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Problem 1 : Solution/marking scheme Two Problems in Mechanics (
points)
Part A. The Hidden Disk (3.5 points)
A1 (0.8 pt) Find an expression for b as a function of the quantities (1), the angle φ and the tilting angle Θ of the base.
Solution A1: [0.8]
Geometric solution: use that torque with respect to point of contact is 0 ⇒ cen- ter of gravity has to be vertically above point of contact.
sin φ = D b
sin Θ = D r 1
Here D may be called another name. Solve this:
sin φ = r^1 b
sin Θ ⇒ b = r^1 sin Θ sin φ
Alternative: Torque and forces with respect to another point: [0.8] Correct equation for torque 0. Correct equation for force 0. Correct solution 0.
A2 (0.5 pt) Find the equation of motion for ϕ. Express the moment of inertia IS of the cylinder around its symmetry axis S in terms of T , b and the known quantities (1). You may assume that we are only disturbing the equilibrium position by a small amount so that ϕ is always very small.
Solution A2: [0.5]
Write some equation of the form ϕ¨ = −ω^2 ϕ 0. Writing an equation of the form ϕ = A cos ωt is also correct. Two solutions:
- Kinetic energy: 12 IS ϕ˙^2 and potential energy: −bM g cos ϕ. Total energy is con- served, and dierentiation w.r.t. time gives the equation of motion.
- Angular equation of motion from torque, τ = IS ϕ¨ = −M gb sin ϕ. Correct equation (either energy conservation or torque equation of motion) 0. Final answer T = 2π
IS
M gb
⇒ IS = M gbT^
2 4 π^2
(Derivation:
⇒ ϕ¨ = − bM gI S
sin ϕ ' − bgMI S
ϕ
so that ω^2 = bgM IS
)
A3 (0.4 pt) Find an expression for the distance d as a function of b and the quantities (1). You may also include r 2 and h 2 as variables in your expression, as they will be calculated in subtask A.5.
Solution A3: [0.4]
Some version of the center of mass equation, e.g.
b = dM^2 M 1 + M 2
correct solution: d =
bM πh 2 r^22 (ρ 2 − ρ 1 ) 0.
A4 (0.7 pt) Find an expression for the moment of inertia IS in terms of b and the known quantities (1). You may also include r 2 and h 2 as variables in your expression, as they will be calculated in subtask A.5.
Solution A4: [0.7]
correct answer for moment of inertia of homogeneous disk
I 1 =^1 2
πh 1 ρ 1 r^41 0.
Mass wrong -0. Factor 1/2 wrong in formula for moment of inertia of a disk -0. Correct answer for moment of inertia of `excess' disk:
I 2 =
2 πh^2 (ρ^2 −^ ρ^1 )r
Using Steiner's theorem:
IS = I 1 + I 2 + d^2 πr 22 h 2 (ρ 2 − ρ 1 ) 0.
correct solution:
IS =^1 2
πh 1 ρ 1 r^41 +^1 2
πh 2 (ρ 2 − ρ 1 )r 24 + b
2 M 2
πr 22 h 2 (ρ 2 − ρ 1 )
Balancing the forces, correct equation
gE = ω ss^2 R 0.
Correct solution ωss =
gE /R 0.
B2 (0.2 pt) Assuming that on Earth gravity is constant with acceleration gE , what would be the angular oscillation frequency ωE that a person on Earth would measure?
Solution B2: [0.2]
Realize that result is independent of gE 0.
Correct result: ωE =
k/m 0.
B3 (0.6 pt) What angular oscillation frequency ω does Alice measure on the space station?
Solution B3: [0.6]
some version of the correct equation for force
F = −kx ± mω ss^2 x 0.
getting the sign right F = −kx + mω ss^2 x 0.
Find correct dierential equation
m¨x + (k − mω ss^2 )x = 0 0.
Derive correct result ω =
k/m − ω^2 ss 0.
Using gE /R instead of ω^2 ss is also correct.
B4 (0.8 pt) Derive an expression of the gravity gE (h) for small heights h above the surface of the Earth and compute the oscillation frequency ω˜E (linear approximation is enough). The radius of the Earth is given by RE.
Solution B4: [0.8]
gE (h) = −GM/(RE + h)^2 0.
linear approximation of gravity:
gE (h) = −
GM
R^2 E^ + 2h
GM
R^3 E^ +^...^ 0.
Realize that gE = GM/R^2 E :
gE (h) = −gE + 2hgE /RE +... 0.
Opposite sign is also correct, as long as it is opposite in both terms. Realize what this means for force, i.e. that the constant term can be eliminated by shifting the equilibrium point: F = −kx + 2xmgE /RE 0.
Find correct dierential equation
m¨x + (k − 2 mgE /RE )x = 0 0.
correct result ω ˜E =
k/m − 2 gE /RE 0.
No points are deducted if student answers with ω˜E /(2π) because oscillation frequency might also be interpreted as inverse period.
B5 (0.3 pt) For what radius R of the space station does the oscillation frequency ω match the oscillation frequency ω˜E on the surface of the Earth? Express your answer in terms of RE.
Solution B5: [0.3]
Write down equation ω^2 ss = 2gE /RE 0.
Solve R = RE / 2 0.
If GM/R^2 E rather than gE is used, give only 0.1pt.
B6 (1.1 pt) Calculate the horizontal velocity vx and the horizontal displacement dx (relative to the base of the tower, in the direction perpendicular to the tower) of the mass at the moment it hits the oor. You may assume that the height H of the tower is small, so that the acceleration as measured by the astronauts is constant during the fall. Also, you may assume that dx H.
- dx With the notation from the calculation of vx
dx = (α − φ)R 0.
φ = arccos
1 − H
R
α = ωsst where t is the fall time of the mass, which is given by
t =
R^2 − (R − H)^2
ωss(R − H) 0.
(see solution to B7) Writing ξ ≡ H/R this means
dx =
[ √
1 − (1 − ξ)^2 1 − ξ
− arccos(1 − ξ)
]
R 0.
which is a valid end answer to the problem. It is possible, but not necessary, to approximate this for small ξ:
arccos(1 − ξ) ≈
2 ξ
1 + ξ 12
which after insertion into the equation for dx and approximation of small ξ yields the same result as in Solution one:
dx =
2 H^3
R
If this end answer misses the factor 2/3, deduct 0.1 points. -0.
Solution three Inertial frame with geometry trick This is an alternative solution to obtain dx The mass travels the distance l, and during the fall the space station rotates by φ, see Figure 2. According to the intersecting chord theorem,
l^2 = H(2R − H) 0.
The rotated angle is φ = ωsst where
t = l R − H
is the fall time. Thus
φ =
H(2R − H)
R − H 0.
d R =^ φ^ −^ arcsin^
l R =
H(2R − H)
R − H −^ arcsin^
x(2 − x) 0.
Figure 1: Notation for solution two
Figure 2: Notation for solution three.
Denote x ≡ H/R and y ≡
x(2 − x). Since
arcsin y ≈ y + y
3 6
one gets
d R ≈^ y(1 +^ x)^ −^ y^ −^ y
(^3) /6 = y(x − y (^2) /6) ≈ 2 xy/ 3 ≈ 2 x√ 2 x/3 =^2 3
2 H^3
R
Final answer 0.
B7 (1.3 pt) Find a lower bound for the height of the tower for which it can happen that dx = 0.
Solution B7: [1.3]
The key is to use a non-rotating frame of reference. If the mass is released close enough
Figure 3: Plot of f (H/R) and H/R
Figure 4: Plot of g(x) and x
- introduce new variable x := 1 − H/R, so that the equation becomes
x = cos(
1 − x^2 /x) =: g(x)
- g(x) is then smaller than x up to the rst solution. In particular it is negative in some region (see gure 4). Finding the third zero thus gives a lower bound for the solution: (^) √ 1 − x^2 x
= 5π/ 2
x = 1/
25 π^2 /4 + 1 ⇒ H = R(1 − 1 /
25 π^2 /4 + 1) ≈ 0. 874
Note: the actual result is H/R = 0. 871.. .. Use the same points for the numerical answer as was mentioned in solution one. 0. If the student plots f rather than g, nd solution to f = 1: is equivalent to the solution above. Give same number of points. It is also possible to use cos
1 −x^2 x
= sin(1/x).
B8 (1.7 pt) Alice pulls the mass a distance d downwards from the equilibrium point x = 0, y = 0, and then lets it go (see gure 4).
- Give an algebraic expression of x(t) and y(t). You may assume that ωssd is small.
- Sketch the trajectory (x(t), y(t)), marking all important features such as amplitude.
Solution B8: [1.7]
Note: we did not specify the overall sign of the Coriolis force. Give same amount of points if using opposite convention, but it has to be consistent! Otherwise: subtract 0.1pt for each instance of inconsistency. -0. Students are allowed to express everything in terms of√ ω, they don't need to write k/m − ω ss^2 explicitly. Deduct 0.1pt however if they use k/m instead of ω.. -0.
Realize that y(t) is standard harmonic oscillation:
y(t) = A cos ωt + B 0.
Give correct constants from initial conditions
y(t) = −d cos ωt 0.
Correct expression for vy(t): vy(t) = −dω sin ωt 0.
Coriolis force in x-direction
Fx(t) = 2mωssvy(t) = − 2 mωssdω sin ωt 0.
Realize that this implies that x(t) is also a harmonic oscillation... 0.
... but with a constant movement term superimposed: vt 0.
getting the correct amplitude:
A =^2 ωssd ω
Correct answer with correct initial conditions:
x(t) =^2 ωssd ω
sin ωt − 2 ωssdt 0.
Sketch:
x
y
-d
d
A
B 4 πω^ ss^ d ω C
