Mechanik Solution 9

Mechanik HS 2018 Solution 9. Prof. Gino Isidori

Exercise 1. Rolling cones

In the following exercise we consider a circular cone of height h, density ρ, mass m and opening angle α.

(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.

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: Z ZZZ 2 2 3 Iz = ρ (x + y )dV = ρ r dz dr dφ

Z 2π Z R Z h π = ρ dφ r3dr dz = ρhR4 (S.1) 0 0 hr/R 10 3 = mh2 tan2 α . 10 where we used R = h tan α and m = πhR2ρ/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: Z ZZZ 2 2 2 2 2 I1 = ρ (y + z )dV = ρ (r sin φ + z )r dz dr dφ Z 2π Z R Z h = ρ dφ rdr (r2 sin2 φ + z2)dz 0 0 hr/R (S.2) π = ρ hR2(R2 + 4h2) 20 3 = mh2 tan2 α + 4 . 20

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

(b) The kinetic energy is given by: 1 1 1 T = I ω2 + I ω2 + I ω2 . (S.3) 2 1 1 2 2 2 2 3 3 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: ˙ vB = φh cos α . (S.4)

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

vB = ωR cos α . (S.5)

Equating the two expressions yields: h ω = φ˙ . (S.6) R 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:

ω3 = ω cos α , ω2 = 0 , ω1 = ω sin α . (S.7)

We are now ready to insert all these expressions into (S.3) to ﬁnd: 1 T = I ω2 + I ω2 + I ω2 2 1 1 2 2 3 3 1 3 1 3 = mh2(tan2 α + 4)ω2 + mh2 tan2 α ω2 2 20 1 2 10 3 3 3 (S.8) = mh2ω2 sin2 α(tan α2 + 4) + mh2ω2 sin2 α 40 20 3 sin4 α = mh2ω2 + 6 sin2 α . 40 cos2 α

Using eq. (S.6), we can rewrite:

3 h2 R2 R2 3 T = mh2φ˙2 sin2 α + 6 cos2 α = mh2φ˙2 1 + 5 cos2 α . (S.9) 40 R2 h2 h2 40

(c) Now the axis of rotation is from the cones apex to the point where it touches the plane. ˙ ˙ The touching point moves with vA = h · φ = ωR cos α. We therefore have ω = φ/ sin α and we can write: h ω = ω sin α = φ˙ , ω = 0 , ω = ω cos α = φ˙ . (S.10) 1 2 3 R

2 Putting things together, we obtain: 1 3 1 3 T = mh2 tan2 α + 4 ω2 + mh2 tan2 α ω2 2 20 1 2 10 3 (S.11) 3 R2 R2 h2 3 R2 = mh2φ˙2 + 4 + 2 · = mh2φ˙2 6 + . 40 h2 h2 R2 40 h2

Exercise 2. Scattering on ice

Consider a point-like particle of mass m hitting a bar of length l and mass M. Both objects are ﬁxed to the plane described by z = 0 and the particle hits the bar with the impact parameter d, see Fig.3. After the elastic scattering, the particle is at rest.

(a) Use conservation laws to determine the motion of the bar after the scattering. (b) Determine the ratio m/M for which the collision is elastic and discuss its dependence of v and d.

l/2 y M v m x d

Figure 3: The initial conﬁguration of Exercise 2.

Solution.

(a) The collision is elastic, so energy and momenta (both linear and angular) are conserved. Conservation of linear momentum yields:

Mvs = mv . (S.12)

From the conservation of angular momentum, we ﬁnd:

Isω = mvd . (S.13)

The inertia can be computed with:

Z l/2 2 1 2 Is = 2 ρ r dV = Ml . (S.14) 0 12 Therefore we ﬁnd for the angular velocity: mvd ω = 12 . (S.15) Ml2

Thus the bar rotates with the angular frequency ω and moves with the velocity vs = mv/M.

3 (b) To compute the mass ratio m/M, we consider the kinetic energy of the puck before collision:

mv2 T = . (S.16) 1 2 After the collision, this energy is completely transferred to the bar, which has a kinetic and rotational energy: 1 T = Mv2 + I ω2 . (S.17) 2 2 s s Equating the two expressions leaves us with: 1 mv2 = Mv2 + Ml2ω2 . (S.18) s 12

We have computed the expressions for ω and vs already, so we can quickly insert them to ﬁnd: m 1 = . (S.19) M 1 + 12(d/l)2

First, it should be noted that the velocity of the puck does not matter; the result is completely independent of v. It does however depend on the impact parameter d. The term 12d2/l2 in the denominator of our result accounts for the rotational energy: When the bar is hit right in the middle (d = 0), then the equation becomes m = M. Note that in this case, no rotation of the bar occurs. When the puck strikes the bar on the outermost edge (d = l/2), the mass ratio needs to be M = 4m for the collision to be elastic.

4 Exercise 3. Earth’s nutation

Earth is not a perfect sphere but to a good approximation described as a rotational ellipsoid with the half-axes being:

a = b = 6378 km , c = 6357 km . (1)

Since earth’s rotation axis is not aligned with its ﬁgure axis, it performs a swaying motion called nutation. Compute the period of nutation under the assumption that the density is homogeneous. Compare your result to the observed value Tobs = 433 d. Is this the number you ﬁnd? If not, what might be the reasons?

Solution. The derivation of the nutation frequency can be found in the lecture notes. One solves the Euler equations (5.15) for the symmetric top I2 = I1 and ﬁnds ω3 = const. Then, one obtains the equations of motion for the other two angular velocities,

ω˙ 1 = Ωω2 , (S.20) ω˙ 2 = −Ωω1 . where

I1 − I3 Ω = ω3 . (S.21) I1 This is equation (5.18). The angular velocity Ω is the nutation frequency. We now need the expressions I1 and I3. We start at: Z Z 2 2 2 I3 = dV ρ r = ρ dx dy dz(x + y ) . (S.22)

It is useful to make the substitutions

x˜ = x/a y˜ = y/b z˜ = z/c (S.23) transforming the ellipsoid into the unit sphere. The integral becomes: Z 2 2 2 2 ∂(x, y, z) I3 = ρ a x˜ + b y˜ dx˜ dy˜ dz˜ . (S.24) ∂(˜x, y,˜ z˜) The Jacobian evaluates to:

∂(x, y, z) J = = abc . (S.25) ∂(˜x, y,˜ z˜) We then ﬁnd: Z 1 Z 2π Z π 2 2 2 2 2 2 2 2 2 I3 = abc ρ (a r cos θ sin φ + b r sin θ sin φ)r sin φ dφ dθ dr 0 0 0 Z 1 Z 2π Z π = abc ρ (a2 cos2 θ + b2 sin2 θ)r4 sin3 φ dφ dθ dr (S.26) 0 0 0 4π = abc ρ a2 + b2 . 15 We can now replace the density with the mass: 3M ρ = . (S.27) 4πabc

5 With this we ﬁnd: 1 2 I = M(a2 + b2) = M(a2) . (S.28) 3 5 5 By symmetry, we have: 1 I = M(a2 + c2) . (S.29) 1 5 Plugging these results into eq (S.21), we obtain:

2 2 a − c Ω = ω3 . (S.30) a2 + c2

Finally, we can insert the rotational frequency of earth, ω3 = 2π/d and insert the values for a and c to obtain: 2π T = ≈ 304 d . (S.31) Ω The reason why this number is oﬀ from the measured value is the fact that earth is neither a perfect ellipsoid, nor homogeneous in mass. In fact, it is not even a rigid body: It has a liquid core and water on the surface.