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Question Number 58171 by pierre last updated on 19/Apr/19
a particle of mass m kg is moving along  a smooth wire that is fixed in a plane.  The polar equation of the wire  is   r = ae^(3θ) . The particle moves with a cons  tant velocity of 6. At time  t = 0 , the par  ticle is at the point with polar equation  (a,θ)  a)Find the transverse and radial compo  nents of the acceleration of the particle  in terms of a and t.  b) the resultant force on the particle is  F. Show that the magnitude of F at time  t is 360mae^(18t)
aparticleofmassmkgismovingalongasmoothwirethatisfixedinaplane.Thepolarequationofthewireisr=ae3θ.Theparticlemoveswithaconstantvelocityof6.Attimet=0,theparticleisatthepointwithpolarequation(a,θ)a)Findthetransverseandradialcomponentsoftheaccelerationoftheparticleintermsofaandt.b)theresultantforceontheparticleisF.ShowthatthemagnitudeofFattimetis360mae18t
Answered by tanmay last updated on 19/Apr/19
a)a_r ^→ =r^(..) −r(θ^(..) )^2 →radial accelaration        a_T ^→ =2r^. θ^. +rθ^(..) →transvers accelaration  now r^. =(dr/dt)=(d/dt)(ae^(3θ) )=a×e^(3θ) ×3=3ae^(3θ) ×(dθ/dt)  r^. =3ae^(3θ) ×θ^.   r^(..) =(d/dt)(r^. )=(d/dt)(3ae^(3θ) ×θ^. )  r^(..) =3a[e^(3θ) ×θ^(..) +e^(3θ) ×3×(dθ/dt)×θ^. ]  r^(..) =3ae^(3θ) θ^(..) +9ae^(3θ) (θ^(..) )^2 →3rθ^(..) +9r(θ^(..) )^2   so a_r ^→ =r^(..) −r(^ θ^(..) )^2   a_r ^→ =3rθ^(..) +9r(θ^(..) )^2 −r(θ^(..) )^2 →3rθ^(..) +8r(θ^(..) )^2 ←radial accelaration  a_T ^→ =2r^. θ^. +rθ^(..)   a_T ^→ =2(3ae^(3θ) θ^. )θ^. +(ae^(3θ) )θ^(..)   a_T ^→ =6ae^(3θ) θ^. +ae^(3θ) θ^(..)   a_T ^→ =6rθ^. +r^ θ^(..) →transverse acc  pls check...
a)ar=r..r(θ..)2radialaccelarationaT=2r.θ.+rθ..transversaccelarationnowr.=drdt=ddt(ae3θ)=a×e3θ×3=3ae3θ×dθdtr.=3ae3θ×θ.r..=ddt(r.)=ddt(3ae3θ×θ.)r..=3a[e3θ×θ..+e3θ×3×dθdt×θ.]r..=3ae3θθ..+9ae3θ(θ..)23rθ..+9r(θ..)2Extra \left or missing \rightar=3rθ..+9r(θ..)2r(θ..)23rθ..+8r(θ..)2radialaccelarationaT=2r.θ.+rθ..aT=2(3ae3θθ.)θ.+(ae3θ)θ..aT=6ae3θθ.+ae3θθ..aT=6rθ.+rθ..transverseaccplscheck

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