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 16.1, Problem 16.43P

Disk A has a mass mA = 4 kg, a radius rA = 300 mm, and an initial angular velocity ω0 = 300 rpm clockwise. Disk B has a mass mB = 1.6 kg, a radius rB = 180 mm, and is at rest when it is brought into contact with disk A. Knowing that μk = 0.35 between the disks and neglecting bearing friction, determine (a) the angular acceleration of each disk, (b) the reaction at the support C.

Chapter 16.1, Problem 16.43P, Disk A has a mass mA = 4 kg, a radius rA = 300 mm, and an initial angular velocity 0 = 300 rpm

Fig. P16.43 and P16.44

(a)

Expert Solution
Check Mark
To determine

Find the angular acceleration of each disk (αAandαB).

Answer to Problem 16.43P

The angular acceleration of each disk (αAandαB) are 9.16rad/s2_ and 38.2rad/s2_.

Explanation of Solution

The mass of the disk A (mA) is 4kg.

The mass of the disk B (mB) is 1.6kg.

The initial angular velocity of the disk A (ω0) is 300rpm.

The coefficient of the kinetic friction (μk) is 0.35.

The radius of the disk A (rA) is 300mm.

The radius of the disk B (rB) is 180mm.

Calculation:

Consider the acceleration due to gravity (g) as 9.81m/s2.

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

rA=300mm×103m1mm=0.3m

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

rB=180mm×103m1mm=0.18m

Calculate the mass moment of inertia of the disk A (IA):

IA=12mArA2

Substitute 4kg for mA and 0.3m for rA.

IA=12×4×(0.3)2=0.18kgm2

Calculate the mass moment of inertia of the disk B (IB):

IB=12mBrB2

Substitute 1.6kg for mB and 0.18m for rB.

IB=12×1.6×(0.18)2=25.92×103kgm2

Calculate the load of the disk A (WA):

WA=mAg

Substitute 4kg for mA and 9.81m/s2 for g.

WA=4×9.81=39.24N

Calculate the load of the disk B (WB):

WB=mBg

Substitute 1.6kg for mB and 9.81m/s2 for g.

WB=1.6×9.81=15.696N

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

VECTOR MECHANICS FOR ENGINEERS W/CON >B, Chapter 16.1, Problem 16.43P , additional homework tip  1

Here, F is the magnitude of the friction force, RB is the reaction force of the disk B, N is the vertical force of the disk B, and αB is the angular acceleration of the disk B.

Refer to Figure 1.

Calculate the vertical forces by applying the equation of equilibrium:

Sum of vertical forces is equal to 0.

Fy=0NWB=0N=WB

Substitute 15.696N for WB

N=15.696N

Calculate the magnitude of the friction force (F):

F=μkN

Substitute 15.696N for N and 0.35 for μk.

F=0.35×15.696=5.4936N

Calculate the horizontal forces by applying the equation of equilibrium:

Sum of horizontal forces is equal to 0.

Fx=0FRB=0RB=F

Substitute 5.4936N for F

RB=5.4936N

Calculate the angular acceleration of the disk B (αB):

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

MB=IGα+madFrB=I¯BαB

Substitute 5.4936N for F, 0.18m for rA, and 25.92×103kgm2 for I¯A.

(5.4936×0.18)=25.92×103×αBαB=5.4936×0.1825.92×103αB=38.2rad/s2

Show the free body diagram of the disk A as in Figure 2.

VECTOR MECHANICS FOR ENGINEERS W/CON >B, Chapter 16.1, Problem 16.43P , additional homework tip  2

Here, RAx is the horizontal reaction force of the disk A, RAy is the vertical reaction force of the disk A, and αA is the angular acceleration of the disk A.

Refer to Figure 2.

Calculate the horizontal forces by applying the equation of equilibrium:

Sum of horizontal forces is equal to 0.

Fx=0FRAx=0RAx=F

Substitute 5.4936N for F

RAx=5.4936N

Calculate the vertical forces by applying the equation of equilibrium:

Sum of vertical forces is equal to 0.

Fy=0RAyWAN=0RAy=WA+N

Substitute 39.24N for WA and 15.696N for N.

RAy=39.24+15.696=54.936N

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

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

MA=IGα+madFrA=I¯AαA

Substitute 5.4936N for F, 0.3m for rA, and 0.18kgm2 for I¯A.

(5.4936×0.3)=0.18×αAαA=5.4939×0.30.18αA=9.16rad/s2

Hence, the angular acceleration of each disk (αAandαB) are 9.16rad/s2_ and 38.2rad/s2_.

(b)

Expert Solution
Check Mark
To determine

Find the reaction at the support C (CyandMC).

Answer to Problem 16.43P

The reaction at the support C (CyandMC) is 54.9N_ and 2.64Nm_.

Explanation of Solution

The mass of the disk A (mA) is 4kg.

The mass of the disk B (mB) is 1.6kg.

The initial angular velocity of the disk A (ω0) is 300rpm.

The coefficient of the kinetic friction (μk) is 0.35.

The radius of the disk A (rA) is 300mm.

The radius of the disk B (rB) is 180mm.

Calculation:

Refer to part (a).

Show the free body diagram of the support C as in Figure 3.

VECTOR MECHANICS FOR ENGINEERS W/CON >B, Chapter 16.1, Problem 16.43P , additional homework tip  3

Here, Cx is the horizontal force of the support C, Cy is the vertical force of the support C, and MC is the moment of the support C.

Refer to Figure 3.

Calculate the horizontal forces by applying the equation of equilibrium:

Sum of horizontal forces is equal to 0.

Fx=0Cx=0

Calculate the vertical forces by applying the equation of equilibrium:

Sum of vertical forces is equal to 0.

Fy=0Cy54.936=0Cy=54.936N

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

Sum of moments about point C is equal to 0.

MC=0MC[5.4936×(0.3+0.18)]=0MC2.64=0MC=2.64Nm

Substitute 5.4936N for F, 0.18m for rA, and 25.92×103kgm2 for I¯A.

(5.4936×0.18)=25.92×103×αBαB=5.4936×0.1825.92×103αB=38.2rad/s2

Calculate the time required for the disk to come to rest (t):

Substitute 0.08m for rA, 0.06m for rB, 12πrad/s2 for (ωB)0, 12.5rad/s2 for αA, and 33.3rad/s2 for αB in Equation (1).

12.5t×0.08=[12π33.3t]×0.061t=2.26192tt+2t=2.2619

3t=2.2619t=2.26193t=0.75397s

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

ωA=αAt

Substitute 12.5rad/s2 for αA, and 0.75397s for t.

ωA=12.5×0.75397s=9.425rad/s×602πrpm1rad/s=90rpm

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

ωB=(ωB)0αBt

Substitute 12πrad/s2 for (ωB)0, 33.3rad/s2 for αB, and 0.75397s for t.

ωB=12π(33.3×0.75397)=12.5919rad/s×602πrpm1rad/s=120rpm

Hence, the reaction at the support C (CyandMC) is 54.9N_ and 2.64Nm_.

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Chapter 16 Solutions

VECTOR MECHANICS FOR ENGINEERS W/CON >B

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