Problem 39:
For the system shown to the right, the disk of mass m rolls without slip and the inner hub has radius ρ/2.
a) Find the equations of motion (in terms of the given parameters—do not substi-tute in numerical values yet);
b) If the applied moment takes the form:
M (t) = (2N · m) sin(4 t), find the steady-state amplitude of the translation of the center of the disk when:
k = 16N/m, b = 2N/(m/s), m = 2kg, ρ = 0.125m c) Determine the steady state amplitude
of the friction force.
b 2 k
M (t) (m, ρ) k
θ x
z
ˆ
ˆı
C G
Solution:
a) We identify the three coordinates x, z, and θ as shown in the figure above. These are related as:
x = −ρ θ, z = 3 2x.
An appropriate free-body diagram for this system is shown to the right. Since the disk is assumed to roll without slip, the equation of motion can be directly obtained with an-gular momentum balance about the contact point C
XMC = IC αD,
−b ˙x ˆı
−2 k z ˆı
−k x ˆı M (t) ˆk
frˆı N ˆ
−m g ˆ
which yields
ρ (k x + b ˙x) +3 ρ
2 (2 k z) + M (t)
ˆk= 3 m ρ2 2 θ ˆ¨k.
Using the above coordinate transformations this equation can be written as 3 m
2 x + b ˙x +¨ 11 k
2 x = −M (t) ρ .
b) For the numerical values given above (with consistent units), this equation reduces to 3 ¨x + 2 ˙x + 88 x = −8 sin(4 t),
from which we can identify the appropriate parameters as:
ωn=r 88
3 , ζ = 1
√264, r =r 6 11.
Therefore, the amplitude of the translational oscillations becomes X = F
kM(r, ζ) = 8 88
1 r
1 − 116
2
+
2√2641 q
6 11
2
= 1
√26.
Likewise, the phase shift of the response is:
tan φ = 2 ζ r
1 − r2 =2√2641 q
6 11
1 − 116
= 1 5,
so that φ = 0.20rad = 11.3◦.
c) In the development of the equation of motion, the friction force was eliminated by summing moments about C. Using linear momentum balance we can reintroduce the friction force as
XF =
fr− k x − 2 k z − b ˙x ˆı +
N − m g
= m ¨x ˆı = m aG.
Therefore, solving for fr we find that
fr= m ¨x + b ˙x + 4 k x.
With x(t) represented as x(t) = X sin(ω t − φ), where X and φ are found above, the friction force becomes
fr(t) =
4 k − m ω2
X sin(ω t − φ) + b ω
X cos(ω t − φ).
The magnitude of the friction force is then found to be
f
= X
r
4 k − m ω22
+ b ω2
.
For these parameter values kfk = 6.47N.
Problem 40:
For the mechanical system shown to the right, the uniform rigid bar has mass m and pinned at point O. For this system:
a) find the equations of motion (in terms of the given parameters—do not substi-tute in numerical values yet);
b) if c = 0.25N/(m/s), k = 32N/m, m = 2kg, and ℓ = 0.25m, find the amplitude of the force transmitted to the ground through combination of the spring and damper when ω = 4 rad/s.
c) if c = 0, m = 2kg, and ℓ = 0.25m, find the value of the stiffness k so that the bar’s amplitude of oscillation is less than π/6rad for all forcing frequencies greater than 20rad/s.
ℓ
a) In addition to θ, we define the additional coordinate z, which measures the deflection at the left end of the bar, with θ and z related as:
z = ℓ 2 θ
A free-body diagram for this system is shown to the right. Applying angular mo-mentum balance on the bar eliminates the appearance of the reaction force and leads to:
Solving for z in terms of θ, the equation of motion becomes:
m ℓ2
12 θ +¨ c ℓ2
4 ˙θ + k ℓ2 4 θ = ℓ
2f (t).
b) In standard form, this equation of motion can be written as:
θ +¨ 3 c
The amplitude of the moment transmitted to the ground can be written as:
The amplitude of the force transmitted to the ground is then FT = MT/(ℓ/2), or:
FT = pk2+ (c ω)2
Substituting in the numerical values given in the problem statement, we find that:
FT = 1.50 N
c) The amplitude of the steady-state vibrations can be written as:
Θ = M0
Substituting in the numerical values given in the problem statement, we find that:
Θ = 8
k −2 ω32 Therefore, if the amplitude of vibration is less than π/6:
8
This inequality has two solutions:
k ≥ 48
π +2 ω2
3 , k ≤ 2 ω2 3 −48
π.
Since this condition must be satisfied for all ω ≥ 20rad/s, we take the second inequality and find that:
k ≤ 2 (20)2 3 −48
π = 251.
Problem 41:
For the system shown to the right, the disk of mass m rolls without slip and x measures the displacement of the disk from the unstretched position of the spring.
a) find the equations of motion;
b) if the forcing takes the form:
f (t) =
f0, 0 ≤ t < t0, f0/2, t0≤ t,
find the response of the system with zero initial conditions.
k 3k
f (t) m
x z
θ ˆı ˆ
Solution:
a) We define the three coordinates as shown as the figure, related as:
x = −r θ, z = −2 r θ, z = 2 x.
A free-body diagram for this system is shown to the right. Notice that the force in the upper spring depends on z, rather than x, while the friction force has an un-known magnitude fr. Because the disk is assumed to roll without slip, we are unable to specify the value of fr, but instead can relate the displacement and rotation of the disk through the coordinate relations above.
−k x ˆı
−3k z ˆı
f (t) ˆı frˆı
G C
The equations of motions can be developed directly with angular momentum balance about the contact point, so that:
XMC = IC θ ˆ¨k,
(3k z) 2r + (k x) r − f(t) r ˆk = 3 m r2 2 θ ˆ¨k.
Finally, writing this equation in terms of a single coordinate, we obtain:
3 m r2 2
θ + 13 k r¨ 2
θ = r f (t).
In standard form:
θ +¨ 26 k 3 m
θ = 2 f (t) 3 m r.
b) We use the convolution integral to determine the response, so that:
θ(t) = Z t
0 F (τ ) h(t − τ) dτ,
and for this system:
Because the forcing function changes abruptly at t = t0, the solution must be written separately for 0 < t ≤ t0, and t > t0:
Evaluating these integrals, the solution becomes:
x(t) = 2 f0
The system shown to the right is subject to base excitation. Find the steady-state re-sponse of the system in terms of z, with:
m = 2.0kg, b = 4.0N/(m/s),
a) We define the addition coordinate x which measures the absolute displacement of the mass with respect to the ground, so that:
x = z + u(t).
Notice that the collection of springs can be replaced by a single equivalent spring, with:
keq= 1
1
2 k1+k12 = 2 k1k2
2 k1+ k2
= 4N/m.
The new equivalent system is shown to the
right. u(t)
z
keq b
m
An appropriate free-body diagram is shown to the right. In terms of the identified coor-dinates, the acceleration of the mass center is:
aG= ¨x ˆ = (¨z + ¨u) ˆ, with ¨u(t) = −(u0ω2) sin(ω t).
−keqz ˆ −b ˙z ˆ
Therefore, linear momentum balance on the mass yields:
XF = m aG,
− keqz − b ˙z
ˆ = m ¨x ˆ
Writing this in terms of z, the equation of motion is:
m ¨z + b ˙z + keqz = −m ¨u(t),
and in standard form:
¨
z + 2 ζ ωn ˙z + ωn2 z = u0 ω2 sin(ω t),
with:
ωn =r keq
m, ζ = b
2pkeq m.
Therefore the steady state response of this system becomes:
z(t) = Z sin(ω t − ψ),
with Z = u0Λ(r, ζ), and:
Λ = r2
p(1 − r2)2+ (2 ζ r)2, tan ψ = 2 ζ r
1 − r2, and r = ω ωn
For the numerical values of this problem:
ωn =√
2, ζ = 1
√2, r =√ 2,
so that:
Λ = 2
√5, tan ψ = 2
−1
Recall that the phase shift ψ must be positive, so that is tan ψ is negative, then ψ is in the second quadrant, so that ψ = 3.03rad. Finally:
z(t) = 1
√5 sin(2 t − 2.03).
Problem 43:
The disk shown in the figure rolls without slip and is subject to a time-varying moment M (t) = sin(t).
a) Find the governing equations of mo-tion;
b) Find the frequency of oscillation for the unforced response, i.e. M (t) = 0;
c) What is the steady-state amplitude of the forced response?
d) Determine the amplitude of the fric-tional force during the steady-state mo-tion.
(m, r) G
M (t) ˆk x
k b
g
Solution:
a) We assume that (ˆı, ˆ, ˆk) represent the standard orthonormal basis, while the transla-tional displacement of G is xˆı and the angular displacement is θˆk. Linear and angular momentum balance on the disk yield:
m¨x = −kx − b ˙x − fµ, mr2
2 θ¨ = −fµr − M(t),
where fµ is the unknown frictional force. If the disk rolls without slipping, we find the kinematical relation x = −rθ, and, eliminating fµ from the above balance laws, we find the governing equation of motion can be expressed as:
3m
2 x + b ˙x + kx =¨ M (t) r .
b) For M (t) = 0, the frequency of oscillation is the damped natural frequency ωd = ωnp1 − ζ2, where:
ωn=r 2k 3m =r 2
3, ζ = b
√6km = 1
√2,
so that the damped natural frequency is ωd=√1
3 = 0.577.
c) In nondimensional form, the system is represented as:
¨
x + 2ζωn˙x + ω2nx = F0sin(t),
where ζ and ωn are given above, and:
F0= 2 3m= 2
3.
Recall that the magnification factor for a harmonically driven system is:
M = 1
p(1 − r2)2+ (2ζr)2,
where r = ωωn is the frequency ratio. For this system we find r =q
3
2, so that M =√213. As a result, the steady-state amplitude is:
A = F0
From the expression for angular momentum balance, we find:
fµ = m
13, so that the amplitude of the frictional force is:
|fµ| = 4
√13.
Problem 44:
In the figure, the disk has mass m, radius r, and moment of inertia IG = mr22 about the mass center G, subject to an applied moment of the form:
M(t) = M0sin(ωt)ˆk.
If the disk rolls without slip (µ is sufficiently large):
a) find the equations of motion for this system;
b) for m = 2kg, b = 0.5(N · s)/m, and k = 8N/m, find the steady-state amplitude of the rotation of the disk?
(m, r)
In addition to x, defined in the figure, we define θ as the angular displacement of the disk from the unstretched position in the ˆk direction. If the disk rolls without slip, x and θ are related as:
x = −rθ.
a) The frictional force, which is unknown, is defined as f = fˆı, while the forces due to the spring and damper are:
Fspring= −kx ˆı, Fdamper= −b ˙x ˆı.
Using linear and angular momentum balance on the disk, we find that:
XF = (f − kx − b ˙x)ˆı = m¨x ˆı = maG, XMG = (M0 sin(ωt) + f r) ˆk = mr2
2 θ ˆ¨k= IGαβ.
Eliminating the unknown frictional force, and using the kinematic constraint, we find the equation of motion is:
3m 2
x + b ˙x + kx = −¨ M0
r sin(ωt).
b) For the given values of the parameters, we find that:
ωn2= 8
3, ζ = 1 8√
6, F0= −M0
3r .
Thus, the steady-state amplitude may be easily found as:
X = F0
ωn2M,
= −M3r0
q 8 3− ω22
+ ω62 .
Problem 45:
The mass m = 2kg is supported by an elas-tic cantilever beam attached to a founda-tion which undergoes harmonic mofounda-tion of the form:
u(t) = 4 sin(ωt)m,
If the beam has length l = 20cm, while AE = 16N:
a) find the equations of motion in terms of z, the relative displacement between the mass and the foundation (assume the beam has zero mass);
b) what is the amplitude of the result-ing motion in terms of the forcresult-ing fre-quency ω?
u(t) = 4 sin(ω t)m z
m
x
ˆ
ˆı
Solution:
a) The equivalent spring for this cantilever beam is:
keq=AE
l = 16N
0.2m= 80N/m.
The acceleration of the block with respect to the ground is aG = (¨u + ¨z)ˆ, so that, in terms of z, the equations of motion become:
m¨z + keqz = −m ¨u,
¨
z + 40 z = 4 ω2 sin(ω t).
b) For this undamped system, the amplitude of the resulting steady-state motion is:
X = 4ω2 40
1 q
1 −ω402
2 ,
= 4ω2
|40 − ω2|.
Problem 46:
The unbalanced rotor shown in the figure is pinned to a frame and supported by a spring and damper. If the total mass is m while the mass center G is located at an eccentricity of ε from the the center of rotation O,
a) find the damped natural frequency;
b) what is the steady-state amplitude of vibration when the rotor spins at this angular speed.
m ε O
G
k b
x ˆ
ˆı
Solution:
We define x(t) as the vertical displacement of the geometric center of the rotor as measured from static equilibrium. As a result, the mass center G is described by the position z(t) = x(t) + ε sin(ωt). Note that ε measures the eccentricity of the mass center, not the location of the mass imbalance. Consequently, the governing equations of motion can be written:
m ¨z = −k x − b ˙x, m ¨x − ε ω2 sin(ω t) = −k x − b ˙x, or in more standard form:
¨ x + b
m˙x + k
mx = εω2sin(ωt).
a) In the above system we find ωn = q
k
m and ζ = 2√bkm, so that the damped natural frequency can be written:
ωd= ωnp1 − ζ2= rk
m− b2 4m2.
b) For an arbitrary forcing frequency the amplitude of oscillation is A = ε Λ, where:
Λ = ω2
p(ωn2− ω2)2+ (2ζωωn)2,
which, with ω = ωd, reduces the amplitude to:
A = ε 1 − ζ2 ζp4 − 3ζ2,
with ζ defined above.
Problem 47:
The system shown in the figure is supported by a foundation that undergoes harmonic mo-tion of the form:
u(t) = 4 sin(ωt)m,
If the mass and stiffness are m = 2kg, and k = 8N/m:
a) find the equations of motion in terms of z, the relative displacement between the mass and the base (assume the spring has zero unstretched length);
b) what is the amplitude of the resulting motion in terms of the frequency ratio r;
c) for what forcing frequencies is the re-sulting amplitude of the steady-state motion Z ≤ 8?
u(t) = 4 sin(ω t)m k z
g m
x
ˆ
ˆı
Solution:
a) We construct a free-body diagram as shown. The only forces acting on the mass arise from the gravitational force and the spring force.
However, the acceleration of the mass with respect to the ground is written as:
aG=
¨ u + ¨z
ˆ.
Therefore, linear momentum balance takes the form:
XF = m aG,
− k z − m g
ˆ = m
¨ z + ¨u
ˆ.
−k z ˆ
−m g ˆ
Substituting in the expression for u(t), and taking the component in the ˆ direction, we obtain the governing equation of motion:
m ¨z + k z = −m g − m ¨u,
¨
z + (4s−2)z = −9.81m/s2+ 4ω2sin(ωt)m,
Notice that each term has units of acceleration, that is, m/s2. In what follows we will dispense with the explicit inclusion of the units. For this system ωn= 2, ζ = 0, and the amplitude of the forcing is F = 4ω2. Thus we have frequency-squared excitation.
b) Using the above values for the natural frequency and the damping ratio, we find that
the forced response can be written as z(t) = Z sin(ωt + φ), where the amplitude Z is:
Z = U · Λ(r, ζ) = U r2
p(1 − r2)2+ (2ζr)2,
= 4r2
|1 − r2|.
c) If Z < 8 then this implies that:
Z < 4r2 p(1 − r2)2.
Defining Zcras:
Zcr= 4r2 p(1 − r2)2, We may solve for r to yield:
r =
sZcr2 ± 4Zcr
Zcr2 − 16 ,
=
(r Zcr
Zcr− 4,
r Zcr
Zcr+ 4 )
.
Thus for Zcr= 8, we find that Z < 8 in the range:
0 < r <r 2
3, or r >√ 2.
Problem 48:
The single-degree-of-freedom system shown is subject to a harmonic force. If the natural frequency is ωn= 4rad/s and m = 1kg:
a) determine the spring and damping con-stants when the system is critically damped;
b) determine the amplitude of the total force transmitted to the ground un-der steady-state oscillations when ω = 1rad/s.
x
m
k b
F (t) = sin(ω t)N
Solution:
The equation of motion for this system takes the form:
¨ x + b
m˙x + k
mx = F (t) m .
a) In terms of the system parameters, the damping ratio and natural frequency can be written as:
ζ = b 2√
km, ωn=r k m.
So, for a critically damped system ζ = 1, and solving for k and b we find:
k = 16N/m, b = 8N · s/m.
b) The amplitude of the total force transmitted to the ground, which is defined as Ft, is Ft= Xωn2p1 + (2ζr)2, where X is the amplitude of the response and r = ωω
n = 14 is the frequency ratio. Thus, for this system:
Ft = F0
A constant force is applied to the undamped single degree-of-freedom system for a dura-tion of t1, at which point it is removed, that is:
F(t) =
F0ˆı, 0 ≤ t < t1, 0, t ≥ t1.
If the system starts with zero initial condi-tions, determine the resulting displacement of the mass x(t). You may either use the convolution integral or you may try to solve this explicitly.
The equation of motion for this system is:
¨ x + 4k
mx = F (t) m , which has an impulse response of the form:
h(t) = sin ωnt mωn
, ωn = 2r k m.
Therefore, using the convolution integral, the response of the system is:
x(t) =
2.2 Unsolved Problems
Problem 50:
For the system shown to the right the block of mass m = 100kg is supported by springs and dampers with coefficients
k1= 1200N/m, k2= 3600N/m b1= 100N/(m/s), b2= 120N/(m/s), and is subject to a time dependent force
F(t) = F0 sin(ω t) ˆ, with F0= 100N and ω = 5rad/s.
a) Find the equations of motion for the displacement of the block;
b) Write these equations as a set of first-order equations (as you would for nu-merical simulation within MATLAB);
c) What is the steady-state response of the forced response;
d) What is the magnitude of the force transmitted to the ground?
k1
k2
b1 b2
m F(t)
Problem 51:
The unbalanced rotor shown in the figure is pinned to a frame and supported by a spring and damper. The total mass is measured as m = 200kg. When the system is operated at ω = 25rad/s the phase φ of the response with respect to the rotation of the unbalanced disk is measured to be π/2rad and the steady-state vibration amplitude is X = 2.00cm.
When the rotation rate of the disk is much larger than this value the amplitude reduces to X = 0.50cm. Find the stiffness and damp-ing coefficient for the foundation, as well as the distance ρ between the center of rotation C and the mass center G.
m ρ C
G
x
k b
Problem 52:
For the system shown to the right, the disk of mass m and radius r rolls without slip. If x measures the displacement of the disk from the unstretched position of the spring.
a) find the equations of motion;
b) if the forcing takes the form:
f (t) =
f0t, 0 ≤ t < t0, 0, t0≤ t,
find the response of the system with zero initial conditions. If you choose to use the convolution integral you do not actually have to solve the integrand, just set it up.
b 2 k
f (t) m
Problem 53:
In the figure, the disk has mass m, radius ℓ, and moment of inertia IG = m ℓ22 about the mass center G. The disk is assumed to roll without slip and is subject to a harmonic moment of the form
M (t) = M0 sin(ω t).
a) Find the equations of motion for this system.
b) With
m = 4kg, ℓ = 0.10m, M0= 2 N · m,
the frequency ratio of the system is r = 1 when ω = 2 π. If, in addi-tion, the amplitude of the response is Θ = π/2rad, find the damping coeffi-cient c and the stiffness k for this sys-tem.
k
k c
M (t)
(m, ℓ) ℓ/2
G g
Problem 54:
In the figure shown to the right, the machine of mass m = 100kg rests on a regid support and is subject to a harmonic force
f (t) = F0sin(ω t).
If the support is redesigned as an elastic foun-dation, find the stiffness k and damping coef-ficient c so that the natural frequency of the new system is ωn = 10 π, and when the sys-tem operates with a frequency ratio r = 3 the amplitude of the force transmitted to the ground is 20% of the applied forcing ampli-tude.
f (t)
Original
k c
f (t)
Redesigned
Problem 55:
The base of the seismograph shown to the right is subject to harmonic ground motion of the form
u(t) = u0 sin(ω t).
a) Find the equations of motion for the relative displacement z between the mass m and the base;
b) With
k = 0.25N/m, c = 0.40N/(m/s), m = 0.50kg,
Determine the steady-state amplitude of the response for u0 = 0.005m and ω = 2 πrad/s.
k c
m
u(t) z
Problem 56:
For the system shown to the right the bar of length ℓ has mass m and is subject to a time-dependent moment of the form
M(t) = M0 sin(ω t) ˆk.
a) Find the equations of motion for this rotation of the bar.
b) What is the steady-state amplitude of the forced response, with
m = 3kg, b = 16N/(m/s), k = 8N/m, ℓ = 1
4m, ω = 4rad/s, M0= 1
16N · m
G
(m, ℓ)
k b
k M(t)
Problem 57:
The block of mass m = 20 kg shown to the right rests on a rigid foundation and is sub-ject to a time-dependent load
F(t) = f0 sin(ω t) ˆ.
a) Design an undamped foundation to achieve isolation ≥ 33% for all forcing frequencies ω > 3 πrad/s;
b) If, for the isolator that you designed, the damping ratio was measured to be ζ = 0.125 (rather than ζ = 0 as as-sumed above), what is the minimum isolation achieved over this frequency range?
f (t) ˆ
m
Problem 58:
In the figure shown to the right, the disk is subject to a time dependent moment of the form
M (t) = M0 sin(ω t).
a) Find the equations of motion for the angular displacement of the disk.
b) With
k = 280N/m, b = 12N/(m/s), m = 4kg,
I = 0.40kg · m2, r = 0.10m M0= 3N · m, ω = 5rad/s, Determine the steady-state response of disk as a function of time.
r
r/2
M (t) ˆk
k b
k
m I
Problem 59:
The unbalanced rotor shown in the figure is pinned to a frame and supported by a spring and damper. The total mass of the structure is m, with a small rotating component (10%
of the total mass) offset by a distance r from the center of rotation C.
a) Find the distance between the center of mass of the system and the center of rotation;
b) what is the damped natural frequency;
c) determine the steady-state amplitude of vibration when the rotor spins at an angular speed of ω = 5rad/s with:
k = 256N/m, b = 12N/(m/s), m = 5kg, r = 10cm
m r
C
k b
Problem 60:
In the figure, the disk has mass m, radius r, and moment of inertia IG = mr22 about the mass center G and is assumed to roll without slip. The attached plate undergoes harmonic motion of the form
u(t) = u0sin(ω t).
a) Find the equation of motion in terms of the displacement between the moving plate and the center of the disk;
b) What are the damping ratio and natu-ral frequency for this system?
c) If the system is critically damped, find the amplitude of the relative displace-ment of the disk for
m = 3kg, r = 0.10m k = 36N/m, c = 3N/(m/s)
u0= 0.05m, ω = 5rad/s
(m, r)
G
u(t)
k c
g
Problem 61:
For the mechanical system shown to the right, the uniform rigid bar is massless and pinned at point O while a force is applied at A of the form
f (t) = t e−σ t. For this system:
a) find the equations of motion;
b) Identify the damping ratio and natural frequency in terms of the parameters m, c, k, and ℓ.
c) With
m = 2kg, ℓ = 30cm, c = 0.25N/(m/s), k = 50N/m,
σ = 2.00s−1,
find the convolution integral for the re-sponse of the system. You need not evaluate the integral.
Assume that the system is in static equilib-rium at θ = 0, and that all angles remain small.
O
ℓ 3
2 ℓ 3
k
f (t) A
m
k c
B
θ
Problem 62:
The unbalanced rotor shown in the figure is pinned to a frame and supported by a spring and damper. The total mass of the structure is m, with a small rotating component (10%
of the total mass) offset by a distance r from the center of rotation C.
a) find the damped natural frequency;
b) what is the steady-state amplitude of vibration when the rotor spins at an an-gular speed of ω = 5rad/s with:
k = 108N/m, b = 9N/(m/s), m = 3kg, r = 7.5cm
m r
C
k b
Problem 63:
For the mechanical system shown to the right, the uniform rigid bar has mass m and length ℓ, and is pinned at point O. A har-monic force is applied at A. For this system:
a) find the equations of motion;
b) Identify the damping ratio and natural frequency in terms of the parameters m, c, k, and ℓ.
c) For:
m = 6kg, ℓ = 25cm, c = 0.50N/(m/s), k = 80N/m,
f0= 2.00N, ω = 10rad/s,
find the steady-state amplitude of the displacement of the block.
Assume that the system is in static equilib-rium at θ = 0, and that all angles remain small.
O
ℓ 2
k
f0sin(ω t) A
4 m
c
B θ
Problem 64:
In the figure, the disk has mass m, radius r, and moment of inertia IG = mr22 about the
In the figure, the disk has mass m, radius r, and moment of inertia IG = mr22 about the