System Dynamics
System Dynamics
3rd Edition
ISBN: 9780073398068
Author: III William J. Palm
Publisher: MCG
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Chapter 3, Problem 3.39P
To determine

(a)

The equation of motion of the scale.

Expert Solution
Check Mark

Answer to Problem 3.39P

mcL12+mL22θ¨=mcgL1sinθ+mgL2cosβθ.

Explanation of Solution

Given:

Friction of the pivot and mass of the scale arm are neglected.

Concept used:

For an objects’ planar motion which rotates only about an axis perpendicular to the plane, the equation of motion can be written down using Newton’s Second Law.

Equation of Motion: Ioω˙=Mo

Where Io = Mass moment of Inertia of the body about point O.

ω = Angular velocity of the mass about an axis through a point O.

Mo = Sum of the moments applied to the body about the point O.

Let the angular displacement be θ.

The angular velocity, ω=dθdt=θ˙

ω˙=dωdt=d2θdt2=θ¨

Hence, the equation of motion of this object can be rewritten by substituting, ω˙=θ¨:

Ioθ¨=Mo

Derivation of Equation of motion:

Free body diagram:

System Dynamics, Chapter 3, Problem 3.39P , additional homework tip  1

To find the equation of motion, the required unknowns are IA (Inertia about point A ) and MA (Sum of moments about pointA ).

The mass moment of Inertia, I about a specified reference axis is given as:

I=r2dm

Where r = distance from the reference axis to mass element

Mass moment of Inertia of a rotating pendulum = r2dm=mr2

Inertia of mass, mc, is mcL12

Inertia of mass, m, is mL22

Total Inertia, IA=mcL12+mL22.

Sum of moments about the point, A:

MA=mcgL1sinθ+mgL2cosβθ

IAθ¨=MAIA=mcL12+mL22MA=mcgL1sinθ+mgL2cosβθmcL12+mL22θ¨=mcgL1sinθ+mgL2cosβθ.

Conclusion:

The equation of motion of the scale is mcL12+mL22θ¨=mcgL1sinθ+mgL2cosβθ.

To determine

(b)

The equilibrium relation between weight and angle.

Expert Solution
Check Mark

Answer to Problem 3.39P

mg=mcgL1sinθL2cosβθ.

Explanation of Solution

Given:

The system is in equilibrium.

Concept used:

For an objects’ planar motion which rotates only about an axis perpendicular to the plane, the equation of motion can be written down using Newton’s Second Law.

Equation of Motion: Ioω˙=Mo

Where Io = Mass moment of Inertia of the body about point O.

ω = Angular velocity of the mass about an axis through a point O.

Mo = Sum of the moments applied to the body about the point O.

Let the angular displacement be θ.

The angular velocity, ω=dθdt=θ˙

ω˙=dωdt=d2θdt2=θ¨

Hence, the equation of motion of this object can be rewritten by substituting, ω˙=θ¨:

Ioθ¨=Mo

Derivation of mg at equilibrium:

Free body diagram:

System Dynamics, Chapter 3, Problem 3.39P , additional homework tip  2

To find the equation of motion, the required unknowns are IA (Inertia about point A ) and MA (Sum of moments about point A ).

The mass moment of Inertia, I about a specified reference axis is given as:

I=r2dm

Where r = distance from the reference axis to mass element

Mass moment of Inertia of a rotating pendulum = r2dm=mr2

Inertia of mass, mc, is mcL12

Inertia of mass, m, is mL22

Total Inertia, IA=mcL12+mL22.

Sum of moments about the point, A:

MA=mcgL1sinθ+mgL2cosβθ

IAθ¨=MAIA=mcL12+mL22MA=mcgL1sinθ+mgL2cosβθmcL12+mL22θ¨=mcgL1sinθ+mgL2cosβθ 1

At equilibrium, the angular acceleration, θ¨ is zero.

Substitute the value of θ¨ in equation, 1:

mcL12+mL22θ¨=mcgL1sinθ+mgL2cosβθ 1θ¨=0mcL12+mL22×0=mcgL1sinθ+mgL2cosβθ0=mgL2cosβθmcgL1sinθmgL2cosβθ=mcgL1sinθmg=mcgL1sinθL2cosβθ.

Conclusion:

When system is at equilibrium, the relation between mg and θ is mg=mcgL1sinθL2cosβθ.

To determine

(c)

The value of weightof the object.

Expert Solution
Check Mark

Answer to Problem 3.39P

22. 713N.

Explanation of Solution

Given:

The mass of counterweight, mc=5kg

L1=0.2m

L2=0.15m

β=30°

θ=20°

Concept used:

For an objects’ planar motion which rotates only about an axis perpendicular to the plane, the equation of motion can be written down using Newton’s Second Law.

Equation of Motion: Ioω˙=Mo

Where Io = Mass moment of Inertia of the body about point O.

ω = Angular velocity of the mass about an axis through a point O.

Mo = Sum of the moments applied to the body about the point O.

Let the angular displacement be θ.

The angular velocity, ω=dθdt=θ˙

ω˙=dωdt=d2θdt2=θ¨

Hence, the equation of motion of this object can be rewritten by substituting, ω˙=θ¨:

Ioθ¨=Mo

Derivation of mg at equilibrium:

Free body diagram:

System Dynamics, Chapter 3, Problem 3.39P , additional homework tip  3

To find the equation of motion, the required unknowns are IA (Inertia about point A ) and MA (Sum of moments about point A ).

The mass moment of Inertia, I about a specified reference axis is given as:

I=r2dm

Where r = distance from the reference axis to mass element

Mass moment of Inertia of a rotating pendulum = r2dm=mr2

Inertia of mass, mc, is mcL12

Inertia of mass, m, is mL22

Total Inertia, IA=mcL12+mL22.

Sum of moments about the point, A:

MA=mcgL1sinθ+mgL2cosβθ

IAθ¨=MAIA=mcL12+mL22MA=mcgL1sinθ+mgL2cosβθmcL12+mL22θ¨=mcgL1sinθ+mgL2cosβθ 1

At equilibrium, the angular acceleration, θ¨ is zero.

Substitute the value of θ¨ in equation, 1:

mcL12+mL22θ¨=mcgL1sinθ+mgL2cosβθ 1θ¨=0mcL12+mL22×0=mcgL1sinθ+mgL2cosβθ0=mgL2cosβθmcgL1sinθmgL2cosβθ=mcgL1sinθmg=mcgL1sinθL2cosβθ 2

Substitute the given values in equation 2 to find the weight, mg.

mg=mcgL1sinθL2cosβθ 2mc=5kg,L1=0.2m,L2=0.15m,β=30°,θ=20°mg=5g×0.2sin20°0.15cos30°20°mg=gsin20°0.15cos10°mg=22.713N.

Conclusion:

The weight of the object is 22. 713N.

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Chapter 3 Solutions

System Dynamics

Ch. 3 - The motor in Figure P3.11 lifts the mass mL by...Ch. 3 - Instead of using the system shown in Figure 3.2.6a...Ch. 3 - Consider the cart shown in Figure P3.13. Suppose...Ch. 3 - Consider the cart shown in Figure P3.13. Suppose...Ch. 3 - Consider the spur gears shown in Figure P3.15,...Ch. 3 - Consider the spur gears shown in Figure P3.15,...Ch. 3 - Derive the expression for the equivalent inertia...Ch. 3 - Prob. 3.18PCh. 3 - The geared system shown in Figure P3.19 represents...Ch. 3 - Prob. 3.20PCh. 3 - Prob. 3.21PCh. 3 - Prob. 3.22PCh. 3 - For the geared system shown in Figure P3.23,...Ch. 3 - For the geared system discussed in Problem 3.23,...Ch. 3 - The geared system shown in Figure P3.25 is similar...Ch. 3 - Consider the rack-and-pinion gear shown in Figure...Ch. 3 - The lead screw (also called a power screw or a...Ch. 3 - Prob. 3.29PCh. 3 - Derive the equation of motion of the block of mass...Ch. 3 - Assume the cylinder in Figure P3.31 rolls without...Ch. 3 - Prob. 3.33PCh. 3 - Prob. 3.34PCh. 3 - A slender rod 1.4 m long and of mass 20 kg is...Ch. 3 - Prob. 3.36PCh. 3 - Prob. 3.37PCh. 3 - The pendulum shown in Figure P3.38 consists of a...Ch. 3 - Prob. 3.39PCh. 3 - A single link of a robot arm is shown in Figure...Ch. 3 - 3.41 It is required to determine the maximum...Ch. 3 - Figure P3.42 illustrates a pendulum with a base...Ch. 3 - Figure P3.43 illustrates a pendulum with a base...Ch. 3 - 3.44 The overhead trolley shown in Figure P3.44 is...Ch. 3 - Prob. 3.45PCh. 3 - The “sky crane” shown on the text cover was a...
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