Vector Mechanics for Engineers: Statics and Dynamics
Vector Mechanics for Engineers: Statics and Dynamics
12th Edition
ISBN: 9781259638091
Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self
Publisher: McGraw-Hill Education
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Chapter 16.1, Problem 16.45P

(a)

To determine

Find the angular acceleration of each cylinder (αA,αBandαC).

(a)

Expert Solution
Check Mark

Answer to Problem 16.45P

The angular acceleration of each cylinder (αA,αBandαC) are 102.7rad/s2_, 20.5rad/s2_, and 10.27rad/s2_.

Explanation of Solution

The weight of the cylinder A (WA) is 5lb.

The weight of the cylinder B (WB) is 5lb.

The weight of the cylinder C (WC) is 20lb.

The initial angular velocity of the cylinder A (ω0) is 720rpm.

The coefficient of the kinetic friction (μk) is 0.25.

The radius of the cylinder A (rA) is 4in..

The radius of the cylinder B (rB) is 4in..

The radius of the cylinder C (rC) is 8in..

Calculation:

Consider the acceleration due to gravity (g) as 32.2ft/s2.

Convert the unit of the radius of the cylinder A (rA):

rA=4in.×1ft12in.=13ft

Convert the unit of the radius of the cylinder B (rB):

rB=4in.×1ft12in.=13ft

Convert the unit of the radius of the cylinder C (rC):

rC=8in.×1ft12in.=23ft

Calculate the mass of the cylinder A (mA):

mA=WAg

Substitute 5lb for WA and 32.2ft/s2 for g.

mA=532.2lbs2/ft

Calculate the mass of the cylinder B (mB):

mB=WBg

Substitute 5lb for WB and 32.2ft/s2 for g.

mB=532.2lbs2/ft

Calculate the mass of the cylinder C (mC):

mC=WCg

Substitute 20lb for WB and 32.2ft/s2 for g.

mC=2032.2lbs2/ft

Calculate the mass moment of inertia of the cylinder A (I¯A):

I¯A=12mArA2

Substitute 532.2lbs2/ft for mA and 13ft for rA.

I¯A=12×532.2×(13)2=8.6266×103lbs2ft

Calculate the mass moment of inertia of the cylinder B (I¯B):

I¯B=12mBrB2

Substitute 532.2lbs2/ft for mB and 13ft for rB.

I¯B=12×532.2×(13)2=8.6266×103lbs2ft

Calculate the mass moment of inertia of the cylinder C (I¯C):

I¯C=12mCrC2

Substitute 2032.2lbs2/ft for mC and 23ft for rC.

I¯C=12×2032.2×(23)2=138.0262×103lbs2ft

Calculate the tangential acceleration of contact point between cylinder B and C (at)BC:

(at)BC=rαB=2rαCαB=2αC

Here, αB is the angular acceleration of the cylinder B and αC is the angular acceleration of the cylinder C.

The friction force at the contact point between cylinder A and C (FAC) as follows:

FAC=μkNAC (1)

Show the free body diagram of the cylinder B as in Figure 1.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16.1, Problem 16.45P , additional homework tip  1

Here, FBC is the friction force at the contact point between cylinder B and C, NBC is the normal force at the contact point between cylinder A and C, Bx is the horizontal force of the cylinder B, and By is the vertical force of the cylinder B.

Refer to Figure 1.

Calculate the moment about point B by applying the equation of equilibrium:

MB=IBαFBCrB=I¯BαB

Substitute 13ft for rB, 2αC for αB, and 8.6266×103lbs2ft for I¯B.

(FBC×13)=8.6266×103×2αC13FBC=17.2532×103αCFBC=17.2532×103αC×3FBC=51.7596×103αC (2)

Show the free body diagram of the cylinder C as in Figure 2.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16.1, Problem 16.45P , additional homework tip  2

Here, FAC is the friction force at the contact point between cylinder A and C and NAC is the normal force at the contact point between cylinder A and C.

Refer to Figure 2.

Calculate the moment about point C by applying the equation of equilibrium:

MC=IGα+madFACrCFBCrC=I¯CαC

Substitute 23ft for rC, 51.7596×103αC for FBC, and 138.0262×103lbs2ft for I¯C.

(FAC×23)(51.7596×103αC×23)=138.0262×103αC23FAC=34.5064×103αC+138.0262×103αCFAC=172.5326×103αC×32FAC=258.7989×103αC (3)

Calculate the normal force at the contact point between cylinder A and C (NAC):

Substitute 0.25 for μk and 258.7989×103αC for FAC in Equation (1).

258.7989×103αC=0.25NACNAC=258.7989×103αC0.25NAC=1.0352αC (4)

Consider that the contact point between cylinder A and C is P.

Calculate the components of forces acting along the line CP:

WCsin30°+FBCFACcos60°NACsin60°=0 (5)

Calculate the angular acceleration of the cylinder C (αC):

Substitute 20lb for WC, 51.7596×103αC for FBC, 258.7989×103αC for FAC, and 1.0352αC for NAC in Equation (5).

[20sin30°+51.7596×103αC(258.7989×103αC×cos60°)(1.0352αC×sin60°)]=010974.1493×103αC=0974.1493×103αC=10

αC=10974.1493×103αC=10.2654rad/s2αC10.27rad/s2

Calculate the friction force at the contact point between cylinder A and C (FAC):

Substitute 10.2654rad/s2 for αC in Equation (3).

FAC=258.7989×103×10.2654=2.6567lb

Calculate the friction force at the contact point between cylinder B and C (FBC):

Substitute 10.2654rad/s2 for αC in Equation (2).

FBC=51.7596×103×10.2654=0.53134lb

Calculate the normal force at the contact point between cylinder A and C (NAC):

Substitute 10.2654rad/s2 for αC in Equation (4).

NAC=1.0352×10.2654=10.6267lb

Show the free body diagram of the cylinder A as in Figure 3.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16.1, Problem 16.45P , additional homework tip  3

Here, Ax is the horizontal force of the cylinder A, and Ay is the vertical force of the cylinder A.

Refer to Figure 3.

Consider that the contact point between cylinder B and C is Q.

Calculate the components of forces acting along the line CQ:

NBCWCcos30°+NACcos60°FACsin60°=0 (6)

Calculate the normal force at the contact point between cylinder B and C (NBC):

Substitute 20lb for WC, 10.6267lb for NAC, and 2.6567lb for FAC in Equation (6).

[NBC20cos30°+(10.6267×cos60°)(2.6567×sin60°)]=0NBC14.3079=0NBC=14.3079lb

Calculate the angular acceleration of the cylinder A (αA):

Calculate the moment about point A by applying the equation of equilibrium:

MA=IAαFACrA=I¯AαA

Substitute 13ft for rA, 2.6567lb for FAC, and 8.6266×103lbs2ft for I¯A.

(2.6567×13)=8.6266×103αA2.65673=8.6266×103αAαA=2.65673×8.6266×103αA=102.7rad/s2

Calculate the angular acceleration of the cylinder A (αA):

αB=2αC

Substitute 10.2654rad/s2 for αC Equation (1).

αB=2×10.2654=20.5rad/s2

Hence, the angular acceleration of each cylinder (αA,αBandαC) are 102.7rad/s2_, 20.5rad/s2_, and 10.27rad/s2_.

(b)

To determine

Find the final angular velocity of each disk (ωA,ωBandωC).

(b)

Expert Solution
Check Mark

Answer to Problem 16.45P

The final angular velocity of each disk (ωA,ωBandωC) are 120rpm_, 120rpm_, and 60rpm_.

Explanation of Solution

The weight of the cylinder A (WA) is 5lb.

The weight of the cylinder B (WB) is 5lb.

The weight of the cylinder C (WC) is 20lb.

The initial angular velocity of the cylinder A (ω0) is 720rpm.

The coefficient of the kinetic friction (μk) is 0.25.

The radius of the cylinder A (rA) is 4in..

The radius of the cylinder B (rB) is 4in..

The radius of the cylinder C (rC) is 8in..

Calculation:

Refer to part (a).

The convert the unit of the initial angular velocity of the disk A (ω0):

(ω0)=720rpm×2π60rad/s21rpm=24πrad/s2

Calculate the angular velocity of cylinder A (ωA):

ωA=ω0αAt

Substitute 24πrad/s2 for ω0 and 102.7rad/s2 for αA.

ωA=24π102.7t (7)

Calculate the tangential velocity of cylinder A (vt)AC:

(vt)AC=rAωA

Substitute (24π102.7t) for ωA and 13ft for rA.

(vt)AC=13×(24π102.7t)=8π34.23t

Calculate the angular velocity of cylinder C (ωC):

ωC=αCt

Substitute 10.27rad/s2 for αC.

ωC=10.27t (8)

Calculate the tangential velocity of cylinder C (vt)CA:

(vt)CA=2rAωC

Substitute 10.27t for ωC and 13ft for rA.

(vt)CA=2×13×10.27t=6.846t

Calculate the time taken when tangential velocities are equal:

(vt)AC=(vt)CA

Substitute (8π34.23t) for (vt)AC and 6.846t for (vt)CA.

8π34.23t=6.846t6.846t+34.23t=8π41.076t=8π

t=8π41.076t=0.6119s

Calculate the final angular velocity of the disk A (ωA):

Substitute 0.6119s for t in Equation (7).

ωA=24π(102.7×0.6119)=12.56rad/s×602πrpm1rad/s=120rpm

Calculate the final angular velocity of the disk C (ωC):

Substitute 0.6119s for t in Equation (8).

ωC=(10.27×0.6119)=6.28rad/s×602πrpm1rad/s=60rpm

Calculate the final angular velocity of the disk B (ωB):

ωB=αBt

Substitute 20.5rad/s2 for αB, and 0.6119s for t.

ωB=20.5×0.6119=12.54×602πrpm1rad/s=120rpm

Hence, the final angular velocity of each disk (ωA,ωBandωC) are 120rpm_, 120rpm_, and 60rpm_.

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Chapter 16 Solutions

Vector Mechanics for Engineers: Statics and Dynamics

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