## Assignment 3, solutions

#### Problem 1:

A 100 kg load is placed on a 40 kg stretcher to be carried by two persons, one on each end.  The stretcher is 2 m long and the load is placed 0.66 m from the left end.  How much force must each person exert to carry the stretcher?

Solution:

• Concepts:
Equilibrium
• Reasoning:
Equilibrium implies no net force and no net torque.
• Details of the calculation:

No net force:
F1 + F2 - mstretcherg - mloadg = 0.
F1 + F2 = (392 + 980) N = 1372 N.
No net torque about any point:
Choose point A.  F1d1 - mstretcher g dstretcher - mload g dload = 0.
F1 = (392*1 + 980*0.6)/2 N = 490 N.
F2 = 882 N.

#### Problem 2:

A stationary space station can be approximated as a hollow spherical shell of mass 6 tons (6000 kg) and inner and outer radii of 5 m and 6 m.  To change its orientation, a uniform fly wheel of radius 10 cm and mass 10 kg located at the center of the station is spun quickly from rest to 1000 rpm.
(a)  How long (in minutes) will it take the station to rotate by 10o?
(b)  What energy (in Joules) is needed for the whole operation?

(Moment of Inertia of hollow spherical shell: [2M(Ro5 - Ri5 )]/[5(Ro3 - Ri3)] where Ro is outer radius and Ri is inner radius)

Solution:

• Concepts:
Conservation of angular momentum
• Reasoning:
The system consists of the shell and the flywheel.  No external torques act on the system, so the total angular momentum of the system is conserved.  If a component of L is zero before the flywheel starts to rotate, then it is zero afterwards.
• Details of the calculation:
(a)  The flywheel receives an angular impulse, so the shell receives an angular impulse of equal magnitude and opposite direction.
L = Iω.
For the disk I = ½mr2 = 0.05 kgm2.  L = (0.05 kgm2)1000*2π/60s = 5.24 kgm2/s.
For the shell I = [2M(Ro5 - Ri5 )]/[5(Ro3 - Ri3)] = 122664 kgm2
ω = dθ/dt = L/I = 4.27*10-5/s.
t = (10*π/(180*4.25*10-5))s = 4089 s ≈ 68 min.

(b)  The energy needed is the final kinetic energy of the flywheel and the shell.
E = ½Iwωw2 + ½Isωs2 = 274.2 J.

#### Problem 3:

Two identical billiard balls of radius R and mass M, rolling with CM velocities ±vi, collide elastically, head-on.  Assume that after the collision they have both reversed motion and are still rolling.
(a)  Find the impulse which the surface of the table must exert on each ball during its reversal of motion.
(b)  What impulse is exerted by one ball on the other?

Solution:

• Concepts:
Force and torque
• Reasoning:
Each ball reverses the direction  of its momentum and angular momentum.
• Details of the calculation:
(a) Take the left ball.  Let k point into the page and j point down.
Before the collision: p = mviL = Iωk = (Iv/R) k = (2mRv/5)k.
After the collision: p = -mviL = -(2mRv/5) k.
(Fball + Ftable)Δt = -2mvi, (Rj × Ftable)Δt  = -(4mRv/5) k.
FtableΔt = (4mv/5)i is the impulse the surface of the table exerts on each ball.
The impulse the table exerts on the left ball is (4mv/5) i, and the impulse the table exerts on the right ball is -(4mv/5)i.

(b)  FballΔt = -(2mv + 4mv/5)i = -(14mv/5) i is the impulse the right ball exerts on the left ball.
The impulse the left ball exerts on the right ball is (14mv/5) i.

#### Problem 4:

A rigid, symmetrical spaceship is shaped in the form of a cone with a uniform density.  The height of the cone is h, the radius of the base is r, and the total mass is m1.  Being suspended in outer space without any external forces acting on it, the space ship has a center of mass velocity v and angular momentum L not quite parallel to the symmetry axis.  Thus it experiences precession.
(a)  Calculate the principal moments of inertia of the spaceship about its CM in terms of h, r, and m1.
(b)  Show that the symmetry axis rotates in space about the fixed direction of the angular momentum L.

Solution:

• Concepts:
Rigid body motion, principal moments of inertia
• Reasoning:
We are asked to find the principal moments of inertia about the CM of a symmetrical top.  It is easier to calculate the principal moments of inertia about the tip and then use the parallel axes theorem.
• Details of the calculation:
(a)  Use cylindrical coordinates, with the origin at the origin of the primed system.
I3' = ∫(x2 + y2) dm = ∫ρ2μdV.
Here ρ is a cylindrical coordinate and μ is the mass per unit volume.
m = μ∫0hdz π(rz/h) = μπr2h/3.
I3' = μ∫0dφ ∫0rdρ ∫hρ/rhdz ρ3 = 2πμ∫0rdρ ρ3(h - hρ/r)
= 2πμh[r4/4 - r4/5] = πμhr4/10 = 3mr2/10.
I1' = I2' from symmetry.
I2' = ∫(x2 + z2) dm = ∫(ρ2cos2φ + z2)μdV = ∫ρ2cos2φ μdV +  ∫z2μdV.
I2' = μ∫0cos2φdφ ∫0rdρ ∫hρ/rhdz ρ3 + μ2π∫0rdρ ∫hρ/rhdz ρ z2
= πμhr4/20 + πμh3r4/5 = (3/5)m(r2/4 + h2) = I1'.

The CM is located a ρ = 0 and zCM, where
zCM = (μ/m)∫0dφ ∫0rdρ ∫hρ/rhdz ρ z = 3h/4.
Using the parallel axes theorem we therefore have
I3 = I3' = 3mr2/10,
I1 = I1' - m(3h/4)2 = (3/20)mr2 + (3/80)mh2 = I2.

(b)  L1 = I1ω1, L2 = I1ω2, L3 = I3ω3
We can chose the orientation of our axes so that L2 = 0, ω2 = 0.
In the body-fixed coordinate system L, ω, and the x3-axis lie in the same plane.
Consider a point P on the x3-axis of the top, displaced by r from the CM.
With respect to space-fixed coordinates, the velocity of point P is v = ω × r.
At every instant the velocity of points on the x3-axis is perpendicular to the plane containing L, ω, and the x3-axis, so the x3-axis must rotate about L.
The direction of L is fixed in space.

#### Problem 5:

Consider a right triangular lamina of areal density ρ, with one edge of length a along the x-axis and another edge of length b along the y-axis, as shown in the diagram.

(a)  Find the center of mass (X,Y,Z) in this coordinate system.
(b)  Find the components of the inertia tensor in this coordinate system.
(c)  For a = b, transform the inertia tensor to principal axes, giving the angle between the principal axes and those shown in the diagram, and find the three moments of inertia in the principal axes system.
(d)  Use the parallel axis theorem to find the principal moments of inertia about the center of mass.

Solution:

• Concepts:
The center of mass, the inertia tensor
• Reasoning:
We chose a coordinate system.  Then
RCM = (1/M)Σmiri --> (1/M)∫rdm = (1/M)∫ρ(r) rdV
Iij = Σkmkl(xk)l2δij - (xk)i(xk)j] --> ∫dm[Σlxl2δij - xixj] = ∫ρ(r)dV[Σlxl2δij - xixj] .
• Details of the calculation:
(a)  MRCM = i ∫dm x + j ∫dm y
= i ρ∫0b dy ∫0a-ay/b dx x  + j ρ∫0a dx ∫0b-bx/a dy y
= (ρab/6)(a i + b j) = ⅓(a i + b j), since M = ρab/2.

(b)  I11 = ∫dm y2 = ρ∫0a dx ∫0b-bx/a dy y2 = ρab3/12.
I22 = ∫dm x2 = ρa3b/12.  I33 = ∫dm (x2 + y2) = (a2 + b2) ρab/12.
I12 = I21 = -∫dm xy = -ρ∫0b y dy ∫0a-ay/b dx = -ρa2b2/24.  I13 = I23 = I31 = I32 = 0.

(c)  For b = a we have
I11 = I22 = ρa4/12,  I33 = ρa4/6, I12 = I21 = -ρa4/24, I13 = I23 = I31 = I32 = 0.
The inertia tensor in the given coordinate system is

To diagonalize this inertia tensor and find the principal moments of inertia Ix', Iy', and Iz', and the corresponding principal axes (x', y', z') we use Li = ΣjIijΩj = IΩi when the body is rotating about one of its principal axes.  We write

Here I = ρa2λ/12.  Since the matrix is block-diagonal we have immediately
Iz' = I33 = ρa4/6.  The z'-axis is the z-axis.
To find Ix' and Iy' we solve (a2 - λ)2 - a4/4 = 0,  a2 - λ =  ±a2/2,  λx' = a2/2,  λy' = 3a2/2.
Ix' = ρa4/24,  Iy' = ρa4/8.
Solving for the corresponding principal axes (the direction of Ω) we find
(a2 - λx'1 - (a2/2)Ω2 = 0, Ω1 - Ω2 = 0,  Ω1 = Ω2, The components along of Ω along the x and y-axis are equal to each other, the principal axis points in the (i + j) direction, it makes an an angle of +45o with the x-axis.
(a2 - λy'1 - (a2/2)Ω2 = 0, -Ω1 - Ω2 = 0,  Ω1 = -Ω2, The components along of Ω along the x and y-axis are equal in magnitude and have opposite signs.  The principal axis points in the (-i + j) direction, it makes an an angle of +135o with the x-axis.

(d)  Parallel axes theorem:  I'ii = Iii + Ma2.
Iz'CM = Iz' - M(xCM2 + yCM2) = ρa4/6 - (ρa2/2)(2a2/9) = ρa4/18.
Ix'CM = Ix' = ρa4/24,
Iy'CM = Iy' - M(xCM2 + yCM2) = ρa4/8 - (ρa2/2)(2a2/9) = ρa4/72.