Vector mechanics PHY1021: Jan 2011 exam-Hints and tips

Lecturer: Dr. Stavroula Foteinopoulou

1 (i)W < 0 if force causes a !KE < 0 contribution.Friction is not possible to do negative work in translational motion.W = !KEW =

! B

AFxdx =

! B

Amaxdx =

! B

Amdvx

dtdx =

! B

Avxdvx = KB ! KA

1 (ii)minimum velocity at the bottom of the circle vA is:

0 ! 1

2mv2

A = mg(2L) "vA = 5.6 ms!1

Force at the top of the circle is:

FRod = mg = 4.9 NFirst find velocity at this point vB from work energy theorem, when mass

has completed a quarted of a circle.

1

2mv2

B ! 1

2mv2

A = !mgL " vB =#

2gL

Force of rod, when the mass has compelted a quarter of a circle is:

FRod =mv2

B

L" FRod = 9.8 N

2(i)

1

(a)v = (4i + 4j) ms!1

(b)a = dvdt

= (2i + 4j) ms!2

(c) a = a

|a| = 1"20

(2i + 4j)

(d) F = ma = (2i + 4j) N(e) v1 = |v(2s)| = 4

!2 ms!1

v2 = |v(4s)| =!

320 ms!1

From work-energy theorem:W = 1

2m(v2

2 " v21) = 144 J

(f) v1 = v(2s) = (4i + 4j) ms!1

v2 = v(4s) = (8i + 16j) ms!1

Magnitude of impulse is:|J| = |mv2 " mv1| = 12.65 Kgms!1

Direction- find angle ! between J and x-axis.cos(!) = J·i

|J| # ! = 71.60

2 (ii)The location of the support should be the center of gravity, which is the

same as the center of mass (constant g), and has coordinate xCM :

xCM = 1

M1+M2+M3(M1

L4

+ mL2

+ M23L4

)# xCM = 1.714 m

2

3

[3]

zero [2]

Position A: cylinder on top of inclinePositon B: cylinder at the bottom of inclineConservation of mechanical energy:KA + UA = KB + UB ! UA = KB !Mgh = 1

2Mv2

CM + 1

2I!2

since vCM = !R and I = 1

2MR2 we have:

vCM =!

4

3gh ! vCM = 4.43ms!1

2vCM = 8.85ms!1

We have:

3

! = fsR = I|a|/R So Since:I = 1

2MR2

we get:fs = 1

2M |a|

But:M |a| = Mgsin" -fs

which together gives:fs = 1

3Mgsin" = 6.54 N

For rolling without slipping to occur friction must be static. So:fs ! µsN = µsMgcos"So we get for the coe!cient of friction µs,µs " 0.192

4(i)

(a) This is the only way to conserve momentum along the y-direction(perpendicular to x-direction).(The momentum along the y-direction is initially zero).

(b)Conservation of momentum in the y-direction:mvf1sin450 = mvf2sin450 # vf1 = vf2

Accordingly, conservation of momentum in the x-direction becomes:mv = (mvf1 + mvf2)cos450

The two above equations yield:

vf1 = vf2 = vf = v!2

(c)

4

Ki ! Kf = 1

2mv2 ! 21

2mv2

f = 1

2mv2 ! m( v!

2)2 = 0

Which means initial and final kinetic energy are equali.e. collision is ellastic.

4(ii)(a) ! = RF = 1N · m(b) W = !!" = 4# J(c) Work energy theorem for rotational motion:1

2I$2 ! 0 = W

which gives $ = 7.1rads"1

(d) $ = $0 + %t with $0 = 0, and % = !I

which givest = 1s

5 (i)(a) The work of a force must be indepedent of the pathfor the force to be conservative.

(b)K + U = constantK: Kinetic energy U: Potential energy

(c) No, only when only conservative forces are present.In the case that non-conservative forces are presentmechanical energy is still conservedif the latter contribute zero work(e.g. friction in rolling without slipping motion, normal force)

(d)

5

Work energy theorem:

WA!B = KB ! KA

Conservative force:WA!B = UA ! UB

which indicates that work depends only on start and end pointbut not on the particular path taken. So:KB ! KA = UA ! UB

which gives:UA + KA = UB + KB

which expresses conservation of mechanical energy

5 (ii)Fr = !dU

dr=

= 12U0

R0[(R0

r)13 ! (R0

r)7]

Fr = 0 which occurs ar: r = R0

6

6(i)proper time:Time interval between two events measured in a reference framewhere the events took place at the same location.proper length:The length of an object in a reference frame where its end points are not

moving, or the distance between two points in a reference frame where theyare stationary.

7

6(ii)

O represents the event (at x = 0, t = 0).The black lines represent x = ct and x = !ct!s2 = c2!t2 ! !x2

6(iii)(a) Lorentz scalar:A quantity that is invariant under the Lorentz transformation.(The norm (or norm squared) of a Minkowskian four-vector,is a Lorentz scalar, as all scalar products between four-vectors).(b)P = (E/c,p)So: P · P = E2

c2! p2

But:P · P is a Lorentz scalar so:E2

c2! p2 is a Lorentz scalar.

Using the enery-momentum relationship we get that m0c2 is a Lorentzscalar,

and since the speed of light in vacuum is invariant

8

(Einstein’s second postulate),the rest mass m0, must also be an invariant.

6(iii)m = !m0 ! ! = 2! = 1

q

1! v2

c2

Sov = 0.866 c

9

Charles Williams

Typo - should be 2011

PHYSICS EXAMINATION PROBLEMS

SOLUTIONS AND HINTS FOR STUDENT SELF-STUDY

Module Code PHY3129

Name of module Device Physics

Date of examination January 2011

1.(i) Luminescence definitions, operation of electroluminescent device - see lecture

notes.

ac operation avoids formation of a space charge that cancels the accelerating

electric field within the particle

The non-linear (threshold type) dependence of optical emission upon drive voltage

favors matrix addressing: there is minimal emission from half-selected cells.

1(ii) Rate of injection of angular momentum

!

"!J

e

ˆ M 1

The absorbed transverse component is

!

" =#!J

e

ˆ M 1 #ˆ M 1 $

ˆ M 2( ) ˆ M 2( )

Using

!

a " b" c( ) = b a # c( ) $ c a #b( ),

the torque is rewritten as

!

" =#!J

e M22d2

M2 $ M2 $ˆ M 1( ) per unit volume of layer 2,

which has the same vector form as the Landau Lifshitz damping if

!

ˆ M 1 is parallel to

B.

Critical current density is

!

Jc =" eµ0M2

2d2

! so Ic = 1.2 mA, and Oersted field at

perimeter is 24 mT.

2.(i) Director definition, elastic deformations, HAN cell – see lecture notes

tdecay = 65 ms, trise = 0.28 ms. These values are well-suited to a video rate

application. A pixel can be switched on quickly and then holds its state until

refreshed at 50 Hz rate.

2(ii) Operation of bipolar transistor – see lecture notes.

The common base circuit has low input and high output impedances and is

commonly used for impedance matching.

Base transport factor, emitter efficiency – see lecture notes. Ic/Ib = 33.

3.(i) Hard disk drive operation – see lecture notes.

Relaxation time obeys and Arrhenius relation

!

" = "0 exp µB kBT( ) where the

magnetic moment µ is the product of the magnetization and the particle volume,

and !0 = 1/fFMR. Setting ! = 10 years, critical diameter = 9.6 nm.

Reduced grain sizes require larger anisotropy for thermal stability, which then

requires larger write fields.

3(ii) Operation of MESFET – see lecture notes.

ID = 0 when Ld = 0.15 µm, so Nd = 2.8 " 1022 m#3

Begin from

!

Q = AeNdLd and differentiate.

4(i) Derivation of guiding equation – see lecture notes.

($1)min = 80.8°, 12 allowed modes (counting both possible polarization states)

Intermodal dispersion (see lecture notes) yields a pulse duration of 675 fs. Shorter

pulse duration may be achieved by reducing the core thickness so that the

waveguide supports only a single mode.

4(ii) MOSFET and CMOS – see lecture notes

In 3 years the rate of heat dissipation will increase by a factor of 4 under constant

voltage scaling.

PHY3148

The nature of the material and hence the

examination means Solutions and Hints are not

applicable to this module.

The level of detail expected in each answer is

that given in the lectures. The answer is expected

to provide a focussed and logical response to the

question asked, rather than a random selection of

more-or-less relevant facts regurgitated from the

notes. For more guidance contact the lecturer.

1

PHYSICS EXAMINATION PROBLEMS SOLUTIONS AND HINTS FOR STUDENT SELF-STUDY

Module Code PHYM432

Name of module Relativity and Cosmology

Date of examination January 2011

1.

2. First law of thermodynamics, connect a change in volume to a change in total energy for an

ideal cosmological fluid.

Substitute and take the time derivative to get the fluid equation

Use conservation of energy for a test particl of unit mass on the surface of a unifomr ideal

sphere enclosing mass M at density !.

2

3. Constant comes from GR and is

!

"kc2. The energy equation:

In the LI, the metric tensor in the Minkowski metric, transform to the accelerated frame:

Of the 16 terms, only the diagonal terms are non-zero

Eqn of motion in accelerated frame is

non-zero terms are for the equation with

!

" =1. Using the formula gives

with a diagonal metric

!

g11 = g11[ ]"1 so

4. Singularity at

!

r = RS = 2GM c2 , which is a coordinate singularity and at

!

r = 0 , which is a gravitational singularity.

!

RS = 2GM c2 = 2969 m