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 18, Problem 18.147RP

Three 25-lb rotor disks are attached to a shaft that rotates at 720 rpm. Disk A is attached eccentrically so that its mass center is 1 4 in . from the axis of rotation, while disks B and C are attached so that their mass centers coincide with the axis of rotation. Where should 2-lb weights be bolted to disks B and C to balance the system dynamically?

Chapter 18, Problem 18.147RP, Three 25-lb rotor disks are attached to a shaft that rotates at 720 rpm. Disk A is attached

Fig. P18.147

Expert Solution & Answer
Check Mark
To determine

Where should 2 lb weights be bolted to disks B and C to balance the system dynamically.

Answer to Problem 18.147RP

The 2 lb weights be bolted to disks B and C to balance the system dynamically are 614in.(belowshaft)_ and 318in.(aboveshaft)_.

Explanation of Solution

Given Information:

The weight (W) of the disk is 25 lb.

The shaft rotates at 720 rpm.

The mass center of the disk A is 14in.

Calculation:

Draw the free body diagram of the system as shown in Figure (1).

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 18, Problem 18.147RP

The angular velocity of the shaft is constant. So, the rate of change of angular momentum is zero.

Express the effective force (FA) of disk A:

FA=mrAω2j

Here,ω is angular velocity.

Express the effective force (FB) of disk B:

FB=mrBω2j

Express the effective force (FC) of disk C:

FC=mrCω2j

The sum of the effective forces of the discs should be equal to zero for the system to be in dynamic balance.

FA+FB+FC=0

Substitute mrAω2j for FA, mrBω2j for FB, and mrCω2j for FC.

mrAω2j+mrBω2j+mrCω2j=0mrBω2j+mrCω2j=mrAω2jmω2(rB+rC)=mrAω2jrB+rC=mmrA

Substitute 14in. for rA, 25 lb for m, and 2 lb for m.

rB+rC=25lb2lb(14in.)rB+rC=258rB+rC=3.125in. (1)

According to principle of impulse momentum, the total momentum is zero.

Express the moment about O:

a(mrAω2j)+b(mrBω2j)+c(mrCω2j)=0

Here, a is distance between O and A, b is distance between O and B and c distance between O and C .

Substitute 4i for a, 10i for b, and 16i for c.

4i(mrAω2j)+10i(mrBω2j)+16i(mrCω2j)=010i(mrBω2j)+16i(mrCω2j)=4i(mrAω2j)mω2(10rB+16rC)=4(mrAω2)10rB+16rC=4mmrA

Substitute 14in. for rA, 25 lb for m, and 2 lb for m.

10rB+16rC=425lb2lb(14in.)10rB+16rC=12.5in. (2)

Multiply Equation (1) by 10 and add to Equation (2).

10rB10rC+10rB+16rC=12.5in.31.25in.6rC=18.75rC=18.756rC=3.125in.rC=258in.rC=318in.

The 2 lb weight is be placed above the shaft for disk C.

Substitute 3.125in. for rC in Equation (1).

rB3.125in.=3.125in.rB=6.25in.rB=254in.rB=614in.

The 2 lb weight is be placed below the shaft for disk B.

Thus, the 2 lb weights be bolted to disks B and C to balance the system dynamically are 614in.(belowshaft)_ and 318in.(aboveshaft)_.

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Chapter 18 Solutions

Vector Mechanics for Engineers: Statics and Dynamics

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