Columbia UniversityDepartment of Physics

QUALIFYING EXAMINATION

Monday, January 13, 20141:00PM to 3:00PMClassical Physics

Section 1. Classical Mechanics

Two hours are permitted for the completion of this section of the examination. Choose4 problems out of the 5 included in this section. (You will not earn extra credit by doing anadditional problem). Apportion your time carefully.

Use separate answer booklet(s) for each question. Clearly mark on the answer booklet(s)which question you are answering (e.g., Section 1 (Classical Mechanics), Question 2, etc.).

Do NOT write your name on your answer booklets. Instead, clearly indicate your ExamLetter Code.

You may refer to the single handwritten note sheet on 812” × 11” paper (double-sided) you

have prepared on Classical Physics. The note sheet cannot leave the exam room once theexam has begun. This note sheet must be handed in at the end of today’s exam. Pleaseinclude your Exam Letter Code on your note sheet. No other extraneous papers or booksare permitted.

Simple calculators are permitted. However, the use of calculators for storing and/or recov-ering formulae or constants is NOT permitted.

Questions should be directed to the proctor.

Good Luck!

Section 1 Page 1 of 6

1. A solid cylinder of mass M and radius R, rotating with angular speed ω0 about an axisthrough its center, is set on a horizontal surface for which the kinetic friction coefficientis µk.

(a) Calculate the linear acceleration of the center of mass, and the angular accelerationof rotation about the center of mass.

(b) The cylinder is initially slipping completely. Eventually, the cylinder is rollingwithout slipping. Calculate the distance the cylinder moves before slipping stops.

(c) Calculate the work done by the friction force on the cylinder as it moves from whereit was set down to where it begins to roll without slipping.

Note: the cylinder has rotational inertia Icm = 12MR2

Section 1 Page 2 of 6

2. In the figure below, a 20 kg block is held in place via a pulley system. The person’supper arm is vertical; the forearm makes angle θ = 300 with the horizontal. Forearmand hand together have a mass of 2 kg, with a center of mass at distance d1 = 10 cmfrom the contact point of the forearm bone and the upper-arm bone (humerus). Thetriceps muscle pulls vertically on the forearm at distance d2 = 4.9 cm from the contactpoint. Distance d3 is 30 cm. What are the forces on the forearm from (a) the tricepsmuscle and (b) the humerus? Take g = 9.8 m/s2.

Section 1 Page 3 of 6

3. A spider is hanging by a silk thread from a tree in NYC. Find the orientation and thevalue of the equilibrium angle that the thread makes with the vertical (i.e. with thedirection of gravity), taking into account the rotation of the Earth.

Assume that the latitude of NYC is θ ≈ 410 and the radius of the Earth is R ≈ 6, 400km.

Section 1 Page 4 of 6

4. A swing of mass m made from an arc section of radius of R is suspended from a pivot byropes at both ends. A hoop, also of mass m, and radius a rolls without slipping on theswing. The swing and the hoop move without dissipative friction subject to a constantgravitational force Fg = −mgj. Find all possible frequencies of oscillation of the systemfor small displacements from equilibrium assuming a/R 1.

Section 1 Page 5 of 6

5. Consider the diatomic molecule oxygen (O2) which is rotating in the xy plane about thez axis. The z axis passes through the center of the molecule and is perpendicular to itslength. At room temperature, the average separation between the two oxygen atoms is1.21 × 10−10m (the atoms are treated as point masses).

(a) Calculate the moment of inertia of the molecule about the z axis.

(b) If the angular velocity of the molecule about the z axis is 2.00 × 1012 rad/s, whatis its rotational kinetic energy? The molar mass of oxygen is 16 g/mol.

Section 1 Page 6 of 6

2014 Quals Question: Mechanics (Dodd)

A solid cylinder of mass M and radius R, rotating with angular speed ω0 about an axis through its center, is set on a horizontal surface for which the kinetic friction coefficient is μk.

a). Calculate the linear acceleration of the center of mass, and the angular acceleration of rotation about the center of mass.

b). The cylinder is initially slipping completely. Eventually, the cylinder is rolling without slipping. Calculate the distance the cylinder moves before slipping stops.

c). Calculate the work done by the friction force on the cylinder as it moves from where it was set down to where it begins to roll without slipping.

Note: the cylinder has rotational inertia 21cm 2 .I MR=

Solution:

As the cylinder is placed on the surface, we have the following forces acting:

a). The friction force is k k k, so .f n Mg a gµ µ µ= = =

The magnitude of the angular acceleration is k k2

2 .(1/2)MgR gfR

I RMRµ µ

= =

b). Setting 0( )v at R t Rω ω α= = = − once the cylinder reaches the rolling condition, and solving for t gives

0 0 0

k k k,

2 3R R Rta R g g gω ω ωα µ µ µ

= = =+ +

and

2 2 22 0 0

kk k

1 1 ( ) .2 2 3 18

R Rd at gg g

ω ωµ

µ µ

⎛ ⎞= = =⎜ ⎟

⎝ ⎠

c). The final kinetic energy is 2 2(3/4) (3/4) ( ) ,Mv M at= so the change in kinetic energy is

22 2 2 20

k 0 0k

3 1 1 .4 3 4 6

RK M g MR MRg

ωµ ω ω

µ

⎛ ⎞Δ = − = −⎜ ⎟

⎝ ⎠

So the work done by the friction force is

W =16MR2ω0

2.

1 Mechanics Problem

In Figure 1, a 20 kg block is held in place via a pulley system. The person’s upper arm is vertical; theforearm makes angle θ = 30 with the horizontal. Forearm and hand together have a mass of 2 kg, with acenter of mass at distance d1 = 10 cm from the contact point of the forearm bone and the upper-arm bone(humerus). The triceps muscle pulls vertically on the forearm at distance d2 = 4.9 cm from the contactpoint. Distance d3 is 30 cm. What are the forces on the forearm from (a) the triceps muscle and (b) thehumerus? Take g = 9.8 m/s2.

Figure 1: A person holding a weight.

1.1 Solution

Let’s assume the triceps applies an upward force Ft and the force Fh from the humerus is downward. Forpart a, look at the torques around the pivot point:

τnet = 0 = (4.9cmFt + 10cmFg − 30cmT ) sin30. (1)

This yields Ft = 1160N, so Ft points upwards indeed. For part b, along the vertical we have

Ft + T − Fg − Fh = 0, (2)

so Fh = 1336.4 N and Fh indeed points downward.

2 Qualitative Problem

Find, by using dimensional analysis, the ratio of the muon and neutron lifetimes. The muon decays throughµ− → νµe

−νe and the neutron through d→ ue−νe. Assume the form of the weak hamiltonian to be

Hwk =GF√

2ψ1Oψ2ψ3Oψ4, (3)

with O some operator. GF ≈ 10−5/m2p with mp the mass of the proton. The muon, electron, up and down

quark masses are 105, 0.5, 2.3 and 4.8 MeV, respectively. You can neglect the neutrino masses.

2.1 Solution

The decay width Γ will be proportional to G2F , but it also has dimension energy (or mass). So we must have

Γ ∼ G2Fm

5, where m is the characteristic mass of the process, i.e. the mass difference between the initialand final states. For muon decay, that is ≈ 105 MeV, and for neutron decay ≈ 2.5 MeV. The ratio of the

muon and neutron lifetimes is thus ∼ (100)5

(2.5)5 ≈ 108

CLASSICAL PHYSICS – MECHANICS

A spider on a thread. SOLUTION.

Since the spider is stationary, the thread will be deflected from the verticaldue to the centrifugal force, but not the Coriolis force. The force on the spiderfrom the thread is

~Fs = −~Fg − ~Fc = Mgz +M~Ω × (~Ω × ~R). (1)

Here, the z axis points vertically through the spider away from the center of theEarth, and ~R = Rz. We also define an orthogonal x axis through the spider,such that the Earth’s angular speed is

~Ω = Ω(z sin θ − x cos θ), (2)

i.e. x points 90 south of z. We find

~Ω × ~R = ΩR cos θy (3)

and~Ω × (~Ω × ~R) = −Ω2R cos θ(x sin θ + z cos θ). (4)

Then from eq. (1) we obtain

~Fs = Mz(g − Ω2R cos2 θ) − (MΩ2R/2)x sin(2θ). (5)

Thus the angle is south of the vertical, and its value is

α = tan−1

[(Ω2R/2) sin(2θ)

g − Ω2R cos2 θ

]≈ tan−1

[Ω2R sin(2θ)

2g

]. (6)

Substituting the numerical values, we obtain

α ≈ 1.7 mrad ≈ 0.1. (7)

Here we used the rate of the Earth’s rotation Ω = (2π rad)/(24 h) ≈ 7.3× 10−5

rad/s.

1

ProblemA swing of mass m made from an arc section of radius of R is suspendedfrom a pivot by ropes at both ends. A hoop, also of mass m, and radius arolls without slipping on the swing. The swing and the hoop move withoutdissipative friction subject to a constant gravitational force Fg = −mgj.Find all possible frequencies of oscillation of the system for smalldisplacements from equilibrium assuming a/R << 1.

R

SolutionsRepresent the motion of the system in terms of two angles that representthe angular displacements of the swing and hoop from equilibrium, φ and θ,respectively. The rolling of the hoop can be described by an angularvelocity ωhoop which will be taken to be positive for the hoop rollingcounter-clockwise.. The rolling of the hoop depends on the differencebetween φ and θ since if θ and φ increase or decrease together, the hoopremains in the same position with respect to the swing. Then, the rolling

without slipping condition is aωhoop = R(φ− θ

). The kinetic energy of the

system is then,

T =1

2m(R2φ2 + (R− a)2θ2

)+

1

2Ihoop ω

2, (1)

1

with Ihoop = ma2. Substituting a2ω2 = R2(φ− θ

)2and taking R− a→ R,

T = mR2(φ2 + θ2 − θφ

)(2)

The potential energy of the system can be expressed

U = mgR (1− cosφ) +mg(R− a) (1− cos θ) . (3)

Again taking R− a→ R the complete Lagrangian for the system is

L = mR2(φ2 + θ2 − θφ

)−mgR (2− cosφ− cos θ) . (4)

Writing out the Euler-Lagrange equations

d

dt

(∂L

∂φ

)− ∂L

∂φ= mR2

(2φ− θ

)+mgR sinφ = 0, (5)

d

dt

(∂L

∂θ

)− ∂L

∂θ= mR2

(2θ − φ

)+mgR sin θ = 0. (6)

Taking the small displacement limit, sin θ → θ and sinφ→ φ, we obtaincoupled harmonic oscillator equations of motion,

2φ− θ = − gRφ, (7)

2θ − φ = − gRθ. (8)

There are many ways to solve for the frequencies of such a system. Takinga “brute force” but straight-forward way, assume normal mode solution(s)such that φ = c1e

iωt and θ = c2eiωt. For convenience, define γ = g

Rω2 . Then,

−ω2 (2c1 − c2) = − gRc1 → 2c1 − c2 = γc1 (9)

−ω2 (2c2 − c1) = − gRc2 → 2c2 − c1 = γc2 (10)

Re-writing this as an eigenvalue equation,[2− γ −1−1 2− γ

] [c1c2

]= 0 (11)

We obtain the characteristic equation by requiring

Det

[2− γ −1−1 2− γ

]= 3− 4γ + γ2 = 0→ γ = 1, 3. (12)

So, the two possible frequencies are ω =

√g

Rand ω =

√g

3R.

2

Quals Problem #1

January 2014

Problem

Consider the diatomic molecule oxygen (O2) which is rotating in the xy plane about the z axis. The zaxis passes through the center of the molecule and is perpendicular to its length. At room temperature,the average separation between the two oxygen atoms is 1.21 × 10−10 m (the atoms are treated as pointmasses). (a) Calculate the moment of inertia of the molecule about the z axis. (b) If the angular velocity ofthe molecule about the z axis is 2.00× 1012 rad/s, what is its rotational kinetic energy? The molar mass ofoxygen is 16 g/mol.

1

Suggested Solution to Quals Problem #1

January 2014

Solution

(a) I =∑

mir2i = m(d/2)2+m(d/2)2 = md2/2 = (2.66×10−26 kg)(1.21×10−10 m)2/2 = 1.95×10−46 kg·m2

(b) K = Iω2/2 = (1.95× 10−46 kg ·m2)(2.00× 1012 rad/s)2/2 = 3.89× 10−22 J.

2