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Question Number 24526 by ajfour last updated on 20/Nov/17

Commented by ajfour last updated on 20/Nov/17

A thin rod rotates in horizontal plane with a constant angular velocity π›š.From its farther end a needle emerges at a constant relative velocity u along the rod. Find the initial absolute acceleration of the tip of the needle taking O to be the origin.

Athinrodrotatesinhorizontalplanewithaconstantangularvelocityω.Fromitsfartherendaneedleemergesataconstantrelativevelocityualongtherod.FindtheinitialabsoluteaccelerationofthetipoftheneedletakingOtobetheorigin.

Answered by ajfour last updated on 20/Nov/17

Answered by sma3l2996 last updated on 20/Nov/17

we have : OA^(β†’) =re_r ^β†’ with e_r ^β†’ =(r^β†’ /r) and e_ΞΈ ^β†’ itβ€²s normal vector (e_r ^β†’ βŠ₯e_ΞΈ ^β†’ ) so v^β†’ (A)=((dOA^(β†’) )/dt)=(dr/dt)e_r ^β†’ +r(de_r ^β†’ /dt)=ue_r ^β†’ +rΟ‰e_ΞΈ ^β†’ a^β†’ (A)=((dv^β†’ (A))/dt)=u(de_r ^β†’ /dt)+Ο‰((dr/dt)e_(ΞΈ ) ^β†’ +r(de_ΞΈ ^β†’ /dt)) =uΟ‰e_ΞΈ ^β†’ +Ο‰(ue_ΞΈ ^β†’ βˆ’rwe_r ^β†’ ) a^β†’ (A)=βˆ’rΟ‰^2 e_r ^β†’ +2uΟ‰e_ΞΈ ^β†’

wehave:OAβ†’=reβ†’rwitheβ†’r=rβ†’randeβ†’ΞΈitβ€²snormalvector(eβ†’rβŠ₯eβ†’ΞΈ)sovβ†’(A)=dOAβ†’dt=drdteβ†’r+rdeβ†’rdt=ueβ†’r+rΟ‰eβ†’ΞΈaβ†’(A)=dvβ†’(A)dt=udeβ†’rdt+Ο‰(drdteβ†’ΞΈ+rdeβ†’ΞΈdt)=uΟ‰eβ†’ΞΈ+Ο‰(ueβ†’ΞΈβˆ’rweβ†’r)aβ†’(A)=βˆ’rΟ‰2eβ†’r+2uΟ‰eβ†’ΞΈ

Commented by ajfour last updated on 20/Nov/17

Thanks.

Thanks.

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