ENGINEERING TRIPOS PART IB JUNE 2012

CHAIRMAN PROF P DAVIDSON

MONDAY 4TH JUNE 2012 9 TO 11

PAPER 1 SOLUTIONS - MECHANICS

AUTHORS

PROF JWILLIAMS

DRG CSANYI

Created by Diane Unwin Friday July 13 2012

o 11~ 4shy

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)

Tv1OIMMt1 A~tt 0

(V1 t C lt)~~- 51) - Mt1 sect

_I~I U

~ 1_ tv1o = tA-R~ 4 a 7

e ~ ~ (114- amp) 2r2 R

~~-

~ s-= 114 ~ =-0

amp ~ ampiJ ~-

If) 2 R t tfJ = j 01IA amp4 -B-) 3 (Jtf) J2

212 ~ e~amp ~ 01AA ~J4 - btA 61 ~~

LutlAGlt j2 R amp) -1 ~~4-b1-) + ~ 31

ki~ f) ~ O 6L =0 A= teo 114

e~ ~ fanQY4- -Ef) - ~tt4 rzlaquo

-~------~---

l~kL amp-- 11)+ tJ) lt 11 (- -J2)Jig

lJ ~ (-2- I) 21lt

~ QaLllQ fa --shy

-2 shy

2~ tlP~ shyit

fo J~ ~ WI~+ Wv ~ ~ LJ2-1) j2 2~

~~ + -~lV11 (- lL) 5J2-3 VAC(

+

ampxortV1 J1UtIV-$ Urvvll1tb1A b- ~_~ A------------__- -- _------

Ql) The moment of inertia was computed correctly by almost all including those who chose to compute it about 0 rather than G The biggest stumbling block for part c) was the correct orientation of the centripetal and angular accelerations An alarming number thought that angular acceleration is radial and centripetal acceleration is tangential Few realised that the reactions must in any case be normal marks were awarded to those who did as they were to those who stated that the two reactions are equal in magnitude due to symmetry

Q2) Over half the solutions assumed angular momentum conservation along the axis of the clutch and so quickly got the wrong answer and with no chance for a consistency check proceeded to substitute the results into b) and c) Those who did this correctly were marked generously

Q3) Part a) of this was an easy piece of geometry and most who attempted it got it right In part b) most of the confusion came from identifying which of the two angles to use when determining the potential and kinetic energy terms (for those who did it that way) or by forgetting either the mass or the acceleration terms for those who took moments to write down the equation of motion Part c) was attempted by comparatively few and again was very poorly done most forgot some terms in the force balance very few figured out that they needed to work at the limit of the oscillation by setting the angular velocity to zero

ALLL J 1==~~7~==1

Let w--I

be fi~~ ~~ad ~ 43 f1t1w~

B SVtAa Lv -(1 vt wVt-atLLL

2J 1==77=-===1 fotouu at- 2 ~ gearbox

(0) Lt- is feVvfiVlq dlgt llaquo~) aA VVI ~ CvVVt~-ampJ cl(0lt

kVaJtVV1D 1M~r- ri VV 0 1I1VVlltfu1M Y ~oV ~middotev Veurod ~~

A K I S df)tr ~~ (1 LL~~ ~ So n ~~ f1ictAshy4r--gt2 = 4J Vf -+ J 2 w-5 + 2J 2w-r

f1 DM ~r 6 YVl tvll lS C-6Vl~ I A ~1DVI-- atlt-q)( 15 ~~~r- LJIA1 ~ fvmiddote-ve s ~tJ ~ l-CvVlt1l I L1 otIl[tL1-ltamp lthJvcfi 1fQ ~ Vc0oh6V1- ctk -l1LIL kVIV1~S A- ~ B rvWCU IJlAt1A ~ VVlowevct

0 cr-vr Vvu CcWeck ~oMA)middoto-n 1)- 0 ~OVi1 (f~ +v Gm-t ~oLampC 1IA~ d 111 01111 q ltQ7V1 ~~ Cb IV1 fvt-t 6b i-t7t Ol(- Cf1 hx ~ I J h1- J ao~l)1AJ 6V rfD I~ae-lt Ne +trw1J 1111 (I~ f1V w lit ~tf) ~ Or QAv41 ~ Ie ~ I) I Vct I eLL-f I IAlA hcJ kvG CiA 1~ CD ~A X ) ctXWqr1 fVJ lt4 J -f1 V w ~ ([1

L

~)GU)ltcr ~(vw ~ rT amp = 4T (-J2- wi) 1 c)~ or SlP J

L0lUvL T 1tJ lt+-vv ~~ aA- ~ o +I 1AA-lt rfy6VVI S vIA I 1t-v()

b ellA r-U1 t g1Y14 J( cL1t s lt1 -ftUD 51111 CII 1 e V n T+-I Ll W ledc VIA U ~~ ~ SA~ +iA2N1

~ -v-v q fX tM g w1 lev i ~ T +LAJAt

r11CtT- ~11 ~d 1-0 lcgttA lt euroV 0 2- T [L

ZTU1

T 0 ~O-cv T lttD T J COV1y iILti t1 ~ r ___~~lbw~~_- gearbox

~ v wcgtv1- T =- 2 4- 21 2 lt-----__-L__Jshy

T t -= T(b

T

o 11~ 4shy

amp~ ------------ I~ ~ gte- 2taA-~1 G6 )~e

) tt ~) (n -t) - tit togt 1l ~~ 1 Wl (C) ampr -Ishyt 4 t) 2

Blt 4 VV2 1fY gt

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2 ~ ~

Cr)i t- ViAA t - 2 lt D 1shy

)

Tv1OIMMt1 A~tt 0

(V1 t C lt)~~- 51) - Mt1 sect

_I~I U

~ 1_ tv1o = tA-R~ 4 a 7

e ~ ~ (114- amp) 2r2 R

~~-

~ s-= 114 ~ =-0

amp ~ ampiJ ~-

If) 2 R t tfJ = j 01IA amp4 -B-) 3 (Jtf) J2

212 ~ e~amp ~ 01AA ~J4 - btA 61 ~~

LutlAGlt j2 R amp) -1 ~~4-b1-) + ~ 31

ki~ f) ~ O 6L =0 A= teo 114

e~ ~ fanQY4- -Ef) - ~tt4 rzlaquo

-~------~---

l~kL amp-- 11)+ tJ) lt 11 (- -J2)Jig

lJ ~ (-2- I) 21lt

~ QaLllQ fa --shy

-2 shy

2~ tlP~ shyit

fo J~ ~ WI~+ Wv ~ ~ LJ2-1) j2 2~

~~ + -~lV11 (- lL) 5J2-3 VAC(

+

ampxortV1 J1UtIV-$ Urvvll1tb1A b- ~_~ A------------__- -- _------

Ql) The moment of inertia was computed correctly by almost all including those who chose to compute it about 0 rather than G The biggest stumbling block for part c) was the correct orientation of the centripetal and angular accelerations An alarming number thought that angular acceleration is radial and centripetal acceleration is tangential Few realised that the reactions must in any case be normal marks were awarded to those who did as they were to those who stated that the two reactions are equal in magnitude due to symmetry

Q2) Over half the solutions assumed angular momentum conservation along the axis of the clutch and so quickly got the wrong answer and with no chance for a consistency check proceeded to substitute the results into b) and c) Those who did this correctly were marked generously

Q3) Part a) of this was an easy piece of geometry and most who attempted it got it right In part b) most of the confusion came from identifying which of the two angles to use when determining the potential and kinetic energy terms (for those who did it that way) or by forgetting either the mass or the acceleration terms for those who took moments to write down the equation of motion Part c) was attempted by comparatively few and again was very poorly done most forgot some terms in the force balance very few figured out that they needed to work at the limit of the oscillation by setting the angular velocity to zero

ALLL J 1==~~7~==1

Let w--I

be fi~~ ~~ad ~ 43 f1t1w~

B SVtAa Lv -(1 vt wVt-atLLL

2J 1==77=-===1 fotouu at- 2 ~ gearbox

(0) Lt- is feVvfiVlq dlgt llaquo~) aA VVI ~ CvVVt~-ampJ cl(0lt

kVaJtVV1D 1M~r- ri VV 0 1I1VVlltfu1M Y ~oV ~middotev Veurod ~~

A K I S df)tr ~~ (1 LL~~ ~ So n ~~ f1ictAshy4r--gt2 = 4J Vf -+ J 2 w-5 + 2J 2w-r

f1 DM ~r 6 YVl tvll lS C-6Vl~ I A ~1DVI-- atlt-q)( 15 ~~~r- LJIA1 ~ fvmiddote-ve s ~tJ ~ l-CvVlt1l I L1 otIl[tL1-ltamp lthJvcfi 1fQ ~ Vc0oh6V1- ctk -l1LIL kVIV1~S A- ~ B rvWCU IJlAt1A ~ VVlowevct

0 cr-vr Vvu CcWeck ~oMA)middoto-n 1)- 0 ~OVi1 (f~ +v Gm-t ~oLampC 1IA~ d 111 01111 q ltQ7V1 ~~ Cb IV1 fvt-t 6b i-t7t Ol(- Cf1 hx ~ I J h1- J ao~l)1AJ 6V rfD I~ae-lt Ne +trw1J 1111 (I~ f1V w lit ~tf) ~ Or QAv41 ~ Ie ~ I) I Vct I eLL-f I IAlA hcJ kvG CiA 1~ CD ~A X ) ctXWqr1 fVJ lt4 J -f1 V w ~ ([1

L

~)GU)ltcr ~(vw ~ rT amp = 4T (-J2- wi) 1 c)~ or SlP J

L0lUvL T 1tJ lt+-vv ~~ aA- ~ o +I 1AA-lt rfy6VVI S vIA I 1t-v()

b ellA r-U1 t g1Y14 J( cL1t s lt1 -ftUD 51111 CII 1 e V n T+-I Ll W ledc VIA U ~~ ~ SA~ +iA2N1

~ -v-v q fX tM g w1 lev i ~ T +LAJAt

r11CtT- ~11 ~d 1-0 lcgttA lt euroV 0 2- T [L

ZTU1

T 0 ~O-cv T lttD T J COV1y iILti t1 ~ r ___~~lbw~~_- gearbox

~ v wcgtv1- T =- 2 4- 21 2 lt-----__-L__Jshy

T t -= T(b

T

Tv1OIMMt1 A~tt 0

(V1 t C lt)~~- 51) - Mt1 sect

_I~I U

~ 1_ tv1o = tA-R~ 4 a 7

e ~ ~ (114- amp) 2r2 R

~~-

~ s-= 114 ~ =-0

amp ~ ampiJ ~-

If) 2 R t tfJ = j 01IA amp4 -B-) 3 (Jtf) J2

212 ~ e~amp ~ 01AA ~J4 - btA 61 ~~

LutlAGlt j2 R amp) -1 ~~4-b1-) + ~ 31

ki~ f) ~ O 6L =0 A= teo 114

e~ ~ fanQY4- -Ef) - ~tt4 rzlaquo

-~------~---

l~kL amp-- 11)+ tJ) lt 11 (- -J2)Jig

lJ ~ (-2- I) 21lt

~ QaLllQ fa --shy

-2 shy

2~ tlP~ shyit

fo J~ ~ WI~+ Wv ~ ~ LJ2-1) j2 2~

~~ + -~lV11 (- lL) 5J2-3 VAC(

+

ampxortV1 J1UtIV-$ Urvvll1tb1A b- ~_~ A------------__- -- _------

Ql) The moment of inertia was computed correctly by almost all including those who chose to compute it about 0 rather than G The biggest stumbling block for part c) was the correct orientation of the centripetal and angular accelerations An alarming number thought that angular acceleration is radial and centripetal acceleration is tangential Few realised that the reactions must in any case be normal marks were awarded to those who did as they were to those who stated that the two reactions are equal in magnitude due to symmetry

Q2) Over half the solutions assumed angular momentum conservation along the axis of the clutch and so quickly got the wrong answer and with no chance for a consistency check proceeded to substitute the results into b) and c) Those who did this correctly were marked generously

Q3) Part a) of this was an easy piece of geometry and most who attempted it got it right In part b) most of the confusion came from identifying which of the two angles to use when determining the potential and kinetic energy terms (for those who did it that way) or by forgetting either the mass or the acceleration terms for those who took moments to write down the equation of motion Part c) was attempted by comparatively few and again was very poorly done most forgot some terms in the force balance very few figured out that they needed to work at the limit of the oscillation by setting the angular velocity to zero

ALLL J 1==~~7~==1

Let w--I

be fi~~ ~~ad ~ 43 f1t1w~

B SVtAa Lv -(1 vt wVt-atLLL

2J 1==77=-===1 fotouu at- 2 ~ gearbox

(0) Lt- is feVvfiVlq dlgt llaquo~) aA VVI ~ CvVVt~-ampJ cl(0lt

kVaJtVV1D 1M~r- ri VV 0 1I1VVlltfu1M Y ~oV ~middotev Veurod ~~

A K I S df)tr ~~ (1 LL~~ ~ So n ~~ f1ictAshy4r--gt2 = 4J Vf -+ J 2 w-5 + 2J 2w-r

f1 DM ~r 6 YVl tvll lS C-6Vl~ I A ~1DVI-- atlt-q)( 15 ~~~r- LJIA1 ~ fvmiddote-ve s ~tJ ~ l-CvVlt1l I L1 otIl[tL1-ltamp lthJvcfi 1fQ ~ Vc0oh6V1- ctk -l1LIL kVIV1~S A- ~ B rvWCU IJlAt1A ~ VVlowevct

0 cr-vr Vvu CcWeck ~oMA)middoto-n 1)- 0 ~OVi1 (f~ +v Gm-t ~oLampC 1IA~ d 111 01111 q ltQ7V1 ~~ Cb IV1 fvt-t 6b i-t7t Ol(- Cf1 hx ~ I J h1- J ao~l)1AJ 6V rfD I~ae-lt Ne +trw1J 1111 (I~ f1V w lit ~tf) ~ Or QAv41 ~ Ie ~ I) I Vct I eLL-f I IAlA hcJ kvG CiA 1~ CD ~A X ) ctXWqr1 fVJ lt4 J -f1 V w ~ ([1

L

~)GU)ltcr ~(vw ~ rT amp = 4T (-J2- wi) 1 c)~ or SlP J

L0lUvL T 1tJ lt+-vv ~~ aA- ~ o +I 1AA-lt rfy6VVI S vIA I 1t-v()

b ellA r-U1 t g1Y14 J( cL1t s lt1 -ftUD 51111 CII 1 e V n T+-I Ll W ledc VIA U ~~ ~ SA~ +iA2N1

~ -v-v q fX tM g w1 lev i ~ T +LAJAt

r11CtT- ~11 ~d 1-0 lcgttA lt euroV 0 2- T [L

ZTU1

T 0 ~O-cv T lttD T J COV1y iILti t1 ~ r ___~~lbw~~_- gearbox

~ v wcgtv1- T =- 2 4- 21 2 lt-----__-L__Jshy

T t -= T(b

T

2~ tlP~ shyit

fo J~ ~ WI~+ Wv ~ ~ LJ2-1) j2 2~

~~ + -~lV11 (- lL) 5J2-3 VAC(

+

ampxortV1 J1UtIV-$ Urvvll1tb1A b- ~_~ A------------__- -- _------

Ql) The moment of inertia was computed correctly by almost all including those who chose to compute it about 0 rather than G The biggest stumbling block for part c) was the correct orientation of the centripetal and angular accelerations An alarming number thought that angular acceleration is radial and centripetal acceleration is tangential Few realised that the reactions must in any case be normal marks were awarded to those who did as they were to those who stated that the two reactions are equal in magnitude due to symmetry

Q2) Over half the solutions assumed angular momentum conservation along the axis of the clutch and so quickly got the wrong answer and with no chance for a consistency check proceeded to substitute the results into b) and c) Those who did this correctly were marked generously

Q3) Part a) of this was an easy piece of geometry and most who attempted it got it right In part b) most of the confusion came from identifying which of the two angles to use when determining the potential and kinetic energy terms (for those who did it that way) or by forgetting either the mass or the acceleration terms for those who took moments to write down the equation of motion Part c) was attempted by comparatively few and again was very poorly done most forgot some terms in the force balance very few figured out that they needed to work at the limit of the oscillation by setting the angular velocity to zero

ALLL J 1==~~7~==1

Let w--I

be fi~~ ~~ad ~ 43 f1t1w~

B SVtAa Lv -(1 vt wVt-atLLL

2J 1==77=-===1 fotouu at- 2 ~ gearbox

(0) Lt- is feVvfiVlq dlgt llaquo~) aA VVI ~ CvVVt~-ampJ cl(0lt

kVaJtVV1D 1M~r- ri VV 0 1I1VVlltfu1M Y ~oV ~middotev Veurod ~~

A K I S df)tr ~~ (1 LL~~ ~ So n ~~ f1ictAshy4r--gt2 = 4J Vf -+ J 2 w-5 + 2J 2w-r

f1 DM ~r 6 YVl tvll lS C-6Vl~ I A ~1DVI-- atlt-q)( 15 ~~~r- LJIA1 ~ fvmiddote-ve s ~tJ ~ l-CvVlt1l I L1 otIl[tL1-ltamp lthJvcfi 1fQ ~ Vc0oh6V1- ctk -l1LIL kVIV1~S A- ~ B rvWCU IJlAt1A ~ VVlowevct

0 cr-vr Vvu CcWeck ~oMA)middoto-n 1)- 0 ~OVi1 (f~ +v Gm-t ~oLampC 1IA~ d 111 01111 q ltQ7V1 ~~ Cb IV1 fvt-t 6b i-t7t Ol(- Cf1 hx ~ I J h1- J ao~l)1AJ 6V rfD I~ae-lt Ne +trw1J 1111 (I~ f1V w lit ~tf) ~ Or QAv41 ~ Ie ~ I) I Vct I eLL-f I IAlA hcJ kvG CiA 1~ CD ~A X ) ctXWqr1 fVJ lt4 J -f1 V w ~ ([1

L

~)GU)ltcr ~(vw ~ rT amp = 4T (-J2- wi) 1 c)~ or SlP J

L0lUvL T 1tJ lt+-vv ~~ aA- ~ o +I 1AA-lt rfy6VVI S vIA I 1t-v()

b ellA r-U1 t g1Y14 J( cL1t s lt1 -ftUD 51111 CII 1 e V n T+-I Ll W ledc VIA U ~~ ~ SA~ +iA2N1

~ -v-v q fX tM g w1 lev i ~ T +LAJAt

r11CtT- ~11 ~d 1-0 lcgttA lt euroV 0 2- T [L

ZTU1

T 0 ~O-cv T lttD T J COV1y iILti t1 ~ r ___~~lbw~~_- gearbox

~ v wcgtv1- T =- 2 4- 21 2 lt-----__-L__Jshy

T t -= T(b

T

ALLL J 1==~~7~==1

Let w--I

be fi~~ ~~ad ~ 43 f1t1w~

B SVtAa Lv -(1 vt wVt-atLLL

2J 1==77=-===1 fotouu at- 2 ~ gearbox

(0) Lt- is feVvfiVlq dlgt llaquo~) aA VVI ~ CvVVt~-ampJ cl(0lt

kVaJtVV1D 1M~r- ri VV 0 1I1VVlltfu1M Y ~oV ~middotev Veurod ~~

A K I S df)tr ~~ (1 LL~~ ~ So n ~~ f1ictAshy4r--gt2 = 4J Vf -+ J 2 w-5 + 2J 2w-r

f1 DM ~r 6 YVl tvll lS C-6Vl~ I A ~1DVI-- atlt-q)( 15 ~~~r- LJIA1 ~ fvmiddote-ve s ~tJ ~ l-CvVlt1l I L1 otIl[tL1-ltamp lthJvcfi 1fQ ~ Vc0oh6V1- ctk -l1LIL kVIV1~S A- ~ B rvWCU IJlAt1A ~ VVlowevct

0 cr-vr Vvu CcWeck ~oMA)middoto-n 1)- 0 ~OVi1 (f~ +v Gm-t ~oLampC 1IA~ d 111 01111 q ltQ7V1 ~~ Cb IV1 fvt-t 6b i-t7t Ol(- Cf1 hx ~ I J h1- J ao~l)1AJ 6V rfD I~ae-lt Ne +trw1J 1111 (I~ f1V w lit ~tf) ~ Or QAv41 ~ Ie ~ I) I Vct I eLL-f I IAlA hcJ kvG CiA 1~ CD ~A X ) ctXWqr1 fVJ lt4 J -f1 V w ~ ([1

L

~)GU)ltcr ~(vw ~ rT amp = 4T (-J2- wi) 1 c)~ or SlP J

L0lUvL T 1tJ lt+-vv ~~ aA- ~ o +I 1AA-lt rfy6VVI S vIA I 1t-v()

b ellA r-U1 t g1Y14 J( cL1t s lt1 -ftUD 51111 CII 1 e V n T+-I Ll W ledc VIA U ~~ ~ SA~ +iA2N1

~ -v-v q fX tM g w1 lev i ~ T +LAJAt

r11CtT- ~11 ~d 1-0 lcgttA lt euroV 0 2- T [L

ZTU1

T 0 ~O-cv T lttD T J COV1y iILti t1 ~ r ___~~lbw~~_- gearbox

~ v wcgtv1- T =- 2 4- 21 2 lt-----__-L__Jshy

T t -= T(b

T

No-w S1tI1CvHt S 1- fttt~ STI~ = J2Wf~ D

1 -1~ LfTd Y2Wf

b 0

n T

~ ~=O J--~=77=---l 4J

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gearbox

TlAeve tVtoIJ$1- bz ~ ~ V t Vt -eM+ VD+Ov~middotVI q ~(n~ C-oVSiShVl1 rt- Ol r iVq Ie 1111l w1lU( b)- 1MAtVl1 OtoY [YI-evt) $lot 1 X B X~ t-o CA S Y~+- LtJVIAutA i ~ wi lIUcy W4-tA -tvt-ltl _

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I bull 4 r -)2 = (l-J1-X) VUf - CD ~---------~ ~ -----_ -- ------------~~-

L-f ~q~ tU- WAL STcgtIgte ~ lt51N1 f+-e 0( ~ c IV+cN V T) ~ T- Xw -(f)

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rr-(ttAAAq Wto~ ~t- A

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11 amp ~ CtM~~ c= ~ rf b -t

b - t1

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~~ b-~~ b

VIr) -= Wlea IJ

~ ~ G-fit) ( 1- Q9f-J

VQ) ~ (b -ttJ(-CltPf9) tVlC1J

J 1(tJ) ~ (b -c-) 3-m 19

(f~

V(v-) ~ (10-0) ~amp

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gtIwl- V 11

(ampL ~-(el_Wl1 =~ I(M 2(b-~)41YV cl(b~~)

~~ b= 0 W = Vt It ~ -gshy

----

E

b

~CGIr ct-tc1fvampoVV (Ylo CU~i ~+~H_VIt)

I~ ~ +1 V1 cCca (cJt b) tl =-~ Z J3 -) ot0 3 =i yen J3

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lJ3 U + VYI (l) 0 - i 2C a

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s =- ~ I1S3

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t

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lID Cq~ =- Vi p +- (l x A13

aA -= ~B + ak )( ( ~ ~2 - J~ a et)

~ = t (~y tAl ~2~ - lAO ~2 ~ - ~~

cap ~~ +- OtT e

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lt

ClQ ttlI C( v laquo +- C5vi e J3

D r--=--- b

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w tr1 poundtG19 f)c ~

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fBco D~ A emiddot)mw~ shyAt i W1~

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b

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Le c= U 1 WlW X-= 1L V1tJ ltc 13

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t 1)) YVYIT2 Q~GlfJ +tvA B C1J

JampM 0- -Vn+ )11--1lt- 1-tltl1- ~ (A +c6L 1 ~(L1R OlJY1-~~-lt

0+ A +n tV ()1 hllv t tlAA pO i U t tAO J Vt1 OLmiddott- Cty1 ( Lt

V-ct()bl h tr-

R1rrvlV i II C1

~ 11 ~ ke ~ 4r ~7 + ~ T (2wfY- +- ~ 2J(2wf yshy (2 + 2-t 4)middot f-~)~ = r-~l-

1~ 0lt61shy

6V ~~ IlpQ

UM ltf- CVAt ~ +cv ClAgt poundr6I f111 IAJ~(er) T

CjT ct ~ 41 ( Yt--wf) D

4-T i~1 4shy

33 T

- fashy

3

ltvop v$-eo f - ()CA hAaA- -tx 1 (

It-A li1IC A Nt

OA 1) - ~A tlY -rl)

~~ LA g ~ b U - centgt)

b ~ = Co - ~) 19 ~

rr-(ttAAAq Wto~ ~t- A

VV~ t~ g 1- ~b~ t- 11 ~ =-0

11 amp ~ CtM~~ c= ~ rf b -t

b - t1

~~ ~~~C cent

I ) VL vI

IA I~1- T VI1)~ r M b C T vVl Ccl (Y

- -

- wi- 11 = - g if 0IMdb ~ at

F _ 1J

_--__ ft1

6+ 6f =- c A ( f) -tgt) k tx51 b (fJ-cent)

~~ b-~~ b

VIr) -= Wlea IJ

~ ~ G-fit) ( 1- Q9f-J

VQ) ~ (b -ttJ(-CltPf9) tVlC1J

J 1(tJ) ~ (b -c-) 3-m 19

(f~

V(v-) ~ (10-0) ~amp

~~ ~ -m (b (p)L t ~ centJ ~ 5 2rrl b 1 cent~ -rIllt -Cll- iY

gtIwl- V 11

(ampL ~-(el_Wl1 =~ I(M 2(b-~)41YV cl(b~~)

~~ b= 0 W = Vt It ~ -gshy

----

E

b

~CGIr ct-tc1fvampoVV (Ylo CU~i ~+~H_VIt)

I~ ~ +1 V1 cCca (cJt b) tl =-~ Z J3 -) ot0 3 =i yen J3

~

~ (011$ ruv IY1~bett +IV

~~V1I~VV

ht v V-nICJ 1-tl ( ~ 3111 ex V-cloc-d-tt

(Of dlSplCl((Mt(-t) cttOsltiVOIW

Clt)W rV ytil gtIf) C( d a~Vl1~

T ~ - vYl0i~ J3 J 2gt

(01 Hncrt Q h t -+~ Itl-At Vvtv VI n td~t A

lJ3 U + VYI (l) 0 - i 2C a

-f Me)) f - IVlw l r~ + Vtl ( 1 (1

12 I~ ~ )

1 = - yY)OtP- -I

-z ~ A

ritTo f 1Ir- S ic( hL lIlc AAtlltl Rl~ ~A ~ cJxUt C -fJ- -CJLL ~

s =- ~ I1S3

-()------ shy

4- (0) ~l-_~~~middotgt

t

~ k X ~I = ~1 ~

lID Cq~ =- Vi p +- (l x A13

aA -= ~B + ak )( ( ~ ~2 - J~ a et)

~ = t (~y tAl ~2~ - lAO ~2 ~ - ~~

cap ~~ +- OtT e

tdU7 at M-Vl~Amiddot k LV +0 1JC ) ~ c- e3 -= 0

-~ -n~f3 el b 2

lt

ClQ ttlI C( v laquo +- C5vi e J3

D r--=--- b

~~vu~v~ _ 0- ~amp ~---

~

~fI = ~ l( ~amp

~

w tr1 poundtG19 f)c ~

r ~tir~-----=~----------~ ~A 15

v ~lflM IfV-td 1~v~R JY YS-rJi

-CD

~()t ~1V =0 ttt yt~ D t x~

fBco D~ A emiddot)mw~ shyAt i W1~

6middot(

b

Vl- - ttJ WVVl iVVIWXshy+

~ WlOVX ~Vl Sf t= =v l-R lt= ztA- rUM CD TYIA)

Le c= U 1 WlW X-= 1L V1tJ ltc 13

~

dtw - t iViw ~MW l~V 1shy -roi~3130

2 yY1W b W3 2 ~1 lJgt- Cl (( l tOVA amp= 3 bJ3 2

t 1)) YVYIT2 Q~GlfJ +tvA B C1J

JampM 0- -Vn+ )11--1lt- 1-tltl1- ~ (A +c6L 1 ~(L1R OlJY1-~~-lt

0+ A +n tV ()1 hllv t tlAA pO i U t tAO J Vt1 OLmiddott- Cty1 ( Lt

V-ct()bl h tr-

R1rrvlV i II C1

3

ltvop v$-eo f - ()CA hAaA- -tx 1 (

It-A li1IC A Nt

OA 1) - ~A tlY -rl)

~~ LA g ~ b U - centgt)

b ~ = Co - ~) 19 ~

rr-(ttAAAq Wto~ ~t- A

VV~ t~ g 1- ~b~ t- 11 ~ =-0

11 amp ~ CtM~~ c= ~ rf b -t

b - t1

~~ ~~~C cent

I ) VL vI

IA I~1- T VI1)~ r M b C T vVl Ccl (Y

- -

- wi- 11 = - g if 0IMdb ~ at

F _ 1J

_--__ ft1

6+ 6f =- c A ( f) -tgt) k tx51 b (fJ-cent)

~~ b-~~ b

VIr) -= Wlea IJ

~ ~ G-fit) ( 1- Q9f-J

VQ) ~ (b -ttJ(-CltPf9) tVlC1J

J 1(tJ) ~ (b -c-) 3-m 19

(f~

V(v-) ~ (10-0) ~amp

~~ ~ -m (b (p)L t ~ centJ ~ 5 2rrl b 1 cent~ -rIllt -Cll- iY

gtIwl- V 11

(ampL ~-(el_Wl1 =~ I(M 2(b-~)41YV cl(b~~)

~~ b= 0 W = Vt It ~ -gshy

----

E

b

~CGIr ct-tc1fvampoVV (Ylo CU~i ~+~H_VIt)

I~ ~ +1 V1 cCca (cJt b) tl =-~ Z J3 -) ot0 3 =i yen J3

~

~ (011$ ruv IY1~bett +IV

~~V1I~VV

ht v V-nICJ 1-tl ( ~ 3111 ex V-cloc-d-tt

(Of dlSplCl((Mt(-t) cttOsltiVOIW

Clt)W rV ytil gtIf) C( d a~Vl1~

T ~ - vYl0i~ J3 J 2gt

(01 Hncrt Q h t -+~ Itl-At Vvtv VI n td~t A

lJ3 U + VYI (l) 0 - i 2C a

-f Me)) f - IVlw l r~ + Vtl ( 1 (1

12 I~ ~ )

1 = - yY)OtP- -I

-z ~ A

ritTo f 1Ir- S ic( hL lIlc AAtlltl Rl~ ~A ~ cJxUt C -fJ- -CJLL ~

s =- ~ I1S3

-()------ shy

4- (0) ~l-_~~~middotgt

t

~ k X ~I = ~1 ~

lID Cq~ =- Vi p +- (l x A13

aA -= ~B + ak )( ( ~ ~2 - J~ a et)

~ = t (~y tAl ~2~ - lAO ~2 ~ - ~~

cap ~~ +- OtT e

tdU7 at M-Vl~Amiddot k LV +0 1JC ) ~ c- e3 -= 0

-~ -n~f3 el b 2

lt

ClQ ttlI C( v laquo +- C5vi e J3

D r--=--- b

~~vu~v~ _ 0- ~amp ~---

~

~fI = ~ l( ~amp

~

w tr1 poundtG19 f)c ~

r ~tir~-----=~----------~ ~A 15

v ~lflM IfV-td 1~v~R JY YS-rJi

-CD

~()t ~1V =0 ttt yt~ D t x~

fBco D~ A emiddot)mw~ shyAt i W1~

6middot(

b

Vl- - ttJ WVVl iVVIWXshy+

~ WlOVX ~Vl Sf t= =v l-R lt= ztA- rUM CD TYIA)

Le c= U 1 WlW X-= 1L V1tJ ltc 13

~

dtw - t iViw ~MW l~V 1shy -roi~3130

2 yY1W b W3 2 ~1 lJgt- Cl (( l tOVA amp= 3 bJ3 2

t 1)) YVYIT2 Q~GlfJ +tvA B C1J

JampM 0- -Vn+ )11--1lt- 1-tltl1- ~ (A +c6L 1 ~(L1R OlJY1-~~-lt

0+ A +n tV ()1 hllv t tlAA pO i U t tAO J Vt1 OLmiddott- Cty1 ( Lt

V-ct()bl h tr-

R1rrvlV i II C1

- -

- wi- 11 = - g if 0IMdb ~ at

F _ 1J

_--__ ft1

6+ 6f =- c A ( f) -tgt) k tx51 b (fJ-cent)

~~ b-~~ b

VIr) -= Wlea IJ

~ ~ G-fit) ( 1- Q9f-J

VQ) ~ (b -ttJ(-CltPf9) tVlC1J

J 1(tJ) ~ (b -c-) 3-m 19

(f~

V(v-) ~ (10-0) ~amp

~~ ~ -m (b (p)L t ~ centJ ~ 5 2rrl b 1 cent~ -rIllt -Cll- iY

gtIwl- V 11

(ampL ~-(el_Wl1 =~ I(M 2(b-~)41YV cl(b~~)

~~ b= 0 W = Vt It ~ -gshy

----

E

b

~CGIr ct-tc1fvampoVV (Ylo CU~i ~+~H_VIt)

I~ ~ +1 V1 cCca (cJt b) tl =-~ Z J3 -) ot0 3 =i yen J3

~

~ (011$ ruv IY1~bett +IV

~~V1I~VV

ht v V-nICJ 1-tl ( ~ 3111 ex V-cloc-d-tt

(Of dlSplCl((Mt(-t) cttOsltiVOIW

Clt)W rV ytil gtIf) C( d a~Vl1~

T ~ - vYl0i~ J3 J 2gt

(01 Hncrt Q h t -+~ Itl-At Vvtv VI n td~t A

lJ3 U + VYI (l) 0 - i 2C a

-f Me)) f - IVlw l r~ + Vtl ( 1 (1

12 I~ ~ )

1 = - yY)OtP- -I

-z ~ A

ritTo f 1Ir- S ic( hL lIlc AAtlltl Rl~ ~A ~ cJxUt C -fJ- -CJLL ~

s =- ~ I1S3

-()------ shy

4- (0) ~l-_~~~middotgt

t

~ k X ~I = ~1 ~

lID Cq~ =- Vi p +- (l x A13

aA -= ~B + ak )( ( ~ ~2 - J~ a et)

~ = t (~y tAl ~2~ - lAO ~2 ~ - ~~

cap ~~ +- OtT e

tdU7 at M-Vl~Amiddot k LV +0 1JC ) ~ c- e3 -= 0

-~ -n~f3 el b 2

lt

ClQ ttlI C( v laquo +- C5vi e J3

D r--=--- b

~~vu~v~ _ 0- ~amp ~---

~

~fI = ~ l( ~amp

~

w tr1 poundtG19 f)c ~

r ~tir~-----=~----------~ ~A 15

v ~lflM IfV-td 1~v~R JY YS-rJi

-CD

~()t ~1V =0 ttt yt~ D t x~

fBco D~ A emiddot)mw~ shyAt i W1~

6middot(

b

Vl- - ttJ WVVl iVVIWXshy+

~ WlOVX ~Vl Sf t= =v l-R lt= ztA- rUM CD TYIA)

Le c= U 1 WlW X-= 1L V1tJ ltc 13

~

dtw - t iViw ~MW l~V 1shy -roi~3130

2 yY1W b W3 2 ~1 lJgt- Cl (( l tOVA amp= 3 bJ3 2

t 1)) YVYIT2 Q~GlfJ +tvA B C1J

JampM 0- -Vn+ )11--1lt- 1-tltl1- ~ (A +c6L 1 ~(L1R OlJY1-~~-lt

0+ A +n tV ()1 hllv t tlAA pO i U t tAO J Vt1 OLmiddott- Cty1 ( Lt

V-ct()bl h tr-

R1rrvlV i II C1

----

E

b

~CGIr ct-tc1fvampoVV (Ylo CU~i ~+~H_VIt)

I~ ~ +1 V1 cCca (cJt b) tl =-~ Z J3 -) ot0 3 =i yen J3

~

~ (011$ ruv IY1~bett +IV

~~V1I~VV

ht v V-nICJ 1-tl ( ~ 3111 ex V-cloc-d-tt

(Of dlSplCl((Mt(-t) cttOsltiVOIW

Clt)W rV ytil gtIf) C( d a~Vl1~

T ~ - vYl0i~ J3 J 2gt

(01 Hncrt Q h t -+~ Itl-At Vvtv VI n td~t A

lJ3 U + VYI (l) 0 - i 2C a

-f Me)) f - IVlw l r~ + Vtl ( 1 (1

12 I~ ~ )

1 = - yY)OtP- -I

-z ~ A

ritTo f 1Ir- S ic( hL lIlc AAtlltl Rl~ ~A ~ cJxUt C -fJ- -CJLL ~

s =- ~ I1S3

-()------ shy

4- (0) ~l-_~~~middotgt

t

~ k X ~I = ~1 ~

lID Cq~ =- Vi p +- (l x A13

aA -= ~B + ak )( ( ~ ~2 - J~ a et)

~ = t (~y tAl ~2~ - lAO ~2 ~ - ~~

cap ~~ +- OtT e

tdU7 at M-Vl~Amiddot k LV +0 1JC ) ~ c- e3 -= 0

-~ -n~f3 el b 2

lt

ClQ ttlI C( v laquo +- C5vi e J3

D r--=--- b

~~vu~v~ _ 0- ~amp ~---

~

~fI = ~ l( ~amp

~

w tr1 poundtG19 f)c ~

r ~tir~-----=~----------~ ~A 15

v ~lflM IfV-td 1~v~R JY YS-rJi

-CD

~()t ~1V =0 ttt yt~ D t x~

fBco D~ A emiddot)mw~ shyAt i W1~

6middot(

b

Vl- - ttJ WVVl iVVIWXshy+

~ WlOVX ~Vl Sf t= =v l-R lt= ztA- rUM CD TYIA)

Le c= U 1 WlW X-= 1L V1tJ ltc 13

~

dtw - t iViw ~MW l~V 1shy -roi~3130

2 yY1W b W3 2 ~1 lJgt- Cl (( l tOVA amp= 3 bJ3 2

t 1)) YVYIT2 Q~GlfJ +tvA B C1J

JampM 0- -Vn+ )11--1lt- 1-tltl1- ~ (A +c6L 1 ~(L1R OlJY1-~~-lt

0+ A +n tV ()1 hllv t tlAA pO i U t tAO J Vt1 OLmiddott- Cty1 ( Lt

V-ct()bl h tr-

R1rrvlV i II C1

4- (0) ~l-_~~~middotgt

t

~ k X ~I = ~1 ~

lID Cq~ =- Vi p +- (l x A13

aA -= ~B + ak )( ( ~ ~2 - J~ a et)

~ = t (~y tAl ~2~ - lAO ~2 ~ - ~~

cap ~~ +- OtT e

tdU7 at M-Vl~Amiddot k LV +0 1JC ) ~ c- e3 -= 0

-~ -n~f3 el b 2

lt

ClQ ttlI C( v laquo +- C5vi e J3

D r--=--- b

~~vu~v~ _ 0- ~amp ~---

~

~fI = ~ l( ~amp

~

w tr1 poundtG19 f)c ~

r ~tir~-----=~----------~ ~A 15

v ~lflM IfV-td 1~v~R JY YS-rJi

-CD

~()t ~1V =0 ttt yt~ D t x~

fBco D~ A emiddot)mw~ shyAt i W1~

6middot(

b

Vl- - ttJ WVVl iVVIWXshy+

~ WlOVX ~Vl Sf t= =v l-R lt= ztA- rUM CD TYIA)

Le c= U 1 WlW X-= 1L V1tJ ltc 13

~

dtw - t iViw ~MW l~V 1shy -roi~3130

2 yY1W b W3 2 ~1 lJgt- Cl (( l tOVA amp= 3 bJ3 2

t 1)) YVYIT2 Q~GlfJ +tvA B C1J

JampM 0- -Vn+ )11--1lt- 1-tltl1- ~ (A +c6L 1 ~(L1R OlJY1-~~-lt

0+ A +n tV ()1 hllv t tlAA pO i U t tAO J Vt1 OLmiddott- Cty1 ( Lt

V-ct()bl h tr-

R1rrvlV i II C1

D r--=--- b

~~vu~v~ _ 0- ~amp ~---

~

~fI = ~ l( ~amp

~

w tr1 poundtG19 f)c ~

r ~tir~-----=~----------~ ~A 15

v ~lflM IfV-td 1~v~R JY YS-rJi

-CD

~()t ~1V =0 ttt yt~ D t x~

fBco D~ A emiddot)mw~ shyAt i W1~

6middot(

b

Vl- - ttJ WVVl iVVIWXshy+

~ WlOVX ~Vl Sf t= =v l-R lt= ztA- rUM CD TYIA)

Le c= U 1 WlW X-= 1L V1tJ ltc 13

~

dtw - t iViw ~MW l~V 1shy -roi~3130

2 yY1W b W3 2 ~1 lJgt- Cl (( l tOVA amp= 3 bJ3 2

t 1)) YVYIT2 Q~GlfJ +tvA B C1J

JampM 0- -Vn+ )11--1lt- 1-tltl1- ~ (A +c6L 1 ~(L1R OlJY1-~~-lt

0+ A +n tV ()1 hllv t tlAA pO i U t tAO J Vt1 OLmiddott- Cty1 ( Lt

V-ct()bl h tr-

R1rrvlV i II C1

~()t ~1V =0 ttt yt~ D t x~

fBco D~ A emiddot)mw~ shyAt i W1~

6middot(

b

Vl- - ttJ WVVl iVVIWXshy+

~ WlOVX ~Vl Sf t= =v l-R lt= ztA- rUM CD TYIA)

Le c= U 1 WlW X-= 1L V1tJ ltc 13

~

dtw - t iViw ~MW l~V 1shy -roi~3130

2 yY1W b W3 2 ~1 lJgt- Cl (( l tOVA amp= 3 bJ3 2

t 1)) YVYIT2 Q~GlfJ +tvA B C1J

JampM 0- -Vn+ )11--1lt- 1-tltl1- ~ (A +c6L 1 ~(L1R OlJY1-~~-lt

0+ A +n tV ()1 hllv t tlAA pO i U t tAO J Vt1 OLmiddott- Cty1 ( Lt

V-ct()bl h tr-

R1rrvlV i II C1

B

== LJs1LI I) e ( (trClto f) ~ e wS~~

~ Va l ~sn gf c cDtJ) t=D

tv J2w1 )itVamp

---z (~(n tJshycv tv if ~)t ) CVIA )

7

-r~-

~ ~ vr

$vJ~7Vlt-lvt~ lt1l~ 4shy

~o) ~cJ t VVf11A lslt ~~ cV- A ~ i1 66 ltA6t G(lll~~ v~vn~bLlt- tvW(Un-cvt ~~ WOWC(tf0wt

$ CDII~ltVV-0 cbD11 A

= L(1t lAf 1- 1 T (Nr-L 2 2 ~

l TYl1)1 - + oM-( q y) -= _~ yVJ t)l 2 - 2gt 8(l 1

1R YvJ2 IN) - ~ m )1shy 6---middotmiddot---shy

- tt shy

~

liv CoS (j= YVJ

W 1~ltMf)2

~1AW--1~ tAC ~ W ClVNA W2

==

- V LCtI )L U a((~~ -lt ( ~ v 4YV IV I VI 1 -y = lVI ltCl - ~ W ~iV7 B- - ~ iVl Usf)

Sa Iv+-ihl Clt ~v k) Wv~ ltV )l~ tt 1vvti I I~ ttnllt

J

Q4) The solution of this sort of problem is made much simpler by a decent diagram drawn with a ruler There were very few of these and far too many sloppily sketched freeshyhand efforts that bore no relation to the equilateral triangle specified in the problem Most of the more successful attempts used velocity and acceleration diagrams but the idea of the velocity or acceleration image is not understood if when you walk from B to C on the mechanism and you have to turn left to get to G then as you go from b to c on either the velocity of the acceleration diagrams you will also have to turn left to locate g Despite the problem specifying that the mechanism was operating in the horizontal plane a significant number of candidates included gravity

Many of the minority that attempted an analytical treatment got lost in a morass of ~ and ~ s and abandoned ship The idea of taking a dot product between a vector and a unit vector to find the magnitude of the component of the vector in the direction of the

unit vector is apparently unknown - or forgotten

Q5) Some candidates having demonstrated that co == 7sece were unable to successfully

differentiate again with respect to time to derive ciJ =7secetane x () Having drawn a

perfectly fine (and simple) acceleration diagram a lot of candidates reverted to vectors to find the acceleration of the rods centre of mass (say) G instead of simply locating its image g half-way between a and b on the diagram they had just drawn doing this immediately demonstrates that the acceleration of G is parallel to the guide B

Not unusual in part (b) was the statement that BM is a max when SF is zero and that this would be half way along the rod Tempting to think this true perhaps if eis 45deg but not in fact the case because there must be a force at A in the direction of 11 to maintain the constant speed of this end of the rod No candidate appreciated this and included such a force in their FBD though its perfectly possible to get a correct solution for the BM with this omission by integrating twice and setting the BM zero at each end of the rod

Q6) Momentum and energy Generally well done only arithmetic slips marring parts (a) and (b) In part (c) loss of contact with the step occurs when the horizontal component of the reaction falls to zero At this instant the vertical component is still positive and equal to mg4 A good number of complete solutions

-ttSshy

~

liv CoS (j= YVJ

W 1~ltMf)2

~1AW--1~ tAC ~ W ClVNA W2

==

- V LCtI )L U a((~~ -lt ( ~ v 4YV IV I VI 1 -y = lVI ltCl - ~ W ~iV7 B- - ~ iVl Usf)

Sa Iv+-ihl Clt ~v k) Wv~ ltV )l~ tt 1vvti I I~ ttnllt

J

Q4) The solution of this sort of problem is made much simpler by a decent diagram drawn with a ruler There were very few of these and far too many sloppily sketched freeshyhand efforts that bore no relation to the equilateral triangle specified in the problem Most of the more successful attempts used velocity and acceleration diagrams but the idea of the velocity or acceleration image is not understood if when you walk from B to C on the mechanism and you have to turn left to get to G then as you go from b to c on either the velocity of the acceleration diagrams you will also have to turn left to locate g Despite the problem specifying that the mechanism was operating in the horizontal plane a significant number of candidates included gravity

Many of the minority that attempted an analytical treatment got lost in a morass of ~ and ~ s and abandoned ship The idea of taking a dot product between a vector and a unit vector to find the magnitude of the component of the vector in the direction of the

unit vector is apparently unknown - or forgotten

Q5) Some candidates having demonstrated that co == 7sece were unable to successfully

differentiate again with respect to time to derive ciJ =7secetane x () Having drawn a

perfectly fine (and simple) acceleration diagram a lot of candidates reverted to vectors to find the acceleration of the rods centre of mass (say) G instead of simply locating its image g half-way between a and b on the diagram they had just drawn doing this immediately demonstrates that the acceleration of G is parallel to the guide B

Not unusual in part (b) was the statement that BM is a max when SF is zero and that this would be half way along the rod Tempting to think this true perhaps if eis 45deg but not in fact the case because there must be a force at A in the direction of 11 to maintain the constant speed of this end of the rod No candidate appreciated this and included such a force in their FBD though its perfectly possible to get a correct solution for the BM with this omission by integrating twice and setting the BM zero at each end of the rod

Q6) Momentum and energy Generally well done only arithmetic slips marring parts (a) and (b) In part (c) loss of contact with the step occurs when the horizontal component of the reaction falls to zero At this instant the vertical component is still positive and equal to mg4 A good number of complete solutions

-ttSshy

Q4) The solution of this sort of problem is made much simpler by a decent diagram drawn with a ruler There were very few of these and far too many sloppily sketched freeshyhand efforts that bore no relation to the equilateral triangle specified in the problem Most of the more successful attempts used velocity and acceleration diagrams but the idea of the velocity or acceleration image is not understood if when you walk from B to C on the mechanism and you have to turn left to get to G then as you go from b to c on either the velocity of the acceleration diagrams you will also have to turn left to locate g Despite the problem specifying that the mechanism was operating in the horizontal plane a significant number of candidates included gravity

Many of the minority that attempted an analytical treatment got lost in a morass of ~ and ~ s and abandoned ship The idea of taking a dot product between a vector and a unit vector to find the magnitude of the component of the vector in the direction of the

unit vector is apparently unknown - or forgotten

Q5) Some candidates having demonstrated that co == 7sece were unable to successfully

differentiate again with respect to time to derive ciJ =7secetane x () Having drawn a

perfectly fine (and simple) acceleration diagram a lot of candidates reverted to vectors to find the acceleration of the rods centre of mass (say) G instead of simply locating its image g half-way between a and b on the diagram they had just drawn doing this immediately demonstrates that the acceleration of G is parallel to the guide B

Not unusual in part (b) was the statement that BM is a max when SF is zero and that this would be half way along the rod Tempting to think this true perhaps if eis 45deg but not in fact the case because there must be a force at A in the direction of 11 to maintain the constant speed of this end of the rod No candidate appreciated this and included such a force in their FBD though its perfectly possible to get a correct solution for the BM with this omission by integrating twice and setting the BM zero at each end of the rod

Q6) Momentum and energy Generally well done only arithmetic slips marring parts (a) and (b) In part (c) loss of contact with the step occurs when the horizontal component of the reaction falls to zero At this instant the vertical component is still positive and equal to mg4 A good number of complete solutions

-ttSshy