Homework Prob. 2.1 A homogeneous wheel having a radius of R and mass of M rolls without slipping on a horizontal surface. A massless, inextensible cable is attached to the center of mass G of the wheel and to a block B (having a mass of m). The cable is drawn over a massless pulley. A spring of stiffness k is attached between G and a fixed wall. The coordinate x describes the position of G, with x being zero when the spring is unstretched.

a) Use Lagrange’s equations to develop the EOM for this single-DOF system using the generalized coordinate of x. Assume that the cable remains taut for all motion.

b) Using this EOM, determine the value of x corresponding to the equilibrium position of the system. (Note: The system is in equilibrium when

�

˙ x = ˙ ̇ x = 0.) SOLUTION Kinetic energy

�

T = 12MvG

2 + 12IG ˙ θ 2 + 1

2mvB

2 (see defn of θ above)

Using kinematics,

k

R

datum

no slip

M

m

G

x

B

massless, inextensible cable

homogeneous wheel

g

θ

�

vB = vG = ˙ x ; inextensible cable

�

vG = R ˙ θ ; no − slip condition of wheel

Also since wheel is homogeneous,

�

IG = 12MR2 . Combining above gives:

�

T = 12

M˙ x 2 + 12

12

MR2⎛ ⎝ ⎜

⎞ ⎠ ⎟ ˙ x /R( )2 + 1

2m˙ x 2

= 12

32

M + m⎛ ⎝ ⎜

⎞ ⎠ ⎟ ˙ x 2

Potential energy Will choose gravitation datum for B at location where x = 0. Also, since x = 0 when spring is unstretched, the stretch in the spring is given by x. Therefore,

�

U = −mgx + 12kx2

Work Forces N, f and O do no work since they act at points that do not move. Mg does no work since it acts perpendicular to the direction of motion of point G. Fs and mg do work but are included in potential. Therefore, no virtual work and hence the generalized force for x is zero. Lagrange’s EOM’s Using the above:

�

ddt

∂T∂˙ x

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = d

dt32

M + m⎛ ⎝ ⎜

⎞ ⎠ ⎟ ˙ x

⎡ ⎣ ⎢

⎤ ⎦ ⎥ =

32

M + m⎛ ⎝ ⎜

⎞ ⎠ ⎟ ̇ ̇ x

�

∂T∂x

= 0

�

∂U∂x

= −mg + kx

G Mg

B

Fs

N

f

O

mg

Therefore,

�

ddt

∂T∂˙ x

⎛ ⎝ ⎜

⎞ ⎠ ⎟ −

∂T∂x

+ ∂U∂x

= Qx

32

M + m⎛ ⎝ ⎜

⎞ ⎠ ⎟ ̇ ̇ x + kx = mg

Equilibrium state For equilibrium, we have

�

˙ ̇ x = 0 . Therefore,

�

kxst = mg ⇒ xst = mgk

Homework Prob. 2.2 A thin, homogenous bar having a mass of M and length of L is pinned to ground at point O. A particle P of mass m is free to slide on the smooth surface of the bar. A spring of stiffness k and unstretched length of

�

R0 is attached between pin O and particle P. Let r be the radial distance from O to P and θ be the rotation of the bar from a fixed vertical line.

a) Use Lagrange’s equations to develop the EOM’s for this two-DOF system using the generalized coordinates of r and θ. Recall that the potential energy stored in a spring is related to the square of the stretchin the spring, where the stretch is equal to the difference between its actual length and the unstretched length. Also, in writing down the velocity vector of P, you might need to review the polar kinematics expression for velocity.

b) Using these EOM’s, determine the equilibrium values for r and θ. (Note: The system is in equilibrium when

�

˙ r = ˙ ̇ r = ˙ θ = ˙ ̇ θ = 0).

SOLUTION Kinetic energy

�

T = 12mvP

2 + 12IO ˙ θ 2

Using kinematics,

�

v P = ˙ r er + r ˙ θ eθ (polar kinematics) Therefore,

g M

m

O r

θ P

L k

eθ

er

datum

�

vP2 = v P • v P = ˙ r 2 + r2 ˙ θ 2

Also since bar is homogeneous and using parallel axis theorem,

�

IO = IG + Md2

= 112

ML2 + M L2

⎛ ⎝ ⎜

⎞ ⎠ ⎟ 2

= 13ML2

.

Combining above gives:

�

T = 12

m ˙ r 2 + r2 ˙ θ 2( ) + 12

13

ML2⎛ ⎝ ⎜

⎞ ⎠ ⎟ ˙ θ 2

= 12

m˙ r 2 + 12

mr2 + 13

ML2⎛ ⎝ ⎜

⎞ ⎠ ⎟ ˙ θ 2

Potential energy Will choose gravitation datum as horizontal line through point O. Since spring has an unstretched length of R0, then amount of stretch in the spring is given by r - R0: Therefore,

�

U = −mgrcosθ −Mg L2cosθ + 1

2k r − R0( )2

Work Force O does no work since it acts at a point that does not move. Mg and mg do work but are included in potential. Therefore, no virtual work and hence the generalized forces for r and θ are zero. Lagrange’s EOM’s For generalized coordinate x Using the above:

�

ddt

∂T∂˙ r

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = d

dt∂∂˙ r

12

m˙ r 2 + 12

mr2 + 13

ML2⎛ ⎝ ⎜

⎞ ⎠ ⎟ ˙ θ 2

⎡ ⎣ ⎢

⎤ ⎦ ⎥

⎡

⎣ ⎢

⎤

⎦ ⎥

= ddt

m˙ r [ ] = m˙ ̇ r

Mg

O

mg

O

P

�

∂T∂r

= ∂∂r

12

m˙ r 2 + 12

mr2 + 13

ML2⎛ ⎝ ⎜

⎞ ⎠ ⎟ ˙ θ 2

⎡ ⎣ ⎢

⎤ ⎦ ⎥

= mr ˙ θ 2

�

∂U∂r

= ∂∂r

−mgrcosθ −Mg L2cosθ + 1

2k r − R0( )2⎡

⎣ ⎢ ⎤ ⎦ ⎥

= −mgcosθ + k r − R0( )

Therefore,

�

ddt

∂T∂˙ r

⎛ ⎝ ⎜

⎞ ⎠ ⎟ −

∂T∂r

+ ∂U∂r

= Qx

m˙ ̇ r − mr ˙ θ 2 − mgcosθ + k r − R0( ) = 0

For generalized coordinate θ Using the above:

�

ddt

∂T∂ ˙ θ

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = d

dt∂∂ ˙ θ

12

m˙ r 2 + 12

mr2 + 13

ML2⎛ ⎝ ⎜

⎞ ⎠ ⎟ ˙ θ 2

⎡ ⎣ ⎢

⎤ ⎦ ⎥

⎡

⎣ ⎢

⎤

⎦ ⎥

= ddt

mr2 + 13

ML2⎛ ⎝ ⎜

⎞ ⎠ ⎟ ˙ θ

⎡ ⎣ ⎢

⎤ ⎦ ⎥

= 2mr˙ r ̇ θ + mr2 + 13

ML2⎛ ⎝ ⎜

⎞ ⎠ ⎟ ̇ ̇ θ

�

∂T∂θ

= ∂∂θ

12

m˙ r 2 + 12

mr2 + 13

ML2⎛ ⎝ ⎜

⎞ ⎠ ⎟ ˙ θ 2

⎡ ⎣ ⎢

⎤ ⎦ ⎥

= 0

�

∂U∂θ

= ∂∂θ

−mgrcosθ −Mg L2cosθ + 1

2k r − R0( )2⎡

⎣ ⎢ ⎤ ⎦ ⎥

= mgrsinθ + Mg L2sinθ = mgr + Mg L

2⎡ ⎣ ⎢

⎤ ⎦ ⎥ sinθ

Therefore,

�

ddt

∂T∂ ˙ θ

⎛ ⎝ ⎜

⎞ ⎠ ⎟ −

∂T∂θ

+ ∂U∂θ

= Qθ

mr2 + 13

ML2⎛ ⎝ ⎜

⎞ ⎠ ⎟ ̇ ̇ θ + 2mr˙ r ˙ θ + mgr + Mg L

2⎡ ⎣ ⎢

⎤ ⎦ ⎥ sinθ = 0

In summary, the two EOM’s for this problem in terms of the generalized coordinates x and θ are:

�

m˙ ̇ r − mr ˙ θ 2 − mgcosθ + k r − R0( ) = 0

mr2 + 13

ML2⎛ ⎝ ⎜

⎞ ⎠ ⎟ ̇ ̇ θ + 2mr˙ r ˙ θ + mgr + Mg L

2⎡ ⎣ ⎢

⎤ ⎦ ⎥ sinθ = 0

Equilibrium state For equilibrium, we have

�

˙ r = ˙ ̇ r = ˙ θ = ˙ ̇ θ = 0. Therefore,

�

−mgcosθ + k r − R0( ) = 0

mgr + Mg L2

⎡ ⎣ ⎢

⎤ ⎦ ⎥ sinθ = 0

The second equation gives:

�

θst = 0 . Substitution of this into the first equation gives:

�

−mgcosθst + k rst − R0( ) = 0 ⇒

rst = R0 + mgk

Homework Prob. 2.3 A system is made up of a homogeneous wheel (mass m and radius R), block A (mass of 3m) and block B (mass of 2m). The wheel can roll without slipping on A. Springs are attached between A and ground, between B and ground and between B and the centroid G of the wheel, with stiffnesses of k, 2k and 3k, respectively. The absolute coordinate

�

x1 describes the position of A, the relative coordinate

�

x 2 describes the position of B relative to A and θ is the rotation of the wheel measured from a vertical line. All springs are unstretched when

�

x1 = x 2 = θ = 0. A horizontal force F acts at the centroid G of the wheel.

a) Use Lagrange’s equations to develop the EOM’s for this three-DOF system using the generalized coordinates of:

�

x1,

�

x 2 and θ . b) Write these EOM’s in the matrix-vector form of:

�

M[ ]˙ ̇ x + K[ ]x = f where

�

x = x1, x2, θ{ }T . Identify the components of the mass and stiffness matrices [M] and [K] as well as the components of the forcing vector f.

SOLUTION Kinetic energy The kinematics of the problem for the coordinates described above are:

�

vA = ˙ x 1

k

G

x1

smooth no slip R

G B

A

m 2m

3m

2k 3k

x2 θ

F

�

vB = ˙ x 1 + ˙ x 2 (since x2 is measured relative to B)

�

vG = ˙ x 1 + R ˙ θ (since no-slip and

�

R ˙ θ is motion of G relative to A)

Therefore, using the above and

�

IG = 12mR2 gives:

�

T = 12

3m( )vA2 + 1

22m( )vB

2 + 12

m( )vG2 + 1

2IG

˙ θ 2

= 12

3m( ) ˙ x 12 + 1

22m( ) ˙ x 1 + ˙ x 2( )2 + 1

2m( ) ˙ x 1 + R ˙ θ ( )2

+ 12

12

mR2⎛ ⎝ ⎜

⎞ ⎠ ⎟ ˙ θ 2

Potential energy The stretch/compression of each spring is equal to the difference between the absolute motion of the two ends of the spring. Therefore,

�

U = 12k( )x12 + 1

22k( ) x1 + x2( )2 + 1

23k( ) Rθ + x1( ) − x2 + x1( )[ ]2

= 12k( )x12 + 1

22k( ) x1 + x2( )2 + 1

23k( ) Rθ − x2( )2

Work The differential work done by the force F on the wheel is (by definition):

�

dW = F • drG= Fi( ) • d x1 + Rθ( )i[ ]= Fi( ) • dx1 + R dθ( )i= F dx1 + FR dθ = Qx1dx1 + Qx2dx2 + Qθ dθ

Therefore, the generalized forces corresponding to the three generalized coordinates given are:

�

Qx1 = F

Qx2 = 0

Qθ = FR

Lagrange’s EOM’s For generalized coordinate x1 Using the above:

�

ddt

∂T∂˙ x 1

⎛

⎝ ⎜

⎞

⎠ ⎟ = d

dt∂∂˙ x 1

12

3m( ) ˙ x 12 + 1

22m( ) ˙ x 1 + ˙ x 2( )2 + 1

2m( ) ˙ x 1 + R ˙ θ ( )2

+ 12

12

mR2⎛ ⎝ ⎜

⎞ ⎠ ⎟ ˙ θ 2

⎡ ⎣ ⎢

⎤ ⎦ ⎥

⎡

⎣ ⎢

⎤

⎦ ⎥

= ddt

3m( ) ˙ x 1 + 2m( ) ˙ x 1 + ˙ x 2( ) + m( ) ˙ x 1 + R ˙ θ ( )[ ]= 3m( )˙ ̇ x 1 + 2m( ) ˙ ̇ x 1 + ˙ ̇ x 2( ) + m( ) ˙ ̇ x 1 + R ˙ ̇ θ ( )= 6m( )˙ ̇ x 1 + 2m( )˙ ̇ x 2 + mR( )˙ ̇ θ

�

∂T∂x1

= 0

�

∂U∂x1

= ∂∂x1

12k( )x12 + 1

22k( ) x1 + x2( )2 + 1

23k( ) Rθ − x2( )2⎡

⎣ ⎢ ⎤ ⎦ ⎥

= k( )x1 + 2k( ) x1 + x2( )= 3k( )x1 + 2k( )x2

Therefore,

�

ddt

∂T∂˙ x 1

⎛

⎝ ⎜

⎞

⎠ ⎟ −

∂T∂x1

+ ∂U∂x1

= Qx1

6m( )˙ ̇ x 1 + 2m( )˙ ̇ x 2 + mR( )˙ ̇ θ + 3k( )x1 + 2k( )x2 = F

For generalized coordinate x2 Using the above:

�

ddt

∂T∂˙ x 2

⎛

⎝ ⎜

⎞

⎠ ⎟ = d

dt∂

∂˙ x 2

12

3m( ) ˙ x 12 + 1

22m( ) ˙ x 1 + ˙ x 2( )2 + 1

2m( ) ˙ x 1 + R ˙ θ ( )2

+ 12

12

mR2⎛ ⎝ ⎜

⎞ ⎠ ⎟ ˙ θ 2

⎡ ⎣ ⎢

⎤ ⎦ ⎥

⎡

⎣ ⎢

⎤

⎦ ⎥

= ddt

2m( ) ˙ x 1 + ˙ x 2( )[ ]= 2m( ) ˙ ̇ x 1 + ˙ ̇ x 2( )= 2m( )˙ ̇ x 1 + 2m( )˙ ̇ x 2

�

∂T∂x2

= 0

�

∂U∂x2

= ∂∂x2

12k( )x12 + 1

22k( ) x1 + x2( )2 + 1

23k( ) Rθ − x2( )2⎡

⎣ ⎢ ⎤ ⎦ ⎥

= 2k( ) x1 + x2( ) − 3k( ) Rθ − x2( )= 2k( )x1 + 5k( )x2 − 3kR( )θ

Therefore,

�

ddt

∂T∂˙ x 2

⎛

⎝ ⎜

⎞

⎠ ⎟ −

∂T∂x2

+ ∂U∂x2

= Qx2

2m( )˙ ̇ x 1 + 2m( )˙ ̇ x 2 + 2k( )x1 + 5k( )x2 − 3kR( )θ = 0

For generalized coordinate θ Using the above:

�

ddt

∂T∂ ˙ θ

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = d

dt∂∂ ˙ θ

12

3m( ) ˙ x 12 + 1

22m( ) ˙ x 1 + ˙ x 2( )2 + 1

2m( ) ˙ x 1 + R ˙ θ ( )2

+ 12

12

mR2⎛ ⎝ ⎜

⎞ ⎠ ⎟ ˙ θ 2

⎡ ⎣ ⎢

⎤ ⎦ ⎥

⎡

⎣ ⎢

⎤

⎦ ⎥

= ddt

mR( ) ˙ x 1 + R ˙ θ ( ) + 12

mR2⎛ ⎝ ⎜

⎞ ⎠ ⎟ ˙ θ

⎡ ⎣ ⎢

⎤ ⎦ ⎥

= mR( ) ˙ ̇ x 1 + R ˙ ̇ θ ( ) + 12

mR2⎛ ⎝ ⎜

⎞ ⎠ ⎟ ̇ ̇ θ

= mR( )˙ ̇ x 1 + 32

mR2⎛ ⎝ ⎜

⎞ ⎠ ⎟ ̇ ̇ θ

�

∂T∂θ

= 0

�

∂U∂θ

= ∂∂θ

12k( )x12 + 1

22k( ) x1 + x2( )2 + 1

23k( ) Rθ − x2( )2⎡

⎣ ⎢ ⎤ ⎦ ⎥

= 3kR( ) Rθ − x2( )= − 3kR( )x2 + 3kR2( )θ

Therefore,

�

ddt

∂T∂ ˙ θ

⎛ ⎝ ⎜

⎞ ⎠ ⎟ −

∂T∂θ

+ ∂U∂θ

= Qθ

mR( )˙ ̇ x 1 + 32

mR2⎛ ⎝ ⎜

⎞ ⎠ ⎟ ̇ ̇ θ − 3kR( )x2 + 3kR2( )θ = FR

In summary, the three EOM’s for this problem in terms of the generalized coordinates

�

x1,

�

x2 and θ are:

�

6m( )˙ ̇ x 1 + 2m( )˙ ̇ x 2 + mR( )˙ ̇ θ + 3k( )x1 + 2k( )x2 = F

2m( )˙ ̇ x 1 + 2m( )˙ ̇ x 2 + 2k( )x1 + 5k( )x2 − 3kR( )θ = 0

mR( )˙ ̇ x 1 + 32

mR2⎛ ⎝ ⎜

⎞ ⎠ ⎟ ̇ ̇ θ − 3kR( )x2 + 3kR2( )θ = FR

These EOM’s can be written in the following matrix-vector form:

�

6m 2m mR

2m 2m 0

mR 0 32

mR2

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥

˙ ̇ x 1

˙ ̇ x 2

˙ ̇ θ

⎧

⎨ ⎪ ⎪

⎩ ⎪ ⎪

⎫

⎬ ⎪ ⎪

⎭ ⎪ ⎪

+

3k 2k 0

2k 5k −3kR

0 −3kR 3kR2

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥

x1

x2

θ

⎧

⎨ ⎪ ⎪

⎩ ⎪ ⎪

⎫

⎬ ⎪ ⎪

⎭ ⎪ ⎪

=

F

0

FR

⎧

⎨ ⎪ ⎪

⎩ ⎪ ⎪

⎫

⎬ ⎪ ⎪

⎭ ⎪ ⎪

or

�

M[ ] ˙ ̇ x { } + K[ ] x{ } = f{ } Not required Note that the mass and stiffness matrices ([M] and [K], respectively) are symmetric. That is, we see that

�

M[ ] = M[ ]T and

�

K[ ] = K[ ]T . We showed that this is always true for a Lagrangian formulation (see lecture on linearization of EOM’s). Also, the “coupling” (appearance of non-zero off-diagonal terms) of the mass and stiffness matrices are different. Also, although the force F acts on the wheel, F appears in the generalized coordinates for both block A and the wheel. Do you see why?