I¯yω˙1j+I¯zω˙2k (2)

Write the equation mass moment of inertia (I¯y) along y-axis.

I¯y=12mr2

Write the equation mass moment of inertia (I¯z) along z-axis.

I¯z=14mr2

Substitute 12mr2 for I¯y and 14mr2 for I¯z in Equation (2).

H˙A=12mr2ω1ω2i+12mr2ω˙1j+14mr2ω˙2k (3)

Find the position vector (rA/C) of A with respect to C.

rA/C=bicj

Here, b is the horizontal distance and c is the vertical distance.

Write the equation of velocity (vA) of the mass center A of the disk.

vA=ω2k×rA/C

Substitute bicj for rA/C.

vA=ω2k×(bicj)=bω2j+cω2i=cω2i+bω2j

Write the equation of acceleration of the mass center A of the disk.

aA=ω˙2k×rA/C+ω2k×vA

Substitute cω2i+bω2j for vA and bicj for rA/C.

aA=ω˙2k×(bicj)+ω2k×(cω2i+bω2j)=bω˙2j+cω˙2i+cω22jbω22i=(cω˙2bω22)i+(bω˙2+cω22)j

Sketch the free body diagram and kinetic diagram of the system as shown in Figure (1).

VECTOR MECHANICS FOR ENGINEERS W/CON >B, Chapter 18.2, Problem 18.104P

Refer Figure (1),

Apply Newton’s law of motion.

ΣF=ma¯Dxi+Dyj+Exi+Eyj=maA

Substitute (cω˙2bω22)i+(bω˙2+cω22)j for aA.

Dxi+Dyj+Exi+Eyj=m[(cω˙2bω22)i+(bω˙2+cω22)j]=m(cω˙2bω22)i+m(bω˙2+cω22)j (4)

Equate i-vector coefficients in Equation (4).

Dx+Ex=m(cω˙2bω22) (5)

Equate j-vector coefficients in Equation (4).

Dy+Ey=m(bω˙2+cω22) (6)

Find the rate of change of angular momentum about E (H˙E).

H˙E=H˙A+rA/E×maA

Here, rA/E is the position vector of E with respect to A.

Substitute 12mr2ω1ω2i+12mr2ω˙1j+14mr2ω˙2k for H˙A, bicj+lk for rA/E, and (cω˙2bω22)i+(bω˙2+cω22)j for aA.

H˙E=[(12mr2ω1ω2i+12mr2ω˙1j+14mr2ω˙2k)+(bicj+lk)×m((cω˙2bω22)i+(bω˙2+cω22)j)]

Apply matrix multiplication,

H˙E=[12mr2ω1ω2i+12mr2ω˙1j+14mr2ω˙2k+lm(bω˙2+cω22)i+lm(cω˙2bω22)j+bm(bω˙2+cω22)k+cm(cω˙2bω22)k]=[m(12r2ω1ω2blω˙2clω22)i+m(12r2ω˙1+clω˙1blω22)j+m(14r2+b2+c2)ω˙2k] (7)

Take moment about E.

ME=M0k+2lk×(Dxi+Dyj)=2lDyi+2lDxj+M0k= (8)

Here, Cz is the dynamic reaction at C along z-axis, M0 is the couple at O, and Cx is the dynamic reaction at C along x-axis.

The moment at E is equal to the rate of change of angular momentum at E.

Equate Equation (7) and (8).

2lDxj2lDyi+M0k=[m(12r2ω1ω2blω˙2clω22)i+m(12r2ω˙1+clω˙2blω22)j+m(14r2+b2+c2)ω˙2k] (9)

Convert the unit of radius from mm to m.

r=(80mm)(1m1,000mm)=0.08m

Convert the unit of b from mm to m.

b=(120mm)(1m1,000mm)=0.120m

Convert the unit of c from mm to m.

c=(60mm)(1m1,000mm)=0.06m

Convert the unit of l from mm to m.

l=(150mm)(1m1,000mm)=0.15m

Find the couple (M0) which must be applied to shaft DCE to produce that acceleration.

Equate k vector in Equation (9).

M0=m(14r2+b2+c2)ω˙2

Substitute 2.5 kg for m, 0.08 m for r, 0.120 m for b, 0.06 m for c, and 8rad/s2 for ω˙2

M0=(2.5)[14(0.08)2+(0.12)2+(0.06)2](8)=2.5(0.0016+0.0144+0.0036)8=(0.392Nm)k

Thus, the couple which must be applied to shaft DCE to produce that acceleration is (0.392Nm)k_.

(b)

Expert Solution
Check Mark
To determine

Find the corresponding dynamic reactions at D and E.

Answer to Problem 18.104P

The dynamic reaction at D is (21.0N)i+(28.0N)j_.

The dynamic reaction at E is (21.0N)i(4.00N)j_.

Explanation of Solution

Calculation:

Find the component of dynamic reaction (Dx) at D along x-axis.

Equate j vector in Equation (9).

2lDx=m(12r2ω˙1+clω˙2blω22)Dx=m2l(12r2ω˙1+clω˙2blω22) (10)

Substitute 2.5 kg for m, 0.08 m for r, 0.120 m for b, 0.06 m for c, 8rad/s2 for ω˙2, 0.15 m for l, 12rad/s for ω2, 0 for ω˙1,and 12rad/s for ω2.

Dx=2.52(0.15)(12(0.12)2(0)+(0.06)(0.15)(8)(0.12)(0.15)(12)2)=8.3333(0.0722.592)=21N

Find the component of dynamic reaction (Dy) at D along y-axis.

Equate i vector in Equation (9).

2lDy=m(12r2ω1ω2blω˙2clω22)Dy=m2l(12r2ω1ω2+blω˙2+clω22) (11)

Substitute 2.5 kg for m, 0.08 m for r, 0.120 m for b, 0.06 m for c, 8rad/s2 for ω˙2, 0.15 m for l, 50rad/s for ω1, and 12rad/s for ω2.

Dy=2.52(0.15)(12(0.08)2(12)(50)+(0.12)(0.15)(8)+(0.06)(0.15)(12)2)=8.3333(1.92+0.144+1.296)=28N

Find the dynamic reaction at D using the equation:

D=Dxi+Dyj

Substitute 21N for Dx and 28N for Dy.

D=(21N)i+(28N)j

Thus, the dynamic reaction at D is (21.0N)i+(28.0N)j_.

Find the component of dynamic reaction (Ex) at E along x-axis.

Substitute Equation (11) in (5).

m2l(12r2ω˙1+clω˙2blω22)+Ex=m(cω˙2bω22)Ex=mcω˙2mbω22mcω˙22+mbω222(12r2ω˙1)(m2l)=mcω˙22mbω222(m2l)12r2ω˙1=(m2l)(12r2ω˙1+clω˙2blω22)

Substitute 2.5 kg for m, 0.08 m for r, 0.120 m for b, 0.06 m for c,8rad/s2 for ω˙2, 0.15 m for l, 12rad/s for ω2, 0 for ω˙1, and 12rad/s for ω2.

Ex=2.52(0.15)(12(0.12)2(0)+(0.06)(0.15)(8)(0.12)(0.15)(12)2)=8.3333(0.0722.592)=21N

Find the component of dynamic reaction (Ey) at E along y-axis.

Substitute Equation (12) in (6).

m2l(12r2ω1ω2+blω˙2+clω22)+Ey=m(bω˙2+cω22)Ey=mbω˙2+mcω22mbω˙22mcω222(m2l)(12r2ω1ω2)=mbω˙22+mcω222(m2l)(12r2ω1ω2)=(m2l)(12r2ω1ω2+blω˙2+clω22)

Substitute 2.5 kg for m, 0.08 m for r, 0.120 m for b, 0.06 m for c, 8rad/s2 for ω˙2, 0.15 m for l, 50rad/s for ω1, 12rad/s for ω2, and 12rad/s for ω2.

Ey=2.52(0.15)(12(0.08)2(12)(50)+(0.12)(0.15)(8)+(0.06)(0.15)(12)2)=8.3333(1.92+0.144+1.296)=4.00N

Find the dynamic reaction at E using the equation:

E=Exi+Eyj

Substitute 21N for Ex and 4.00N for Ey.

D=(21N)i(4.00N)j

Thus, the dynamic reaction at E is (21.0N)i(4.00N)j_.

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Chapter 18 Solutions

VECTOR MECHANICS FOR ENGINEERS W/CON >B

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