VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS
VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS
11th Edition
ISBN: 9781259633133
Author: BEER
Publisher: MCG
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Chapter 18.2, Problem 18.102P

(a)

To determine

The couple M0.

(a)

Expert Solution
Check Mark

Answer to Problem 18.102P

The couple M0 is (0.212ftlb)j_.

Explanation of Solution

Given information:

The weight (W) of the disk is 6 lb.

The radius (r) of the disk is 3 in..

The angular velocity (ω1) of the disk is 60 rad/s.

The angular velocity of shaft CBD and arm AB (ω2) is 18 rad/s.

The horizontal distance (c) between the center of rod CBD and center of disk is 5 in..

The vertical distance (b) between the center of rod CBD and center of disk is 4 in..

The couple (M0) applied on the shaft is 0.25ftlb.

The time (t) of couple applied is 3 s.

Calculation:

Find the mass (m) of the disk using the equation:

m=Wg

Here, g is the acceleration due to gravity.

Substitute 6 lb for W and 32.2ft/s2 for g.

m=632.2=0.186335lbs2/ft

Write the equation of vector form of angular velocity (Ω) of shaft CBD and AB.

Ω=ω2j

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=ω2

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

ωz=ω1

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

ω=ωxi+ωyj+ωzk

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

ω=(0)i+(ω2)j+(ω1)k=ω2j+ω1k

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

HA=I¯xωxi+I¯yωyj+I¯zωzk

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

HA=I¯x(0)i+I¯yω2j+I¯zω1k=I¯yω2j+I¯zω1k (1)

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

(H˙A)Axyz=I¯yω˙2j+I¯zω˙1k

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

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˙2j+I¯zω˙1k for (H˙A)Axyz, ω2j for Ω, and I¯yω2j+I¯zω1k for H˙A.

H˙A=I¯yω˙2j+I¯zω˙1k+(ω2j)×[I¯yω2j+I¯zω1k]=I¯yω˙2j+I¯zω˙1k+(0+I¯zω1ω2i)=I¯zω1ω2i+I¯yω˙2j+I¯zω˙1k (2)

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

I¯y=14mr2

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

I¯z=12mr2

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

v¯=ω2j×ci=cω2k

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

a¯=ω˙2j×ci+ω2j×v¯

Substitute cω2k for v¯.

a¯=ω˙2j×ci+ω2j×cω2k=cω˙2kcω22i

Find the position vector of D with respect to A.

rAD=ci+bj

Substitute 5 in. for c and 4 in. for b.

rAD=5in.×1ft12in.i+4in.×1ft12in.j=0.41667i+0.33333j

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

H˙D=H˙A+rAD×ma¯

Substitute I¯zω1ω2i+I¯yω˙2j+I¯zω˙1k for H˙A, 0.41667i+0.33333j for rAD, and cω˙2kcω22i for a¯.

H˙D=(I¯zω1ω2i+I¯yω˙2j+I¯zω˙1k)+(0.41667i+0.33333j)×m(cω˙2kcω22i)=(I¯zω1ω2i+I¯yω˙2j+I¯zω˙1k)+0.41667mcω˙2j+00.33333mcω˙2i+0.33333mcω22k=(I¯zω1ω20.33333mcω˙2)i+(I¯y+0.41667mc)ω˙2j+(I¯zω˙1+0.33333mcω22)k

Substitute 12mr2 for I¯z and 14mr2 for I¯y.

H˙D=[(12mr2ω1ω20.33333mcω˙2)i+(14mr2+0.41667mc)ω˙2j+(12mr2ω˙1+0.33333mcω22)k]=[(12r2ω1ω20.33333cω˙2)mi+(14r2+0.41667c)mω˙2j+(12r2ω˙1+0.33333cω22)mk] (3)

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

VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS, Chapter 18.2, Problem 18.102P

Refer Figure (1),

Apply Newton’s law of motion.

ΣF=ma¯Cxi+Cyk+Dxi+Dzk=ma¯

Substitute cω˙2kcω22i for a¯.

Cxi+Cyk+Dxi+Dzk=m(cω˙2kcω22i) (4)

Equate i-vector coefficients in Equation (3).

Cx+Dx=mcω22i (5)

Equate k-vector coefficients in Equation (3).

Cz+Dz=mcω˙2 (6)

Take moment about D.

ΣMD=M0j+2bj×(Cxi+Czk)=M0j2bCxk+2bCzi=2bCzi+M0j2bCxk (7)

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 D is equal to the rate of change of angular momentum at D.

Equate Equation (3) and (7).

2bCzi+M0j2bCxk=[(12r2ω1ω20.33333cω˙2)mi+(14r2+0.41667c)mω˙2j+(12r2ω˙1+0.33333cω22)mk] (8)

Find the angular acceleration (ω2) of shaft CBD and arm AB using the equation:

ω2=(ω2)0+ω˙2t

Substitute 0 for (ω2)0, 18rad/s2 for ω2, and 3 s for t.

18=0+(ω˙2)(3)ω˙2=183ω˙2=6rad/s2

Find the couple M0 using equate the j- vector coefficients in Equation (8).

M0=(14r2+0.41667c)mω˙2

Substitute 3 in. for r, 5 in. for c, 0.186335lbs2/ft for m, and 6rad/s2 for ω˙2.

M0=[14(3in.×1ft12in.)2+0.41667(5in.×1ft12in.)](0.186335)ω˙2=(0.015625+0.17361)(0.186335)(6)=(0.212ftlb)j

Thus, the couple M0 is (0.212ftlb)j_.

(b)

To determine

The dynamic reaction at C and D after the couple has been removed.

(b)

Expert Solution
Check Mark

Answer to Problem 18.102P

The dynamic reaction at C after the couple has been removed is (12.58lb)i+(9.43lb)k_.

The dynamic reaction at D after the couple has been removed is (12.58lb)i(9.43lb)k_.

Explanation of Solution

Calculation:

After the 3 s, the couple (M0) is zero, the acceleration (ω˙2) of shaft CBD and arm is zero.

Find the component of dynamic reaction at C (Cx) along x-axis using equate the k- vector coefficients in Equation (8).

2bCx=(12r2ω˙1+0.33333cω22)m (9)

Substitute 4 in. for b, 3 in. for r, 5 in. for c, 18 rad/s for ω2 and 0.186335lbs2/ft for m.

2(4in.×1ft12in.)Cx=[12(3in.×1ft12in.)2(0)+0.33333(5in.×1ft12in.)(18)2](0.186335)0.6667Cx=(0+45)(0.186335)Cx=8.3850.6667Cx=12.58lb

Find the component of dynamic reaction at D (Dx) along x-axis using equate the k- vector coefficients using Equation (9).

2bDx=(12r2ω˙1+0.33333cω22)m

Substitute 4 in. for b, 3 in. for r, 5 in. for c, 18 rad/s for ω2, and 0.186335lbs2/ft for m.

2(4in.×1ft12in.)Dx=[12(3in.×1ft12in.)2(0)+0.33333(5in.×1ft12in.)(18)2](0.186335)0.6667Dx=(0+45)(0.186335)Dx=8.3850.6667Dx=12.58lb

Find the component of dynamic reaction at C (Cz) along z-axis using equate the i- vector coefficients in Equation (8).

2bCz=(12r2ω1ω20.33333cω˙2)m (10)

Substitute 4 in. for b, 3 in. for r, 60 rad/s for ω1, 18 rad/s for ω2, 5 in. for c, 0 for ω˙2, and 0.186335lbs2/ft for m.

2bCz=(12r2ω1ω20.33333cω˙2)m2(4in.×1ft12in.)Cz=[12(3in.×1ft12in.)2(60)(18)+0.33333(5in.×1ft12in.)(0)](0.186335)0.6667Cz=(33.75+0)(0.186335)Cz=6.28880.6667Cz=9.43lb

Find the component of dynamic reaction at D (Dz) along z-axis using equate the i- vector coefficients in Equation (10).

2bCz=(12r2ω1ω2+0.33333cω˙2)m (10)

Substitute 4 in. for b, 3 in. for r, 60 rad/s for ω1, 18 rad/s for ω2, 5 in. for c, 0 for ω˙2, and 0.186335lbs2/ft for m.

2bDz=(12r2ω1ω2+0.33333cω˙2)m2(4in.×1ft12in.)Dz=[12(3in.×1ft12in.)2(60)(18)0.33333(5in.×1ft12in.)(0)](0.186335)0.6667Dz=(33.75+0)(0.186335)Dz=6.28880.6667Dz=9.43lb

Find the dynamic reactions at C using the equation:

C=Cxi+Czk

Substitute 12.58lb for Cx and 9.43 lb for Cz.

C=(12.58lb)i+(9.43lb)k

Thus, the dynamic reactions at C after the couple has been removed is (12.58lb)i+(9.43lb)k_.

Find the dynamic reactions at D using the equation:

D=Dxi+Dzk

Substitute 12.58lb for Dx and 9.43lb for Dz.

C=(12.58lb)i(9.43lb)k

Thus, the dynamic reactions at D after the couple has been removed is (12.58lb)i(9.43lb)k_.

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Chapter 18 Solutions

VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS

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