VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS
VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS
12th Edition
ISBN: 9781260265521
Author: BEER
Publisher: MCG
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Chapter B, Problem B.1P

A thin plate with a mass m is cut in the shape of an equilateral triangle of side a. Determine the mass moment of inertia of the plate with respect to (a) the centroidal axes AA' and BB’, (b) the centroidal axis CC' that is perpendicular to the plate.

Chapter B, Problem B.1P, A thin plate with a mass m is cut in the shape of an equilateral triangle of side a. Determine the

Expert Solution
Check Mark
To determine

(a)

The mass moment of inertia of the plate about AA and BB' axis.

Answer to Problem B.1P

We got the MI as IAA'=IBB'=ma224

Explanation of Solution

Given information:

Mass m

Sides a

Concept used:

Following formula is used-

  1. dm=ρtdA
  2. MI of stripabout its base dIAA'=x2dm

Calculation:

Diagram,

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter B, Problem B.1P , additional homework tip  1 VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter B, Problem B.1P , additional homework tip  2

(a)               (b)

MI is given as below:

dIAA'=2x2dmIAA'=20a/2x2dm=20a/2x2ptdAIAA'=2pt0a/2x2ydx=2pt0a/2x23(0.5ax)dxIAA'=23pt0a/2(0.5ax2x3)dxIAA'=23pt[0.5ax33x44]0a/2=23pt[0.5a(a/2)33(a/2)44]IAA'=23pt[a4192]IAA'=ma224       (m=ptA=pt12a×a3/2=pt34a2)

The MI about BB’ will be same due to symmetry.

Conclusion:

We got the MI as IAA'=IBB'=ma224

Expert Solution
Check Mark
To determine

(b)

Find the mass moment of inertia of the plate about CC' axis.

Answer to Problem B.1P

We got the MI as ICC'=ma212

Explanation of Solution

Given information:

Mass m

Sides a

Concept used:

Following formula is used-

  1. Parallel axis theorem ICC'=IAA'+IBB'

Calculation:

Diagram,

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter B, Problem B.1P , additional homework tip  3 VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter B, Problem B.1P , additional homework tip  4

(a)                 (b)

MI is given as below:

dIAA'=2x2dmIAA'=20a/2x2dm=20a/2x2ptdAIAA'=2pt0a/2x2ydx=2pt0a/2x23(0.5ax)dxIAA'=23pt0a/2(0.5ax2x3)dxIAA'=23pt[0.5ax33x44]0a/2=23pt[0.5a(a/2)33(a/2)44]IAA'=23pt[a4192]IAA'=ma224       (m=ptA=pt12a×a3/2=pt34a2)

The MI about BB’ will be same due to symmetry

MI about CC’

ICC'=IAA'+IBB'ICC'=ma224+ma224ICC'=ma212 

Conclusion:

We got the MI as ICC'=ma212 

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Chapter B Solutions

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS

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