FURTHER EXAMPLES

2. 4 OTHER SOURCES OF POTENTIAL ENERGY

2.14 FURTHER EXAMPLES

The small angle assumption, where appropriate, is made in these problems. Assuming all systems are linear, the generalized coordinate is measured from the system’s equilibrium position. Thus, the static forces in the spring cancel with the gravity forces, which cause them, and neither are included on the FBDs.

300x$

+ 1200x#

+ 12,000x = 1200y#

+ 12,000y

#

mx$ + cx#

+ kx = cy# + ky -k(x - y) - c(x#

- y# ) = mx$

m

x(t)

k(y – x) c(y˙ – x˙) y(t)

k c

(a) (b)

FIGURE 2.51

(a) SDOF model for simplified suspension system.

Model ignores the stiffness of the tires and the mass of the axle. (b) FBD of the system at an arbitrary instant.

E X A M P L E 2 . 3 0

A mass of 30 kg (shown in Figure 2.52(a)) is hung from a spring of stiffness k2.5 10⁵ N/m, which is attached to an aluminum beam (E71 10⁹N/m², ␳2.7 10³ kg/m³) of moment of inertia I3.5 10^–8m⁴and of length 35 cm. The beam is supported at its free end and by a circular aluminum cable of diameter 1 mm and length 30 cm.

(a) Determine the equivalent stiffness of the assembly.

(b) Write the differential equation governing in the motion of the mass.

S O L U T I O N

The stiffness of the beam is

(a) The equivalent stiffness of the cable is

(b) The beam and cable behave as two springs in parallel, because they have the same displace- ments at their end. The discrete spring is in series with the parallel combination, because

k_c = EA

L =

(71 * 10⁹ N/m²)p(5 * 10^-4)²

0.30 m = 1.86 * 10⁵ N/m k_b = 3EI

L³ =

3(71 * 10⁹N/m²)(3.5 * 10^-8 m⁴)

(0.35m)³ = 1.74 * 10⁵ N/m

the displacement of the mass is the sum of the displacement of the spring and the displace- ment of the end of the beam. The equivalent model is shown in Figure 2.52(b). The equivalent stiffness of the combination is

(c) (b) The differential equation for a SDOF model of the motion of the mass (assuming the beam and the column are massless) is

30x$ (d)

+ 1.48 * 10⁵x = 0

= 1.48 * 10⁵ N/m

= 1

1

2.5 * 10⁵ N/m + 1

(1.74 * 10⁵ N/m) + (1.86 * 10⁵ N/m) k_eq = 1

1

k + 1

k_b + k_c

35 cm 30 cm

Aluminum

30 kg Aluminum

1 mm diameter

I = 3.5 × 10^–8 m⁴

2.5 × 10⁵N/m

30 kg k_cable

k k_beam

(a) (b)

x x

FIGURE 2.52

(a) System of Example 2.30. Mass is suspended from a beam supported by a column. (b) Beam and column are modeled by springs resulting in the equivalent systems model shown.

E X A M P L E 2 . 3 1

A schematic diagram of a compactor is shown in Figure 2.53(a). The compactor is a cylin- der of mass 35 kg, radius 0.9 m, and length 1.5 m. To each end of the cylinder, a viscous damper of damping coefficient c1000 N m/s is connected to the center, while a spring of stiffness k1.4 10⁵N/m is connected to a point 0.2 m from the center.

(a) Derive a mathematical model for the unforced motion of the cylinder if it rolls without slipping.

(b) Derive a mathematical model for the unforced motion of the cylinder when it rolls and slips with a coefficient of friction of 0.25.

S O L U T I O N

(a) The free-body diagram method is used with projections of the diagrams showing the equivalent and effective forces in Figure 2.53(b). When the cylinder rolls without slipping, there is an unknown friction force between the cylinder and the ground. Additionally, a

#

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kinematic relationship exists between the displacement of the mass center and the angular acceleration . When the mass center of the disk has moved a distance x from equi- librium, the spring has also changed in length where r 5 0.2 m and is the angular rota- tion of the disk. Since xR , the change in length of the spring is x. Summing moments on these FBDs using ( Mc)_ext( Mc)_effgives

(a) (b) Substituting given values, noting the moment of inertia of a circular cylinder about the axis of rotation is , leads to

(c) (b) if the disk rolls and slips, the friction force is equal to the maximum allowable friction force equal to mN, and there is no kinematic relationship between the angular acceleration and the

52.5x$

+ 2000x#

+ 4.18 * 10⁵x = 0 I = ¹

2mR² a 1

R² + mb x$ + 2cx#

+ 2ka1 + r

Rb²x = 0 -(2cx#

)R - c2k a1 + r

Rbxd(r + R)x = Iax$

Rb + (mx$ )R g

g

A

^{1 +} R^r

B

u

ru u a = Ra

Iα 2k (1+ r

R) x µmg mg

External forces Effective forces

= mg

N

N F

Ix. 2cx.

2cx.

R mx¨

2k (1+ r R) x

External forces Effective forces x

Cylinder

(a)

(b)

(c)

mx¨

FIGURE 2.53

(a) System of Example 2.31. A compactor is modeled as a cylinder with viscous dampers attached at the center and springs attached at a point above the center. (b) FBDs of the compactor, assum- ing it rolls without slipping. (c) FBDs of the compactor in the case of slipping.

acceleration of the mass center. The appropriate FBDs are shown in Figure 2.53(c). Summing moments about the point contact using the FBDs and , we have

(d) Summing moments about the center of the disk using these FBDs and

, we have

(e) Substituting Equation (e) into Equation (d) leads to

(f) Equation (f ) is derived assuming . The right-hand side is positive if . Upon substitution of given values and taking into account the sign dependence of the right-hand side on Equation (f ) becomes

35x$ (g)

+ 2000x#

+ 3.08x10⁵ = e-77.25 x# 7 0 77.25 x#

6 0 x#

x# 6 0 x#

7 0 m x$

+2cx#

+2ka1 + r

RbR = -mmgR -c2ka1 +

r

Rbxdr + mmgR = Ia (πM_G)_ext = (πM_G)_eff

-(2cx#

)R - c2ka1 + r

Rbxd(r +R)x = Ia+(mx$ )R

(πM_C)_ext = (πM_C)_eff

E X A M P L E 2 . 3 2

Consider the system shown in Figure 2.54(a). A thin rod of mass mis pinned at Oat a dis- tance of from its left end is attached to a viscous damper of damping coefficient c at its left end. Attached to its right end is a cubic block of side dand mass mwhich is initially half submerged in a liquid of mass density .

(a) Determine the value of d such that the equilibrium position is the horizontal con- figuration of the bar.

(b) Determine the equation of motion for small oscillations about the horizontal equi- librium position. Use as the chosen generalized coordinate.

S O L U T I O N

When the system is in equilibrium, the moment of the gravity force must balance with the moment of the buoyant force acting on the block. For the horizontal configuration whose free-body diagram is shown in Figure 2.54(b), summing moments about the pin support

leads to

(a) The buoyant force is equal to the weight of the fluid displaced by the block. For half of the cube to be submerged,

(b) Using Equation (b) in Equation (a) leads to

a 7 (c) 10brd ³

2 = 2

10 mg Qd = a4mg 7r b

1 3

F_B = rd²ad

2b = rd³ 2 -mga2L

10b + F_Ba7L 10b = 0 πM_O = 0,

u

r

3L 10

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(b) When the bar has an angular displacement from its, equilibrium position, the buoyant force acting on the block (assuming small ) becomes

(d) Summing moments about the point of support using the free-body diagrams of Figure 2.54(c),

leads to

(e) After subtracting the equilibrium condition of Equation (a), Equation (d) becomes

. (f)

184mu (g)

$

+ 27cu

#

+ 147rd²u = 210

L F(t) 184

300mL²u

$ +

9 100cL²u

# +

49

100rd²L²u = 7L 10F(t)

= 1 12 mL²u

$ +

2 10 mLu

$a 2 10Lb +

7 10 mLu

$a 7 10Lb F(t)7L

10 - 3 10Lcu

#a 3 10Lb +

2 10mgL -

7

10Lcrd²ad 2 +

7 10Lub d (πM_O)_ext = (πM_O)_eff

F_B = rd²ad 2 + 7

10Lub

u u

c F(t)

Slender bar of mass m m

7L 10 3L

10

(a)

(b)

F_B mg

c( ^3L )

10

F(t)

R

ρd²7L 10

=

m2L 10θ

θ

¨

mL² 1 12 θ¨ m2L

10θ˙² θ˙

m 7L 10θ¨

( )

(c)

FIGURE 2.54

(a) System of Example 2.32. A cube is at the end of a thin bar and is partially submerged in a liquid when acted on by a time dependent force. (b) FBD of the equilibrium position. (c) FBDs at an arbitrary instant. The gravity force and static buoyancy force cancel with each other when deriving the differential equation.

E X A M P L E 2 . 3 3

Use the free-body diagram method to derive the differential equation governing the motion of the system shown in Figure 2.55(a). Use as the clockwise angular displacement of the bar measured from the system’s equilibrium position and as the chosen generalized coordi- nate. Assume small .

S O L U T I O N

FBDs showing the external forces and the effective forces acting on the bar at an arbitrary instant are shown in Figure 2.55(b). The small angle assumption implies that sin

and the springs remain vertical. Thus, a linear differential equation will be derived, and it can be assumed that static spring forces cancel with gravity when deriving the differential equation. Summing moments about the point of support

and using the FBDs, we have

(a) which reduces to

m$ (b) u + 4cu

#

+ 3ku = 0 -ca2L

3 u

#b a2L

3 b - kaL 3ub aL

3b - 2kaL 3ub aL

3b = 1 12mL²u

$

+ maL 6u

$b aL 6b (πM_O)_ext = (πM_O)_eff

cosu L 1,

u L u, u

u

mL²θ 1 12 m Lθ² ¨

6

˙

mLθ 6

˙

=

External forces Effective forces

2kLθ 3 kLθ

3

c2Lθ 3

˙ R_x

R_y

(b)

θ L

3 L 3

L 3

k

c

2k

(a)

FIGURE 2.55

(a) System of Example 2.33. The small angle assumption is used to linearize the differential equationa priori. (b) FBDs of the system at an arbi- trary instant.

E X A M P L E 2 . 3 4

Derive the differential equation governing the motion of the system of Figure 2.56. The system is in equilibrium when the bar is in the vertical position. Use the equivalent systems method using the angular coordinate as the counterclockwise angular displacement of the bar when it is in equilibrium and as the generalized coordinate. Assuming small , the disk rolls without slipping, and there is no friction between the cart and the surface.

u u

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S O L U T I O N

The displacement of the center of the disk is , and the displacement of the cart is with both assuming small . The appropriate equivalent systems model is the tor- sional system whose equation is

(a) The equivalent moment of inertia is obtained using kinetic energy. The kinetic energy of the system at an arbitrary instant is

(b) Noting that, if the disk rolls without slipping, then the moment of inertia of the thin disk is , and the moment of inertia of the slender bar is . Equation (b) becomes

(c) Hence, .

The potential energy at an arbitrary instant is

(d) Thus, . The work done by the viscous damping force is

U = - (e) Lcx#

dx = - Lc(au

#

)d(au) = - Lca²u

# du k_t,eq = k(a² + b²)

V = 1 2kx² +

1 2ky² =

1

2k(a² + b²) u I_eq = ³

2m_d a² + ¹

12mL² + m_cb²

= 1 2a3

2m_da² + 1

12mL² + m_cb²bu

#

2

T = 1 2m_daau

#b² + 1 2a1

2m_d r²b aau

#

r b² + 1 2a 1

12mL²bu

#

2 + 1 2m_c(bu

# )²

I_b = ¹

12mL² I_d = ¹

2m_d r²

v = ^x

# r, T =

1 2 m_dx#

2 + 1 2I_d v² +

1 2I_bu

#

2 + 1 2m_c y#

2

I_eq u

$ + c_t,equ

#

+ k_t,equ = 0 y = bu u

x = au

a

b

m_c k k

Thin disk of mass m_d

radius r

Slender bar of mass m, length L No slip c

θ

FIGURE 2.56

The thin rod connects the disk that rolls without slipping and the cart which moves on a sur- face without friction.

The equivalent viscous damping coefficient is . Hence, the governing differen- tial equation is

a3 (f)

2m_da² + 1

12mL² + m_cb²bu

$ + ca²u#

+ k(a² + b²)u = 0 c_t,eq = ca²

θ

b Slender bar of

mass m₂

a c

k

m₁ y(t)

(a)

x

k(y – x) + c(y˙ – x˙)b a

b a

(^x_a ^.)²

m₂(^{b – a} )

2

(^x¨_a )

m₂(^{b – a} )

2

=

External forces Effective forces

R_x

R_y

kx m₁x¨

m₁

(b)

Iθ¨ FIGURE 2.57

(a) The end of the bar is connected to a spring and viscous damper which is given motion input, perhaps from a cam and fol- lower mechanism.

(b) FBDs of the bar at an arbitrary instant.

E X A M P L E 2 . 3 5

The bar of Figure 2.57(a) is attached to a spring and viscous damper which is attached to a cam and follower system. The cam is designed such that it imparts a displacement y(t) to the spring and viscous damper. The bar is designed to impart a linear motion to the cart.

Derive the differential equation governing the motion using xas the displacement of the cart and as the generalized coordinate. The motion occurs in the horizontal plane.

S O L U T I O N

Assume the displacement of the cart is small. The angular rotation of the bar is related to the displacement of the cart by . The displacement of the end of the bar where the spring is attached is . FBDs showing the external and effective force acting on the bar are shown in Figure 2.57(b). Summing moments about the mass center of the

bar and using these FBDs leads to

(a) + (m₁x$

)a + m₂ab - a 2 bx$

aab - a 2 b kay -

b

axbb + cay# -

b ax#

bb-(kx)a = 1

12m₂L²ax$ ab (πM_G)_ext = (πM_G)_eff

y = bu = ^b

ax x = au

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which is rearranged to

(b)