Mechanics Solutions

5/14/2018 Mechanics Solutions

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Classical MechanicsDepartment of PhysicsUniversity of New Hampshire Comprehensive ExamAugust 2008

Do a I/ of the following 3 problems. Be sure to explain your work (it could lead topartial credit). You can use a calculator, your one sheet of notes, a table ofintegrals, and Abramowitz and Stegun if you wish. There are 50 total points forthis Classical exam, and the point value of each question is shown.

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Question 1 (17 points)

A small 0.1 kg block slides down a frictionless surface through a height h=0.2 mand then sticks to a uniform rod of mass 0.2 kg and length d=O.4 m, as shown inthe figure. The rod then pivots about point O. Find the maximum angle, e ,reached by the rod.

Question 2 (16 points)Consider the potential energy of a system given by V=Va (1 -e xp (-(X -Xa )/ a )/ - V aFor this systema. (4 pts) What is the equilibrium position in terms of Va , a and x a?b. (6 pts) What are the angular frequencies of small oscillations about thisequilibrium position for a particle of mass m?c. (2 pts) Show that the units on your last answer are correct.d. (4 pts) Now consider the potential V(x) = 3;; + 6X4. For what values of x can aparticle in this potential exhibit simple harmonic motion about the equilibriumposition of x=O? Explain your reasoning beginning with the definition of simpleharmonic motion.


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I TL ff"f'- \" - n o . . . : . t { M ] t~til _tj . " - ~ - - LJ---tt--___;._- - - ! - - - i -h- :: :;:, t ; : J - f J Mr:; ~ fLI fJ", v:: ( ;:: L C J- 1M d~ ) 6 1 :

: !

-j

~ : . : .~

k t s d - r r -r/'1/wur.

,\ I W:: '!jr - ,______----t----------+(-hM-:::::-_ ~1. _ .: - + ) - -- - - ~


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r>:I t/(,I 11 . / 'I LJII

'l(\\II)iI!r

( ' v ' - hr e rv a.-f~!0 / r ' - . ...... I}!r -f -- -L

~HJ

,1-, t} 1y T

. . . . . - " fd

' l ~-r:l

!--


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\ ; / . 7 \ "-, \"

/-( t f4' " '

....- .-.-

-I / )'r c ,. j i , t f -1 , . . u: '-'

\ . \'\\


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C.{u~ fCCJl ( )~VV le(-~JLJl -l.l) (,)~


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Question 3 (17 Points)A wedge of mass ml with a inclined face rests on a flat surface as shown. Theangle of the inclined face is 45 degrees. A block of mass m2 rests on theinclined face. To a very good approximation, there is no friction between thewedge and the block. A cord, which may be approximated as massless, runsfrom the block, over a pulley, to a fixed point. The cord is parallel to the inclinedface of the wedge between the pulley and the block; it is horizontal between thefixed point and the pulley. The mass, moment of inertia and friction of the pulleyare negligible. The coefficient of friction between the flat surface and the wedge,u, is large enough to prevent the wedge, and hence, the block, from moving.The whole is in a uniform downward gravitational field, g.

Aj

ml

(a) (2 points) What is the magnitude of the tension on the cord?(b) (2 points) What is the normal force exerted by the block, m2, on the face

of the wedge? (Please express the answer using the unit vectors shown.)(c) (3 points) What force is exerted by the pulley on the wedge? (Pleaseexpress the answer using the unit vectors shown.)(d) (4 points) What is the frictional force exerted by the flat surface on thewedge?(e) (3 points) What is the normal force of the wedge on the flat surface?(f) (3 points) What condition applies to the coefficient of friction between the

flat surface and the wedge, u, for the system to remain static?


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(a) (2 points) What is the magnitude of the tension on the cord?The tension of the cord must balance the component of the gravitational force along theface of the wedge. Thus,

T =m2gsin(45)m2g=-Ji

(b) (2 points) What is the normal force exerted by the block, m2, on the faceof the wedge? (Please express the answer using the unit vectors shown.)

The tension and the inverse of the normal force (the wedge pushing on the block) mustbalance the force of gravity:N ( ~ + ~ ) + 1 - ~ + ~ ) - m 2 g ) ~ OFrom the icomponent. .1 IN-Ji-T-Ji=O

N=T

So, N ~ 7 z ( ~ + ~ )1112g" ":)==-2- 1+]

This the wedge pushing on the block, the the negative is just the block pushing on thewedge.- m2g (" ":)N 2=-T 1 +t

Note, from the j component,A Aj j AN-+T--m2gj=O-Ji -Ji

Aj A2T -Ji:= m2gjT- m2g--Ji

confirming the solution to (a).


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(c) (3 points) What force is exerted by the pulley on the wedge? (Pleaseexpress the answer using the unit vectors shown.)This force comes from the vector sum of the forces from the two sections of cord:

(

A A 1A i jF = = - Ti + T - J 2 - - J 2A Amzg" i j=---1 + mzg- - mzs':- J 2 2 2

=m~g[(1 - - J 2 ) i - ] ](d) (4 points) What is the frictional force exerted by the flat surface on thewedge?The frictional force has to counter the horizontal forces on the wedge from thecomponents of the normal force from the block and the pulley:

F = = _N_l + mzg ( 1 - - J 2 )x - J 2 2=_m;g + m;g ( 1 - - J 2 )

m2g=- - J 2Which is just the tension, as one would expect. Thus the friction has to exert a force of

- m2g,:,Ff=-J21(e) (3 points) What is the normal force of the wedge on the flat surface?Since, overall, there is no force other than gravity in this component, one would expect:

Fy = = - (ml + m2)gThis can be confirmed by adding the three forces on the wedge: gravity, the force fromthe pulley and the normal force from the block.

m2g 1r - ==-mlg--2--N-J2mzg m2g=-mlg------2 2

=-(ml + m2)g(f) (3 points) What condition applies to the coefficient of friction between theflat surface and the wedge, 1 1 , for the system to remain static?We have the horizontal and normal forces, so:


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m2gfl(ml + m2)g > Jim2


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Problem: A particle of mass m and angular momentum L is pulled towards afixed point P with a force that is given by F = -4ar--5p, where r is the particle'sposition relative to point P, r = 1 ; : 1 , P r / r, and a is a constant.

(a) If the particle moves on a circular orbit, find the radius R of that orbit.(b) Suppose that the particle's radial coordinate satisfies

r(t) R +x(t ) ,where R is the radius of a circular orbit found in part (a), and

x(t) R.Furthermore, suppose that at t =0, x = X o and dx] dt =0. Solve for x ( t ) .


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SolutionPart a. The total particle energy is

mv:E =T + Ueff(r),where the effective potential energy is given by

L2 aUeff(r) =--2--4 '2mr rThe radius of the circular orbit corresponds to dUe ff/ d r =0, where

dUef f L2 4a--=--+-dr mr3 r5'The radius of a circular orbit is thus given by

R= (4amYoPart b. When 1 ' ( t ) = R + x( t) with x R, the equation of motion for the

particle can be writtenm d2x dUeff r eo. ]'2 dt2 =~(R+x) c : : : -l--;J,T(R) - x ,

where we have Taylor expanded Ueff about r = R, keeping only the largest non-vanishing term in the expansion. We thus obtain

d2xdt2 = y x ,

2

where L3'1= 2m2a'

The general solution for x( t) is then x( t) = cJ e Y I +C2e-Y1 , where C 1 and C 2 areconstants, The particular solution that satisfies the initial conditions x ( 0) =X o anddx (0 ) = isdt X o (x( t) =- eyr + e-Yt ) ,2


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Problem:A ball of radius r starts from rest at a height h, and rolls down a track and around a loop-the-loop of radius R, as shown in the figure below. The ball has a mass m and radius r.

a) What is the minimum velocity that the particle must have at the top of the loop-the-loop, for it to make it around without falling off?b) What minimum height, h, must the track have for the ball to make it around theloop-the-loop.

m . radius r

h


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Problem with solution.A ball of radius r starts from rest at a height h, and rolls down a track and around a loop-the-loop of radius R, as shown in the figure below. The ball has a mass m and radius r.

a) What is the minimum velocity that the particle must have at the top of the loop-the-loop, for it to make it around without falling off?

When the ball is in the loop-the-loop, it is going around in a circle with radius (R-r). Atthe minimum velocity at the top of the loop, the only force on the ball is gravity. Thusthe condition for making it around the loop is

v 2mg=m---(R - r)or

b) What minimum height, h, must the track have for the ball to make it around theloop- the-loop.

Conservation of energy:Initial energy: E, = mgh.

1 2 1 2 (Final energy: Ef=-mv +-lw +mg2R-r). 2 22 2where w = v! rand 1 ee - mr for a sphere.5 e.s ,

( )

21 2 1 2 :2 Vmgh -mv + -(-mr ) - + mg2(R - r)2 2 5 r7 2gh = = -v + 2g(R - r)10

h = = 2.7(R - r)

Plug in for v2 from part a).7gh = = - g(R - r)+ 2g(R - r)10


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Classical Mechanics Comprehensive ExamJanuary 2007

1. (20 pts total) A particle of mass m moves in an attractive 1/ r (NOT 1 Ir 2] 1 )central force of-+ kF=--e rr

a) (5pts) Taking the motion as confined to a plane, find a Lagrangian usingpolar coordinates. Do not assume that angular momentum is conserved

b) (5pts) Determine the equations of motion using Lagrange's equation.Show that the canonical angular momentum is conserved.

c) (5pts) For an angular momentum, Z O ' determine the radius, to ' for acircular orbit.

d) (5pts) Determine the stability of such an orbit for small perturbations.

2. (15 pts) In the figure below, the spring has a spring constant of 1000 N/m. It iscompressed 15 em, then launches a 200 g block. The horizontal surface is frictionless.The block's coefficient of kinetic friction on the incline is 0.20. What distance d does theblock move horizontally before hitting the ground? Assume that the block is a pointparticle so that the transition from the flat plane to the incline is smooth.

Copvnnlu ) :ZOO,]t 'carson Edu cauon. I n c . . r u h [j ~ h in J ! li~Add is on W e s le : >


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3. (15 pts total) Consider a forced, undamped simple harmonic oscillator that movesaccording to the differential equation:

a. (4 points) Solve the unforced (homogeneous) version of this equation.Show your steps; do NOT just write down the answer.

b. (4 points) Solve the forced (inhomogeneous) version of this equation bystarting with the guess

and solve for A and B in terms of constants and parameters.c. (4 points) Given that initial position and velocity are zero, determine the

full solution for this problem (all constants will be specified).d. (3 points) Do you expect the energy of the oscillator to stay constant,

always increase, always decrease, or both increase and decrease? Explainusing both your answer for x(t) and your understanding of the forcingfunction.


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t), (O-~v ,i) \Solutions for the Comprehensive Exam

January 2007

1. A particle of mass m moves in an attractive 1/r central forceof- kF=--e rra) Taking the motion as confined to a plane, find a

Lagrangian using cylindrical coordinates.

The velocity of the particle is1 7 = i " e + r A . e\f / Its kinetic energy is thenT 1 - -= =mv V2

1 ( . 2 2 ' 2 )= -1n r +r 2SinceF=_dVdranddV kdr rIt followsV =klnrand the Lagrangian is then


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L=T-V=lm(f2 r2ip2)-klnr2

b) Determine the equations of motion using Lagrange'sequation, Show that the canonical angular momentum isconserved,d ( a L ) s:-- =-dt a f dra L .-=lnra fa L ' " k- = n~r~--a r r

.So.. ~ ') klnr = lnr-,~-- r

d ( a L ) o t.dt a ip = a a L ' ) .-. = mr+ tpa a L = 0a

Then

Thus,

and


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2 .mr =lwhere 1 is a constant, the angular momentum.

c) For an angular momentum, 1 0, determine the radius, TO)for a circular orbit.p=;=oSo

") kmrtb ' --=0rFromconservation of angular momentum

2 'mr = 1 0Thus

l 2 ko--=-so

or

d) Determine the stability of such an orbit for smallperturbations .

. , A t2 kmr=mr~ -- r


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But, this can be reduced using conservation of angularmomen turn to., l~ kmr=----

1 1 1 1 3 rFor small perturbations abou t 1 '0r = r o + (jP=(jwhich gives

Expanding to first order in (j gives

Sincel~ k--=- ro

It follows


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! : _ ( j')r (;

Since this is less than zero, the orbit is stable. Smallradial perturbations have a period of

2 2 kOJ = -r 2o


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2. Model: Assume the spring to be ideal that obeys Hooke's law, model the block as a particle, and applythe model of kinetic friction,Visualize: We placed the origin of the coordinate system on the free end of the compressed spring which is incontact with the block, Because the horizontal surface at the bottom of the ramp is frictionless, the spring energyappears as kinetic energy of the block until the block begins to climb up the incline.

So lv e : Although we could find the speedv , of the block as it leaves the spring, we don't need to. We can useenergy conservation to relate tbe initial potential energy of the spring to the energy of the block as it begmsprojectile motion at point 2. However, friction requires us to calculate the increase in thermal energy. The energyequation is

The distance along the slope is Ls =Y2/sin45. The friction force is I,= f . 1 x n , and we can see from the free-bodydiagram that n = mgcos45. Thus

V2 =,,\112- xJ' - 2 gy, - 2J.Lkgy , cos 45 I sin45 j

1./2l lOOON/m '2 -> 1= ----(0.15 m) - 2( 9.8 m / s )(2.0 m) - 2(0.20)( 9.8 m Is' )(2.0 m) cos 45 0 I sin45 = 8.09J0.20 kg m IskIn

Having found the velocity v,, we can now find - x,) = d using the kinematic equations of projecti le motionJ 2)'3 = Y 2 + v')' (13 - t2) + "2aZ ) (t3 - 12 )

12.0 m = 2.0 m + ( 1 1 , sin 45 0 ) ( t 3 - tJ + '2 ( -9.8 mis' ) (1 3 - Ij

""" ([3 - (2) = 0 s and 1.168sFinally,

1x = x + v - I)+ I...-3 2 2x 2 ")d = (x, - x2) ~ (v, cos 45)(1.168s) +0 m = 6.68m


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CD

,~ tD )::-O~ C D ) ; _0

7\ A e : \-t -t \6 e. - ~t~ C cos(t) - r 1)~ C t )

~ ~ ~ r C f iV - ~ " K f ~ !\coS1 l- \- \Stjf ~-A 1 - S l . ( V ' - r - c t- ~ - : : rLOS~ G -~ < ~ ~


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1 - - - i)~v~- z : LO ~ ~IfV\

cos ~ ..'t0(0)-;_-O - C + - FbV V [ _ I ! ~ )~ L )~(O)~ 1) "70

r - z . - - ; ; , F0~~. \ (- c . o . s+( , 1'". .-.. 2 . . \


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iI /\ ~ ' t A " A C /' " ". / \ \A t - ' \ V,X,/V' ' ~)\) I.> ._ _ _F \: f\(

,/J l/\/\

t,lJv...t'l..U\/I

u

L \j \ C . ~ \l~ -~-~ I 7 / . , ' I - . ..\ ll/\. Vf

;: )\

. . . ; ,\

,/ C U..,!l.s \ c " - lA . - '- t ,


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, ,'. - fis - = I . . J :=; l.~0. . .


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\J ~ -).M. '\ J " \ . . ~ ~L ': ~(' \ , .1.e1.. + - l\'1..c.u~~l. .e) .lr ) .M . . \ t\ . ~ (6

~ e s ltu)~ ~ ~ - '\ 1:.0~ 8 -:::. ='";> t : J : = : 6)1\~ eo~ ~ ~ tU~'> ~J't u . . J 0 ~ H..

. .e - l . U ! ~e~e - ~~~J\..

~ '~~'> o.J;-)~.~ t~~~~


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M . A . ( f \ . - Sl ~7.. ') - = - 'v L rtNJ\ ( T \ . ~ + 2ft ~') =-j,M, (~~ t'll\h.~') ::0 :-:> w .. ~ ( : n : ~ ' ) ~ ( ) - "JAh.~ :::.L

(o.)~~:L '2 . . : : ' - t . R"'" =-:> \< .M.+~"> L'"I W . , . 'R~ kw.." \ : : : : ? : : : ' - : ' 2 . . 1 \ Mf\ .~~ L

l \ ' - ) z ; . - \ _ ' " . = :. - ~ ~(\-t V\~~"l.R~ ~T ~)~ ~ RJ~ - ~ : T ? : ' >V-~ ~ \"- ~ : R " ' V~ r A ~ ), \ L " 1 . . V \V \ 4- ~~R" 'bRl~~M) =D

( , . \ ~ ~ ~ ~ + , v . . . '> 0 = - ' > M"> -~ &'\. N\ ~ - 1-(0.) 1 \ ~ , ) . V I ; . . n: n _ 7 . . " , L . . . . ' " C " ' > + M . \ ~ ~ ~ . . . . (~M.\

- \ ,M A . . " - ~ ~

~~~ 1A'~ \ . Q _ \ ~ \ > \~ \ 0 . . \ = N\~\ ,~~ ~Q.M_ ~ i~ ~ ~~ \~ \J~ +M :::' N \ \ 4 > \ ~ ~~*;" - . 1

~c~~" \'I\:::c-2)N~\ (1..~~V\~~)M= \ )"-S= (2~o\\\~~

~ ~ ~ f ' A , . . t . . . . < C


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(C d ) ' ) =


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c


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- ,,, \MU~

cl' ~2-

V \

v y J / V J 'V iJ(,t '1('\f


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' ~ , ( N v \ . . ~ ~ V V \ . ' ) ( f \ S - \ - 4)-I\_ l\:j - = cl'v\J'. ( \.) - l\~ - )JA. % bv \" ..~~~ ~.~~ ~U;:;~ ~ )\~l>. rv = 0

=:,? 0--~\Jv\. \J .o - Nv \. ~ ~

c :.-

-. =.

'~~ f\AV-v:> ~ ~ )Jv\ = - ( \ - y\) M\.0

( \ - V \ ' j 0 e - 'f tTL,\. (\- \ 1 \ , \= _ . ~ TlAo


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fl : .... \M.~ fL ,SlV\,G,)( .E .: )/2 M X 2. , ~ 12 -1Nl ( i ; ~ . 1

t '~ J b t v V \ _ j :

\

::: - VV\ J ! . _ ~ 6 " , , 8

t - b 8~o~ e - e - I D'{ M77 V I I \

,.,'K ":;::._ V \ I I J L c_osO1 fA'''' .

o f p o ~ k; ('c"'V\~. V\A~\~.(

~. ~:::_ ~ S \'" G _ "::-CCNlSbJ\"\ . . . C \ - ~ ~ ~ ; E t~ ~ - :~ . - J A -::::- V V \ 3",8

C ~ l k lM ) \ _


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CLASSICAL MECHANICSDepartment of PhysicsUniversity of New Hampshire

Comprehensive ExamAugust 23,2005

Instructions: Do all of the following problems. The maximum score for each problemis shown, as wen as for each part of Problems #1 and #3. You have 5 hours for theentire exam including both Classical Mechanics and Electricity and Magnetism. Youare allowed one side of a sheet of notes for Classical Mechanics, which you must turnin with the exam. Otherwise the exam is closed book.

1. (15 points) A set of N identical trucks with distance S between each adjacent pairare lined up at rest on a horizontal smooth track in one dimension. Assume thatcollisions between the trucks are completely inelastic so that the trucks sticktogether following a collision. You may neglect the size of each truck. The endtruck is set in motion with velocity V towards the next.

(a) (5 points) Find the final velocity of the trucks.(b) (10 points) Find how long it takes for the last truck to start.

2. (15 points) A thin uniform horizontal rod of mass M is supported by two verticalstrings at its ends. Find the tension in the remaining string immediately after oneof the strings is severed.

3. (20 points) A particle of mass m moves on the inside frictionless surface of a coneunder the influence of gravity. In a spherical coordinate system with the polar axisparallel to the upward vertical direction, the cone satisfies 8 :::o : At t :: : 0, thevertical component of the particle angular momentum about the vertex is L,

(a) (8 points) Write down the Lagrangian. Find the equations of motion forthe particle.

(b) (6 points) Show that a circular orbit is possible. Find the radius of thecircular orbit in terms of L,m, the acceleration due to gravity g, and o:

(c) (6 points) Show that the circular orbit is stable.


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~lJ


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I


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)

I e

; b L()= 1\)0

( J , . - 1 \ 5 ' < . ) ) ~ I \ _ l ~ LUX\ l~,? : : : . f\ JD ~ L() i\-'~ ~.Q_ eJ~ ~ J~ ~)-A. -rk ~ "u. ~j ,

NO = ~2-MA.uj0~)~ wt


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Mechanics Exam 1 (Practice): Comprehensive Examination SOLUTIONSL \

\'. L "t

.-' L \J \ (\Z ." \)') (J\('.~.,\\ '\_b,\" . ,~~D \ ) : V ' ' ' '

Page 1

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/,- .."'~I\C \ ( \ / . . . .\ ~ ) I\;~,C:)


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\ \S OLUTIONS : C o mprehensive Exam: Mechanics Exam 1 (Practice) Pa


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S OLUTIONS : C o mprehensive Exam: Mechanics Exam 1 (practice)...._"-~

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MECHANICSProblem 1. (17 points) A tiny metal ball of mass In is dropped from a height h. (The gravitationalacceleration is g. ) The ball falls into an empty circular pipe, one quarter of which is cut away, sothat the cross section of the pipe only extends through 2700 of a circle, as shown in the diagram.The pipe is fixed in place, and can neither slide nor roll. The radius of the pipe's cross section is R.The initial location of the ball along the horizontal direction is chosen very precisely, so that whenthe ball first makes contact with the pipe wall it begins to slide down the inside surface of the pipewithout bouncing. (At the initial point of contact, the wall of the pipe is vertical, as in the figure.)The inside surface of the pipe is frictionless.(a) (6 points) What is the minimum height hmin such that the metal ball of mass In reaches thetop-most point on the interior of the pipe without losing contact with the pipe?Suppose now that h = hmin, and that after the ball is dropped it enters the pipe, slides along theinside of the pipe, and then exits the pipe moving towards the left along the dashed-line trajectoryshown in the diagram. The ball's velocity vector at the instant it leaves the pipe is horizontal. Whenthe ball eventually reaches ground level, it strikes a ball of clay of mass M .(b) (4 points) What is the horizontal distance d that the metal ball travels after leaving the pipebefore striking the ball of clay?After the metal ball hits the ball of clay, it becomes embedded in the clay, and the clay starts toslide to the left along the ground. The coefficient of kinetic friction of the clay is }tk. The clayslides a distance L before coming to a stop.(c) (7 points) What is the value of L?

h

m , _,,,,,,

o f

L d


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SOLUTION

Part a. Let vbe the velocity of the ball as it leaves the pipe. Conservation of energy givesmv2mgh= T+mg.2R, (1)

orv2 2g(h - 2R). (2)

In order for the ball to remain in contact with the top of the pipe, the gravitational acceleration cannot exceed the centripetal acceleration, and we must have

v2g


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Problem 2. (17 Points) A cylinder of radius a, mass m and length l rolls withoutslipping in a half-cylindrical cavity of inner radius b > a and length larger than l, Youmay assume the cylinder only rolls and does not move along its length. The opening ofthe cavity faces up and the system is in a constant gravitational field g. Friction preventsthe cylinder from slipping, but energy losses are negligible. Find the equation of motionfor the cylinder. What is the angular frequency for small oscillations of the cylinderabout the bottom of the cavity? Express your solutions in terms of a, b , L , m andphysical constants only.


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SOLUTION:This problem has no friction, a conservative field and generalized coordinates, and so canbe solved using a Lagrangian.The kinetic energy arises from both the motion of the center of mass of the cylinder andthe rotation of the cylinder. Ifwe take 8 as the angle of the center of the cylinder fromthe downward vertical line through the radial center of the cavity, and as the rotation ofthe cylinder, then

T = _ ! _ m ( b - a ) 2 ii + _ ! _ ] 22 2The moment of inertia of the cylinder about its center line is justa a

] = J w22nrdr = 2 .n :! 1 ,f r3dro 02.n:/1 4=--a4where

So

ThenT = _ ! _ m ( b - a ) 2 tP + } _ m a 2 2

2 4The two variable are related since the cylinder does not slip:b8 = = ao. b ,=-8aThen


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aL = -mg(b _ a)sinea eGiving the equation of motion(~b2 - 2ab + a2 ) 8 + g(b - a) sin e =0

For small angles, this reduces to(~b2 - 2ab + a2) 8 + g(b - a)e =0

Which is a simple harmonic oscillator with an angular frequency of2g(b-a)OJ = 3b2 - 4ba + 2a2


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Problem 3. (16 points) A hockey stick of mass m, and length L is at rest on the ice(assumed to be frictionless). A puck with mass m, hits the stick a distance D from themiddle of the stick. Before the collision, the puck was moving with speed v o in adirection perpendicular to the stick, as shown below. The collision is completelyinelastic, and the puck remains attached to the stick after the collision.

(a) (4 points) Find the speed v r of the center of mass of the stick+puck system after thecollision.(b) (2 points) Find the location of the center of mass of the puck-stick system, measuredfrom the point where the puck struck the stick.(c) (10 points) What is the angular velocity, co,of the stick+puck system about the centerof mass after the collision. Assume that the stick is uniform, and has a moment of inertia1 0 about its center.


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SolutionPart a. Linear momentum is conserved.

or

Part b. remmeasured from the location where the puck strikes would be given by:

Part c.Angular momentum is conserved

(3 points)

L, = = Lf (1 point)The initial angular momentum is given by:

The final angular momentum is: L =1 wf p+sThe moment inertia of the stick about the new center of mass is:

using the parallel axis theorem. Adding the moment of inertia of the puck gives

Plugging in for remgivesmpD 2 m D 2 msm1'D211'+.1'= 10 + m .,( ) + m1'( ) =10 + (2 points)m, + " ' m, + mp (ms + mp)

mp1 .pDvoW = = - - -- - -' -- - --[ I aCm, + n11,)+ n1smpD2] (2 points)Or solving for OJ,


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MECHANICSProblem 1. (17 points) A bead of mass m slides on a frictionless circular hoop of radius a, picturedin the diagram below. Throughout this problem you are to neglect gravity. At all times, the hooplies in the xy-plane. The hoop rotates with a constant angular velocity ! about the z axis, which atall times passes through point P on the edge of the hoop as shown in the diagram. The quantity "is the angle between the following two line segments: (1) the line extending from the center of thehoop in the direction opposite from P (the dashed line in the figure), and (2) the line connectingthe center of the hoop to the bead.(a) (5 points) Express the Lagrangian for the bead of mass m in terms of m, a, ", and! .

&

(b) (5 points) Use Lagrange's equation to derive the equation of motion for the bead. Your answershould express d2" / dt2 in terms of one or more of the following quantities: ", m, a, a / dt, and !.(c) (5 points) Assume that 1 " 1 1, show that the bead undergoes small oscillations about" = 0,and derive the frequency of these oscillations.(d) (2 points) Provide a brief physical argument to explain why your answer to part (c) makessense.

#


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yacos(! t ) +acos(! t+ " )a sin( ! t) + a sin( !t + ").

(1)(2)

SOLUTION

Part a. Take the center of the hoop to lie on the positive x axis at t = O.Thenx

Differentiating with respect to time, and noting that there is no potential energy in this problem,we find that 2L = T - V = m ; [! 2 + ( ;,+ ! )2 + 2! (;' + ! ) cos ,,] . (3)Part b. Starting with the Lagrange equation

!! _ ( # L ) _ # L =0dt #;, #" 1 ( 4 )

and using the above expression for L, we find that

( m a 2) d . ( m a 2) .2 dt[2("+!)+2!Cos"]+ 22!("+!)sin"=0. (5)After simplification, we obtain ;;= -!2sin" ( 6 )

Part c. When 1 " 1 1, equation (6) becomes(7)

The general solution to this equation is"= Cj cosl ! t ) +c2sin(! t ) , (8)

where Cj and C2are arbitrary constants. Equation (8) corresponds to sinusoidal oscillations of thebead about " = 0 with angular frequency ! .

Part d. The bead experiences a centrifugal force due to the hoop's rotation, which pushes the beadaway from P and towards " = O . Since the hoop cancels out the component of the force normal tothe hoop, the bead is in equilibrium at " = O . However, if the bead moves a small distance awayfrom" = 0, it experiences a restoring force that causes it to oscillate about" = O .


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Problem 2. (17 points) Consider a particle of reduced mass, u, orbiting ina central force with potential energy U(r ): :: ;: : k r "; where kn > O.(a) (2 points) What does the condition k 1 1 , > 0 tell us about the force?F =_a u =-knrn-1a r

Akn>O so F(r) < 0 so the force is in the -r direction, and so is attractive(b) (7 points) The equation of motion for the particle can be written as.. -a u efJmr = a r .where Ueff is the effective potential energy, Find Ueff,and sketch U eff, for 1 1 , =2, -1, and -3.The total energy of the particle isE =T+U, which is

But . e = /-lr2i J is the angular momentum, so we can write it as:

Thus, the effective potential is:


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\' \ . . . \[cr- Ir

/For n = -1, with k


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aU effWe want values for which r = const, These will have a r =0

(c) (4points) Find the radius at which the particles (with fixed angularmomentum t) can orbit at a fixed radius.

.. -JUeffradial equation: mr = -- a r

a 2u effOrbit is stable if a r 2 >0

1

r o = [ k : 2 ~ r 2(d) (4points) For which values of n is the orbit stable? Discusswhether this result is consistent with your sketches in part (b).

r=ro .

k ( 1 ) n-2 3 f2 0= ni n= ro +--4>/ 1 1 0


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4Multiply by r o .

3 2->0/ - l

Substitute r o from above.

[ 2 ] 3 , e 2kn(n-l)- +->0knu / - l(n -1) + 3> 0n+2>On>-2This is consistent with the sketches, because n=2 and -1 both have localminima, indicating a stable orbit, while n = -3 has a local maximum.


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Problem:

c) What is the period of the pendulum?

A torsion pendulum is suspended from a fix point. It is constrained to moveonly vertically and to rotate only about the vertical axis. The pendulum hasa moment of inertia about the vertical axis of I and a mass of M. Thevertical displacement is of the form

Z = k e 2where Z is the displacement and e is the rotation angle. The only restoringforce is gravity.a) Assuming the linear kinetic energy of the vertical displacement is

negligible compared to the rotational kinetic energy, what is theLagrangian using e as the canonical variable?

b) What is the Lagrange equation for the problem?

d) What is the Lagrangian including the linear kinetic energy?e) Under what conditions is the assumption that the linear kinetic energy

of the vertical displacement is negligible compared to the rotationalkinetic energy? How would this depend upon the amplitude of thependulum?

f) What is the Lagrange equation including the linear kinetic energy?


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Problem with solution:A torsion pendulum is suspended from a fix point. It is constrained to moveonly vertically and to rotate only about the vertical axis. The pendulum hasa moment of inertia about the vertical axis of 1and a mass of M. Thevertical displacement is of the form

Z= ke 2where Z is the displacement and e is the rotation angle. The only restoringforce is gravity.a) Assuming the linear kinetic energy of the vertical displacement is

negligible compared to the rotational kinetic energy, what is theLagrangian using e as the canonical variable?

The kinetic energy isT =!i2 + _ ! _ 1e2

2 2Ifthe linear kinetic energy is negligible, then this reduces toT = ! I B 22The potential energy isV=Mgz= Mgkf )2

so the Lagrangian isL = !1fl-Mgke 22b) What is the Lagrange equation for the problem?

~(~~)- ~~=0Soa~ = I ea e


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If) 2Mgkf) =0

d ( a L ) . .-. =ledt a eand

a L-=-2Mgkea eGivingc) What is the period of the pendulum?Using the ansatze =Acas(wt +a)e = = -Awsin(wt +a)(j = = -Aw2 cas(wt +a)Gives- lAw2 cas( wt +a) + 2MgkA cas( wt +a) = 0and

2 2Mgkw =--Id) What is the Lagrangian including the linear kinetic energy?

T = = _ ! _Mz 2 + _ ! _ 1 8 22 2z = = 2keeSo

T =2Mk2e2e2 + _ ! _ 1 8 22andL = 2Mk2e2e2 + _ ! _ le2 - Mgke22


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e) Under what conditions is the assumption that the linear kinetic energyof the vertical displacement is negligible compared to the rotationalkinetic energy? How would this depend upon the amplitude of thependulum?

or

For the linear kinetic energy to be small compared to the rotational kineticenergy,2Mk2e2 _ ! _ I2

2 Ie 24Mkf) What is the Lagrange equation including the linear kinetic energy?

a~ = 4Mk2e 2e + I ea e! ! _ ( a ~ ) =4Mk2e 28 + 8Mk2e e 2 + 1 8dt afJandaL =4Mk2e e 2 - 2MgkfJa eGiving,4Mk2e 28 + 4Mk2e e 2 + 1 8 + 2Mgke =0


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Problem 3. (16 Points) A cart of mass m runs without friction on a flat track. It ismoving with an initial velocity v0 towards a stationary cart of mass 3m which is alsofree to move without friction on the track. The stationary cart has an ideal, Iosslessspring of spring constant k on the side of the approaching cart. Assume the spring doesnot reach the limits of its length.

(a) (4 points) What are the velocities of the two carts when the spring has reached itsmaximum compression?(b) (4 points) What is the compression of the spring when it reaches its maximumcompression?(c) (4 points) What are the velocities of the two carts after the collision when the cartshave separated?(d) (4 points) For the full relativistic case (vo ~ c), what it the solution to part (a)?


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Solution:(a) When the spring is fully compressed, the relative velocities of the carts is zero.Thus, vI = v2' Conservation of linear momentum then tells us

mVI + 3mv2 =mvo4mvl =mvoVovI =v2 =-4

(b) This is a simple conservation of energy question.The kinetic energy of the first cart is

1 2To =-mvo2The kinetic energy of the two carts together is1 1 =m(; t +~( ;1 2=-mvO8So the energy stored in the spring is1 2 1 2 3 2-mvO --mvO =-mvo288So1 2 3 2-10:: =-mvo2 8x =vo~~ 7(c) Conservation of linear momentum and energy apply.Linear momentum:mVl + 3mv2 = mvoorvI + 3V2 =voEnergy:12321 2-mvl +-mv2 =-mvo222or222vI + 3v2 = voFrom the momentum equation:

1v2 =-(vo -vd3Substituting into the energy equation:2 1 ( )2 2vI + " 3 Vo - vI = = V o


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V o=V --0' 2The v0 solution is, of course, before the collision. SoV ovI =--2And

1V 2 = " 3 ( Vo- V l )

= * ( V O + V ; )= V o2

(d) As in (a), vI = v2' Thus we need only conservation of relativistic momentum.U (3 vismg i =-cmc(3 1 3m c(3 2 mc(3 o- - - ; = = = = = = + = - = = = = = =

~1- (3[ ~1- (3i ~1- (364(31 (30

~l- (3[ = ~1- (3616(3[ (36--2=--21 (31 1- (3016(3[ -16(3[(36 = (36- (36(3[16(3[ -15(3[(36 = (36

2(32 _ (301 - 16 -15(36

(31= (30~16 -15(36

V oV I =-;==~=~16 -15(36

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