(a) The Lagrangian is given by L = 1

Mechanik Solution 9.

HS 2019 Prof. Gino Isidori

Exercise 1. Two masses and one spring

They are free to move without friction on the horizontal plane. Each mass is also attached to the wall by a spring of constant k. Whenx1 =x2= 0, each spring is at rest.

(a) Write the Lagrangian of the system.

(b) Compute the Euler-Lagrange equations.

(c) Find the eigenmodes and eigenfrequencies of the system.

(d) Normalise these modes and define the normal coordinates such that

~ x= ∆

Q+

Q−

∆^TM∆ =1. (1)

Write the Lagrangian with these coordinates.

(e) Starting from a shifted mass 1, x1(t = 0) =aand x2(t = 0) = 0, without initial velocities, write the trajectory of the two masses.

Solution.

(a) The Lagrangian is given by L = 1

2 m( ˙ x

+ ˙ x

) − k(x

− x

x

+ x

) . (S.1) (b) The equations of motion are

m¨ x

= −k(2x

− x

) ,

m¨ x

= −k(2x

− x

) . (S.2) (c) We look for a solution of the form

~

x(t) = Ae ~

ω ∈ R . (S.3) Inserting this in the equations of motion we get

mω

− 2k k k mω

− 2k

A

A

= 0 (S.4)

For this system to have non-trivial solutions, the determinant must vanish (otherwise we can invert the matrix and obtain A ~ = 0), so

(mω

− 2k)

− k

= 0 . (S.5)

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The eigenfrequencies are then given by

ω

= p 3k/m ω

= p

k/m (S.6)

with associated eigenvectors A ~

∝

1

−1

, A ~

∝ 1

1

. (S.7)

The normalisation is then found by imposing A ~

M ~ A

= 1 where M is the mass matrix.

We find that we need to multiply the vectors by 1/ √

2m. Since M is proportional to the identity, both vectors have the same normalisation factor.

(d) The transformation matrix is then given by

∆ = 1

√ 2m

1 1 1 −1

. (S.8)

The normal coordinates are then found to be x

x

= 1

√ 2m

1 1 1 −1

Q

Q

. (S.9)

The Lagrangian with the new coordinates is as expected L = 1

2 ( ˙ Q

+ ˙ Q

− ω

Q

− ω

Q

) . (S.10) (e) We have

Q

(0) = p

m/2 (x

(0) + x

(0)) = a p m/2 , Q

(0) = a p

m/2 , Q ˙

(0) = ˙ Q

(0) = 0 .

(S.11) The trajectory as function of the normal coordinates is then

Q

(t) = a p

m/2 cos(ω

t) , Q

(t) = a p

m/2 cos(ω

t) , (S.12) which translates for the original coordinates as

x

(t) = a

2 [cos(ω

t) + cos(ω

t)] , x

(t) = a

2 [cos(ω

t) − cos(ω

t)] .

(S.13)

Exercise 2. Rolling cones

Consider a circular cone of heighth, densityρ, massm and opening angleα.

(a) Determine the inertia tensor in the representation of the axes of Fig.1.

(b) Compute the kinetic energy of the cone rolling on a level plane.

(c) Repeat the exercise, only this time the cone’s tip is fixed to a point on the z axis such that its longitudinal axis is parallel with the plane.

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Figure 1: The cone with the reference frame for Exercise 1(a).

Solution.

(a) We start with the inertia around the z-axis. To this end, we have to compute:

I

= ρ Z

(x

+ y

)dV = ρ Z Z Z

r

dz dr dφ

= ρ Z

0

dφ Z

0

r

dr Z

hr/R

dz = π 10 ρhR

= 3

10 mh

tan

α .

(S.14)

where we used R = h tan α and m = πhR

ρ/3 in the last step.

By the symmetry of the problem, the other two moments of inerta are identical to each other. So it suffices to compute:

I

= ρ Z

(y

+ z

)dV = ρ Z Z Z

(r

sin

φ + z

)r dz dr dφ

= ρ Z

0

dφ Z

0

rdr Z

hr/R

(r

sin

φ + z

)dz

= ρ π

20 hR

(R

+ 4h

)

= 3

20 mh

tan

α + 4 .

(S.15)

(b) The kinetic energy is given by:

T = 1

2 I

ω

+ 1

2 I

ω

+ 1

2 I

ω

. (S.16)

Now consider the point in the center of the cone’s base. It is moving on a circle around the axis perpendicular to the plane, with the radius h · cos α, as you can see from Fig. 2.

Its velocity is thus:

v

= ˙ φh cos α . (S.17)

Next we apply the rolling condition. From the smaller triangle in the Figure, we can see that:

v

= ωR cos α . (S.18)

Equating the two expressions yields:

ω = ˙ φ h

R . (S.19)

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Figure 2: The rolling cone in part a) of the problem.

The rotation axis of the cone points along the line where it touches the plane. The projections of this vector onto the principle axes are:

ω

= ω cos α , ω

= 0 , ω

= ω sin α . (S.20) We are now ready to insert all these expressions into (S.16) to find:

T = 1

2 I

ω

+ I

ω

+ I

ω

= 1 2

3

20 mh

(tan

α + 4)ω

+ 1 2

3

10 mh

tan

α ω

= 3

40 mh

ω

sin

α(tan α

+ 4) + 3

20 mh

ω

sin

α

= 3

40 mh

ω

sin

α

cos

α + 6 sin

α

.

(S.21)

Using eq. (S.19), we can rewrite:

T = 3

40 mh

φ ˙

h

R

R

h

sin

α + 6 cos

α R

h

= 3

40 mh

φ ˙

1 + 5 cos

α

. (S.22)

(c) Now the axis of rotation is from the cones apex to the point where it touches the plane.

The touching point moves with v

= h · φ ˙ = ωR cos α. We therefore have ω = ˙ φ/ sin α and we can write:

ω

= ω sin α = ˙ φ , ω

= 0 , ω

= ω cos α = ˙ φ h

R . (S.23)

Putting things together, we obtain:

T = 1 2

3

20 mh

tan

α + 4

ω

+ 1 2

3

10 mh

tan

α ω

= 3

40 mh

φ ˙

R

h

+ 4 + 2 R

h

· h

R

= 3

40 mh

φ ˙

6 + R

h

.

(S.24)

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