System Dynamics
System Dynamics
3rd Edition
ISBN: 9780073398068
Author: III William J. Palm
Publisher: MCG
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Chapter 3, Problem 3.33P
To determine

(a)

Equation of motion of the roller in terms of its rotational velocity, ω.

Expert Solution
Check Mark

Answer to Problem 3.33P

24.4648ω˙=fcosϕ.

Explanation of Solution

Given:

Roller radius = R

Inertia of roller = mR22

Mass of roller = m

Weight of roller = 800N

Diameter of roller = 0.4m

The roller does not slip.

Concept used:

The motion of this object is defined by its translational motion in the plane and its rotational motion about an axis perpendicular to the plane. Two force equations describe the translational motion, and a moment equation is needed to describe the rotational motion.

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 acceleration of the mass about an axis through a point O.

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

Using Newton’s laws for plane motion,

fx=maGxfy=maGy

Where, fx and fy are the net forces acting on the object in the x and y directions, respectively. The mass center is located at point G. The quantities aGx and aGy are the accelerations of the mass center in the x and y directions relative to the ?xed x - y coordinate system.

Derivation of Equation of motion:

Free body diagram of the roller:

System Dynamics, Chapter 3, Problem 3.33P , additional homework tip  1

Equations describing the translational motion in the x -direction:

fx=maGx

The acceleration, aGx=v˙.

v is the translational velocity.

v=x˙v˙=x¨

fx=maGxfx=mv˙fx=mx¨fcosϕft=mx¨ 1

Equation describing the rotational motion:

Ioω˙=MoIω˙=Rftft=Iω˙R

Substitute ft in equation 1

fcosϕft=mx¨ 1ft=Iω˙RfcosϕIω˙R=mx¨ 2

Assuming the roller does not slip,

v=Rωv˙=Rω˙v˙=x¨x¨=Rω˙

Substituting this in 2 and simplify:

fcosϕIω˙R=mx¨ 2x¨=Rω˙fcosϕIω˙R=mRω˙fcosϕ=mRω˙+Iω˙Rfcosϕ=ω˙mR+IRmR+IRω˙=fcosϕ 3

Inertia of roller, I=mR22

Substituting inertia in equation 3

mR+IRω˙=fcosϕ 3I=mR22mR+mR22Rω˙=fcosϕmR+mR22Rω˙=fcosϕmR+mR22Rω˙=fcosϕmR+mR2ω˙=fcosϕ3mR2ω˙=fcosϕ 4

Substituting the values m=800g=81.5494kg, R=0.42=0.2m in equation 4

3mR2ω˙=fcosϕ 4m=800g=81.5494kg, R=0.42=0.2m3×81.5494×0.22ω˙=fcosϕ24.4648ω˙=fcosϕ.

Conclusion:

Equation of motion of the roller in terms of its rotational velocity, ω:

24.4648ω˙=fcosϕ.

To determine

(b)

Equation of motion of the roller in terms of its displacement, x.

Expert Solution
Check Mark

Answer to Problem 3.33P

122.324x¨=fcosϕ.

Explanation of Solution

Given:

Roller radius = R

Inertia of roller = mR22

Mass of roller = m

Weight of roller = 800N

Diameter of roller = 0.4m

The roller does not slip.

Concept used:

The motion of this object is defined by its translational motion in the plane and its rotational motion about an axis perpendicular to the plane. Two force equations describe the translational motion, and a moment equation is needed to describe the rotational motion.

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 acceleration of the mass about an axis through a point O.

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

Using Newton’s laws for plane motion,

fx=maGxfy=maGy

Where, fx and fy are the net forces acting on the object in the x and y directions, respectively. The mass center is located at point G. The quantities maGx and maGy are the accelerations of the mass center in the x and y directions relative to the fixed x - y coordinate system.

Derivation of Equation of motion:

Free body diagram of the roller:

System Dynamics, Chapter 3, Problem 3.33P , additional homework tip  2

Equations describing the translational motion in the x -direction:

fx=maGx

The acceleration, aGx=v˙.

v is the translational velocity.

v=x˙v˙=x¨

fx=maGxfx=mv˙fx=mx¨fcosϕft=mx¨ 5

Equation describing the rotational motion:

Ioω˙=MoIω˙=Rftft=Iω˙R

Substitute ft in equation 5

fcosϕft=mx¨ 5ft=Iω˙RfcosϕIω˙R=mx¨ 6

Assuming the roller does not slip,

v=Rωv˙=Rω˙v˙=x¨x¨=Rω˙ω˙=x¨R

Substituting this in 6 and simplify:

fcosϕIω˙R=mx¨ 6ω˙=x¨RfcosϕIR×x¨R=mx¨fcosϕ=mx¨+Ix¨R2fcosϕ=x¨m+IR2m+IR2x¨=fcosϕ 7

Inertia of roller, I=mR22

Substituting inertia in equation 7

m+IR2x¨=fcosϕ 7I=mR22m+mR22R2x¨=fcosϕm+mR22R2x¨=fcosϕm+mR22R2x¨=fcosϕm+m2x¨=fcosϕ3m2x¨=fcosϕ 8

Substituting the values m=800g=81.5494kg, R=0.42=0.2m in equation 8

3m2x¨=fcosϕ 8m=800g=81.5494kg, 3×81.54942x¨=fcosϕ122.324x¨=fcosϕ.

Conclusion:

Equation of motion of the roller in terms of its rotational velocity, ω:

122.324x¨=fcosϕ.

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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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