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Question-29322




Question Number 29322 by ajfour last updated on 07/Feb/18
Commented by ajfour last updated on 07/Feb/18
Find acceleration of sphere as  a function of angle θ.  System is released at θ= θ_0  (<90°) ;  (All surfaces are frictionless).
Findaccelerationofsphereasafunctionofangleθ.Systemisreleasedatθ=θ0(<90°);(Allsurfacesarefrictionless).
Commented by ajfour last updated on 07/Feb/18
Please solve or check my solution   mrW  Sir.
PleasesolveorcheckmysolutionmrWSir.
Commented by ajfour last updated on 07/Feb/18
Commented by ajfour last updated on 07/Feb/18
x=Rcot (θ/2)  v=−((ωR)/2)cosec^2 (θ/2)  ⇒  v^2 =((𝛚^2 R^2 )/4)cosec^4 (θ/2)   ....(i)  a=(dv/dt)= −((αR)/2)cosec^2 (θ/2)           +((ω^2 R)/2)cosec^2 (θ/2)cot (θ/2)     ...(ii)  Nsin θ = ma    ....(iii)  NRcot (θ/2)−((MgLcos θ)/2) =(((ML^2 )/3))α                                            ⇒ α=(3/(ML^2 ))(((maRcot (θ/2))/(sin θ))−((MgLcos θ)/2))                                            .....(iv)  ((MgL)/2)(sin θ_0 −sin θ)=(1/2)mv^2 +(1/2)(((ML^2 )/3))ω^2   using (i):  ((MgL)/2)(sin θ_0 −sin θ)=((mω^2 R^2 )/8)cosec^4 (θ/2)                                            +((ML^2 ω^2 )/6)  ω^2 =((((MgL)/2)(sin θ_0 −sin θ))/(((ML^2 )/6)+((mR^2 )/8)cosec^4 (θ/2)))   ....(A)  Using  eq. (A), (iii), (iv) in eq.(ii)  a=(R/2)cosec^2 (θ/2)((3/(ML^2 )))(((MgLcos θ)/2)−((maRcot (θ/2))/(sin θ)))       +(R/2)cosec^2 (θ/2)cot (θ/2)[((((MgL)/2)(sin θ_0 −sin θ))/(((ML^2 )/6)+((mR^2 )/8)cosec^4 (θ/2)))]    a[1+((3mR^2 cosec^2 (θ/2)cot (θ/2))/(2ML^2 sin θ))]=              (R/2)cosec^2 (θ/2)[((3gcos θ)/(2L))+((((MgL)/2)(sin θ_0 −sin θ)cot (θ/2))/(((ML^2 )/6)+((mR^2 )/8)cosec^4 (θ/2)))]  .
x=Rcotθ2v=ωR2cosec2θ2v2=ω2R24cosec4θ2.(i)a=dvdt=αR2cosec2θ2+ω2R2cosec2θ2cotθ2(ii)Nsinθ=ma.(iii)NRcotθ2MgLcosθ2=(ML23)αα=3ML2(maRcotθ2sinθMgLcosθ2)..(iv)MgL2(sinθ0sinθ)=12mv2+12(ML23)ω2using(i):MgL2(sinθ0sinθ)=mω2R28cosec4θ2+ML2ω26ω2=MgL2(sinθ0sinθ)ML26+mR28cosec4θ2.(A)Usingeq.(A),(iii),(iv)ineq.(ii)a=R2cosec2θ2(3ML2)(MgLcosθ2maRcotθ2sinθ)+R2cosec2θ2cotθ2[MgL2(sinθ0sinθ)ML26+mR28cosec4θ2]a[1+3mR2cosec2θ2cotθ22ML2sinθ]=R2cosec2θ2[3gcosθ2L+MgL2(sinθ0sinθ)cotθ2ML26+mR28cosec4θ2].
Commented by mrW2 last updated on 07/Feb/18
solution is correct sir!
solutioniscorrectsir!

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