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Question-132102




Question Number 132102 by mr W last updated on 14/Feb/21
Commented by mr W last updated on 16/Feb/21
a ball is thrown from point A with  speed u and strikes at a point B  on the semispherical surface with  radius R and returns back to point A.  if the restitution coefficient of  the collision is e, find the minimum  speed u the ball must have and at  which angle 𝛉 the ball should be  thrown.
aballisthrownfrompointAwithspeeduandstrikesatapointBonthesemisphericalsurfacewithradiusRandreturnsbacktopointA.iftherestitutioncoefficientofthecollisionise,findtheminimumspeedutheballmusthaveandatwhichangleθtheballshouldbethrown.
Commented by otchereabdullai@gmail.com last updated on 14/Feb/21
You are the world best no one can   challenge you!
Youaretheworldbestnoonecanchallengeyou!
Commented by ajfour last updated on 15/Feb/21
For u_(min)    I  got   φ=30°  t_1 =(√((2(√3)R)/g)){1±(√((1/e^2 )−1))}  u_(min) ^2 =(((R(√3))/(2t))+((gt)/4))^2 +((3(√3))/2)gR((1/e^2 )−1)
ForuminIgotϕ=30°t1=23Rg{1±1e21}umin2=(R32t+gt4)2+332gR(1e21)
Commented by mr W last updated on 15/Feb/21
do you mean that u_(min)  is always at  φ=30° indepentently from e?  i can′t confirm this sir.
doyoumeanthatuminisalwaysatϕ=30°indepententlyfrome?icantconfirmthissir.
Answered by mr W last updated on 25/Feb/21
Commented by mr W last updated on 12/Feb/21
Commented by mr W last updated on 16/Feb/21
motion from A to B:  t=((R(1+cos φ))/(u cos θ))  R sin φ=−u sin θ×((R(1+cos φ))/(u cos θ))+(g/2)×((R^2 (1+cos φ)^2 )/(u^2  cos^2  θ))  ((sin φ)/(1+cos φ))=((gR)/(2u^2 ))×(1+cos φ)(1+tan^2  θ)−tan θ  let λ=((gR)/(2u^2 ))  ((sin φ)/(1+cos φ))=λ(1+cos φ)(1+tan^2  θ)−tan θ  λ(1+cos φ)tan^2  θ−tan θ+((λ(1+cos φ)^2 −sin φ)/(1+cos φ))=0  tan θ=((1±(√(1+4λ sin φ−4λ^2 (1+cos φ)^2 )))/(2λ(1+cos φ)))  ⇒θ=tan^(−1) [((1±(√(1+4λ sin φ−4λ^2 (1+cos φ)^2 )))/(2λ(1+cos φ)))]    at point B:  U_x =u cos θ  U_y =gt−u sin θ=((gR(1+cos φ))/(u cos θ))−u sin θ  U_⊥ =U_x cos φ+U_y sin φ  U_⊥ =u cos θcos φ+[((gR(1+cos φ))/(u cos θ))−u sin θ]sin φ  ⇒U_⊥ =u cos (φ+θ)+((gR(1+cos φ)sin φ)/(u cos θ))  U_∥ =−U_x sin φ+U_y cos φ  U_∥ =−u cos θ sin φ+[((gR(1+cos φ))/(u cos θ))−u sin θ]cos φ  ⇒U_∥ =−u sin (φ+θ)+((gR(1+cos φ)cos φ)/(u cos θ))    V_⊥ =eU_⊥   ⇒V_⊥ =eu cos (φ+θ)+((egR(1+cos φ)sin φ)/(u cos θ))  V_∥ =U_∥   ⇒V_∥ =−u sin (φ+θ)+((gR(1+cos φ)cos φ)/(u cos θ))    V_x =V_⊥ cos φ+V_∥ sin φ  V_x =[eu cos (φ+θ)+((egR(1+cos φ)sin φ)/(u cos θ))]cos φ+[−u sin (φ+θ)+((gR(1+cos φ)cos φ)/(u cos θ))]sin φ  V_x =eu cos (φ+θ)cos φ+((egR(1+cos φ)sin φ cos φ)/(u cos θ))−u sin (φ+θ)sin φ+((gR(1+cos φ)sin φcos φ)/(u cos θ))  ⇒V_x =u[(1+e)cos (φ+θ)cos φ−cos θ+((λ(1+e)(1+cos φ) sin 2φ)/(cos θ))]  V_y =V_⊥ sin φ−V_∥ cos φ  V_y =[eu cos (φ+θ)+((egR(1+cos φ)sin φ)/(u cos θ))]sin φ−[−u sin (φ+θ)+((gR(1+cos φ)cos φ)/(u cos θ))]cos φ  ⇒V_y =u[(1+e)cos (φ+θ)sin φ+sin θ+((2λ(1+cos φ)((1+e) sin^2  φ−1))/(cos θ))]    return from B to A:  t=((R(1+cos φ))/((1+e)cos (φ+θ)cos φ−cos θ+((λ(1+cos φ)(1+e) sin 2φ)/(cos θ))))×(1/u)  ⇒((tu)/R)=ξ=((1+cos φ)/((1+e)cos (φ+θ)cos φ−cos θ+((λ(1+e)(1+cos φ) sin 2φ)/(cos θ))))  R sin φ=t{u[(1+e)cos (φ+θ)sin φ+sin θ]+((2λ(1+cos φ)(e sin^2  φ−cos^2  φ))/(cos θ))−((gt)/2)}  sin φ=((tu)/R)[(1+e)cos (φ+θ)sin φ+sin θ+((2λ(1+cos φ)(e sin^2  φ−cos^2  φ))/(cos θ))−((λtu)/R)]  sin φ=ξ[(1+e)cos (φ+θ)sin φ+sin θ+((2λ(1+cos φ)((1+e)sin^2  φ−1))/(cos θ))−λξ]  or  sin φ=ξ[e′cos (φ+θ) sin φ+sin θ+((2λ(1+cos φ)(e′sin^2  φ−1))/(cos θ))−λξ]   ...(I)  with  e′=1+e  λ=((gR)/(2u^2 ))  θ=tan^(−1) [((1+(√(1+4λ sin φ−4λ^2 (1+cos φ)^2 )))/(2λ(1+cos φ)))]       ∗)  ξ=((1+cos φ)/(e′cos (φ+θ) cos φ−cos θ+((λe′(1+cos φ) sin 2φ)/(cos θ))))  for a given e ∈(0,1] we can find  solution(s) for φ in (I) for some  values of parameter λ. the maximum  value of λ such that a solution for φ  exists can be determined numerically.  this corresponds to the minimum  speed the ball must have:  u_(min) =(√((gR)/(2λ_(max) )))    examples:  e=0.5 ⇒u_(min) ≈2.7682(√(gR))  e=0.75 ⇒u_(min) ≈1.6082(√(gR))  e=1 ⇒u_(min) ≈0.9098(√(gR))  (= Q65589)    ∗)  with the other solution  θ=tan^(−1) [((1−(√(1+4λ sin φ−4λ^2 (1+cos φ)^2 )))/(2λ(1+cos φ)))]  we can get a solution for the case  that the ball strikes the surface  upwards, i.e. counterclockwise.  but in this case more energy is  needed, i.e. u_(min)  is larger than with  θ=tan^(−1) [((1+(√(1+4λ sin φ−4λ^2 (1+cos φ)^2 )))/(2λ(1+cos φ)))]  see examples with e=0.5 and 0.75.
motionfromAtoB:t=R(1+cosϕ)ucosθRsinϕ=usinθ×R(1+cosϕ)ucosθ+g2×R2(1+cosϕ)2u2cos2θsinϕ1+cosϕ=gR2u2×(1+cosϕ)(1+tan2θ)tanθletλ=gR2u2sinϕ1+cosϕ=λ(1+cosϕ)(1+tan2θ)tanθλ(1+cosϕ)tan2θtanθ+λ(1+cosϕ)2sinϕ1+cosϕ=0tanθ=1±1+4λsinϕ4λ2(1+cosϕ)22λ(1+cosϕ)θ=tan1[1±1+4λsinϕ4λ2(1+cosϕ)22λ(1+cosϕ)]atpointB:Ux=ucosθUy=gtusinθ=gR(1+cosϕ)ucosθusinθU=Uxcosϕ+UysinϕU=ucosθcosϕ+[gR(1+cosϕ)ucosθusinθ]sinϕU=ucos(ϕ+θ)+gR(1+cosϕ)sinϕucosθU=Uxsinϕ+UycosϕU=ucosθsinϕ+[gR(1+cosϕ)ucosθusinθ]cosϕU=usin(ϕ+θ)+gR(1+cosϕ)cosϕucosθV=eUV=eucos(ϕ+θ)+egR(1+cosϕ)sinϕucosθV=UV=usin(ϕ+θ)+gR(1+cosϕ)cosϕucosθVx=Vcosϕ+VsinϕVx=[eucos(ϕ+θ)+egR(1+cosϕ)sinϕucosθ]cosϕ+[usin(ϕ+θ)+gR(1+cosϕ)cosϕucosθ]sinϕVx=eucos(ϕ+θ)cosϕ+egR(1+cosϕ)sinϕcosϕucosθusin(ϕ+θ)sinϕ+gR(1+cosϕ)sinϕcosϕucosθVx=u[(1+e)cos(ϕ+θ)cosϕcosθ+λ(1+e)(1+cosϕ)sin2ϕcosθ]Vy=VsinϕVcosϕVy=[eucos(ϕ+θ)+egR(1+cosϕ)sinϕucosθ]sinϕ[usin(ϕ+θ)+gR(1+cosϕ)cosϕucosθ]cosϕVy=u[(1+e)cos(ϕ+θ)sinϕ+sinθ+2λ(1+cosϕ)((1+e)sin2ϕ1)cosθ]returnfromBtoA:t=R(1+cosϕ)(1+e)cos(ϕ+θ)cosϕcosθ+λ(1+cosϕ)(1+e)sin2ϕcosθ×1utuR=ξ=1+cosϕ(1+e)cos(ϕ+θ)cosϕcosθ+λ(1+e)(1+cosϕ)sin2ϕcosθRsinϕ=t{u[(1+e)cos(ϕ+θ)sinϕ+sinθ]+2λ(1+cosϕ)(esin2ϕcos2ϕ)cosθgt2}sinϕ=tuR[(1+e)cos(ϕ+θ)sinϕ+sinθ+2λ(1+cosϕ)(esin2ϕcos2ϕ)cosθλtuR]sinϕ=ξ[(1+e)cos(ϕ+θ)sinϕ+sinθ+2λ(1+cosϕ)((1+e)sin2ϕ1)cosθλξ]orsinϕ=ξ[ecos(ϕ+θ)sinϕ+sinθ+2λ(1+cosϕ)(esin2ϕ1)cosθλξ](I)withe=1+eλ=gR2u2θ=tan1[1+1+4λsinϕ4λ2(1+cosϕ)22λ(1+cosϕ)])ξ=1+cosϕecos(ϕ+θ)cosϕcosθ+λe(1+cosϕ)sin2ϕcosθforagivene(0,1]wecanfindsolution(s)forϕin(I)forsomevaluesofparameterλ.themaximumvalueofλsuchthatasolutionforϕexistscanbedeterminednumerically.thiscorrespondstotheminimumspeedtheballmusthave:umin=gR2λmaxexamples:e=0.5umin2.7682gRe=0.75umin1.6082gRe=1umin0.9098gR(=Q65589))withtheothersolutionθ=tan1[11+4λsinϕ4λ2(1+cosϕ)22λ(1+cosϕ)]wecangetasolutionforthecasethattheballstrikesthesurfaceupwards,i.e.counterclockwise.butinthiscasemoreenergyisneeded,i.e.uminislargerthanwithθ=tan1[1+1+4λsinϕ4λ2(1+cosϕ)22λ(1+cosϕ)]seeexampleswithe=0.5and0.75.
Commented by mr W last updated on 15/Feb/21
Commented by mr W last updated on 13/Feb/21
Commented by mr W last updated on 13/Feb/21
Commented by mr W last updated on 12/Feb/21
Commented by mr W last updated on 12/Feb/21
Commented by mr W last updated on 14/Feb/21
following diagrams show different  situations for e=0.75.    generally:  for a given e∈(0,1] there exists an  u_(min) . for any u>u_(min)  there are four   possible ways how the ball returns  back to its starting position: two  ways clockwise and two counter−  clockwise.
followingdiagramsshowdifferentsituationsfore=0.75.generally:foragivene(0,1]thereexistsanumin.foranyu>umintherearefourpossiblewayshowtheballreturnsbacktoitsstartingposition:twowaysclockwiseandtwocounterclockwise.
Commented by mr W last updated on 15/Feb/21
Commented by mr W last updated on 13/Feb/21
Commented by mr W last updated on 14/Feb/21
Commented by mr W last updated on 13/Feb/21
Commented by mr W last updated on 14/Feb/21
Answered by mr W last updated on 14/Feb/21
Commented by mr W last updated on 14/Feb/21
background knowledge    what happens when a ball strikes on  a wall and the restitution coefficient  of the collision is e?    U=coming speed  V=leaving speed    U_⊥ =U cos δ_1   U_∥ =U sin δ_1   V_⊥ =V cos δ_2   V_∥ =V sin δ_2     V_⊥ =e×U_⊥  ⇒V cos δ_2 =e×U cos δ_1   V_∥ =U_∥    ⇒ V sin δ_2 =U sin δ_1   tan δ_2 =((tan δ_1 )/e)  δ_2 ≥δ_1     KE_1 =(1/2)mU^2   KE_2 =(1/2)mV^2 =(1/2)mU^2 (sin^2  δ_1 +e^2 cos^2  δ_1 )  ⇒KE_2 =(sin^2  δ_1 +e^2 cos^2  δ_1 )KE_1   we see KE_2 =KE_1  only when e=1,  otherwise KE_2 <KE_1 .    the collision is called elastic if e=1.
backgroundknowledgewhathappenswhenaballstrikesonawallandtherestitutioncoefficientofthecollisionise?U=comingspeedV=leavingspeedU=Ucosδ1U=Usinδ1V=Vcosδ2V=Vsinδ2V=e×UVcosδ2=e×Ucosδ1V=UVsinδ2=Usinδ1tanδ2=tanδ1eδ2δ1KE1=12mU2KE2=12mV2=12mU2(sin2δ1+e2cos2δ1)KE2=(sin2δ1+e2cos2δ1)KE1weseeKE2=KE1onlywhene=1,otherwiseKE2<KE1.thecollisioniscalledelasticife=1.
Commented by Ar Brandon last updated on 13/Feb/21
  Wow ! 😃 Where did you learn your geometry Sir ? I feel you have this unique skill here.
Wow ! 😃 Where did you learn your geometry Sir ? I feel you have this unique skill here.
Commented by Ar Brandon last updated on 14/Feb/21
😃
😃
Commented by mr W last updated on 14/Feb/21
the best teacher is the own interest...
thebestteacheristheowninterest
Answered by ajfour last updated on 15/Feb/21
Commented by ajfour last updated on 15/Feb/21
First just assuming (without  proof) that u is u_(min)  when  least energy is lost upon collision.  Let just before collision   velocity   v_1 ^� =−qi^� −pj^�   and just after  v_2 ^� =−qi^� +epj^�   −△K =(p^2 /2)(1−e^2 )  This is minimum if p is  minimum.  But ball has to reach back ⇒  (e^2 p^2 )_(min) =2(gsin 2φ)R(1+cos 2φ)  say  2φ=θ , and  f(θ)=sin θ(1+cos θ)  f ′(θ)=cos θ+cos^2 θ−sin^2 θ             =2cos^2 θ+cos θ−1 =0  ⇒   cos 2φ=cos θ=−(1/4)±(√((1/(16))+(1/2)))  cos 2φ=cos θ=(1/2)  ⇒  φ=30°    Now from  (e^2 p^2 )_(min) =2(gsin 2φ)R(1+cos 2φ)  ⇒  p_(min) ^2 =((2/e^2 ))(((g(√3))/2))(((3R)/2))  p_(min) =(√((3(√3)gR)/(2e^2 )))  u_(y,min) ^2 =p_(min) ^2 −2(gsin 2φ)R(1+cos 2φ)              =((3(√3)gR)/(2e^2 ))−((3(√3)gR)/2)  u_(y,min) =(√((3(√3)gR(1−e^2 ))/(2e^2 )))  Rsin 2φ=u_x t_1 −(((gcos 2φ)/2))t_1 ^2   u_x =((R(√3))/(2t_1 ))+((gt_1 )/4)  t_1 =((p_(min) −u_(y,min) )/(gsin 2φ))=2(((p_(min) −u_(y,min) )/(g(√3))))  t_1 =(2/(g(√3)))((√((3(√3)gR)/(2e^2 ))) )(1±(√(1−e^2 )) )  u_(min) ^2 =(u_x ^2 +u_y ^2 )    with φ=30°    =(((R(√3))/(2t_1 ))+((gt_1 )/4))^2 +((3(√3)gR(1−e^2 ))/(2e^2 ))
Firstjustassuming(withoutproof)thatuisuminwhenleastenergyislostuponcollision.Letjustbeforecollisionvelocityv¯1=qi^pj^andjustafterv¯2=qi^+epj^K=p22(1e2)Thisisminimumifpisminimum.Butballhastoreachback(e2p2)min=2(gsin2ϕ)R(1+cos2ϕ)say2ϕ=θ,andf(θ)=sinθ(1+cosθ)f(θ)=cosθ+cos2θsin2θ=2cos2θ+cosθ1=0cos2ϕ=cosθ=14±116+12cos2ϕ=cosθ=12ϕ=30°Nowfrom(e2p2)min=2(gsin2ϕ)R(1+cos2ϕ)pmin2=(2e2)(g32)(3R2)pmin=33gR2e2uy,min2=pmin22(gsin2ϕ)R(1+cos2ϕ)=33gR2e233gR2uy,min=33gR(1e2)2e2Rsin2ϕ=uxt1(gcos2ϕ2)t12ux=R32t1+gt14t1=pminuy,mingsin2ϕ=2(pminuy,ming3)t1=2g3(33gR2e2)(1±1e2)umin2=(ux2+uy2)withϕ=30°=(R32t1+gt14)2+33gR(1e2)2e2
Commented by ajfour last updated on 15/Feb/21
Please review it Sir, it dont  seem to taly with your solution  and hence with the original  question, but yet i think it is  a nice question, and its answer  as i have derived, did have a  chance (i suspected) to yield   the same answer...
PleasereviewitSir,itdontseemtotalywithyoursolutionandhencewiththeoriginalquestion,butyetithinkitisanicequestion,anditsanswerasihavederived,didhaveachance(isuspected)toyieldthesameanswer
Commented by mr W last updated on 15/Feb/21
let me some time to follow your  working in detail. i think you  interpreted the question differently  as i did.  the question should ask  with which smallest speed one can  throw the ball into the bowl so that  it can return back to the start point.  e.g.  if e=0.75 the smallest speed is  1.6082(√(gR)) when the ball follows  following track:
letmesometimetofollowyourworkingindetail.ithinkyouinterpretedthequestiondifferentlyasidid.thequestionshouldaskwithwhichsmallestspeedonecanthrowtheballintothebowlsothatitcanreturnbacktothestartpoint.e.g.ife=0.75thesmallestspeedis1.6082gRwhentheballfollowsfollowingtrack:
Commented by mr W last updated on 15/Feb/21
Commented by mr W last updated on 15/Feb/21
for any other cases, i think also your  result, the ball needs more speed  than 1.6082(√(gR)).  if e=1, the smallest speed should be  0.9098(√(gR)).
foranyothercases,ithinkalsoyourresult,theballneedsmorespeedthan1.6082gR.ife=1,thesmallestspeedshouldbe0.9098gR.
Commented by mr W last updated on 15/Feb/21
acc. to your result, for e=1,  t_1 =(√((2(√3)R)/g))   u_(min)  =(((√3)/2)+((√(2(√3)))/4))(√(gR))≈1.3313(√(gR))  >0.9098(√(gR)) (what i got)
acc.toyourresult,fore=1,t1=23Rgumin=(32+234)gR1.3313gR>0.9098gR(whatigot)
Commented by mr W last updated on 15/Feb/21
i think the least energy loss during  collision doesn′t mean the minimum  total energy the ball needs.
ithinktheleastenergylossduringcollisiondoesntmeantheminimumtotalenergytheballneeds.
Commented by ajfour last updated on 16/Feb/21
I know you are right Sir.
IknowyouarerightSir.
Commented by mr W last updated on 16/Feb/21
i think i have found the logic fault  in your solution:  with  But ball has to reach back ⇒  (e^2 p^2 )_(min) =2(gsin 2φ)R(1+cos 2φ)  we only ensure that the ball can reach  back its height (position in y−  direction), the ball may not come  back to its start point (also in x−  direction), so it is something like  this:
ithinkihavefoundthelogicfaultinyoursolution:withButballhastoreachback(e2p2)min=2(gsin2ϕ)R(1+cos2ϕ)weonlyensurethattheballcanreachbackitsheight(positioninydirection),theballmaynotcomebacktoitsstartpoint(alsoinxdirection),soitissomethinglikethis:
Commented by mr W last updated on 16/Feb/21

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