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  • 3 years ago

If you pull on the string so that the radius of the circle decreases to one half its former value, what is the new tension in the string? What equation is used. I've tried m*v^2/.5*r, but this doesn't work.

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  1. sumanth4phy
    • 3 years ago
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    Well the velocity gets doubled when the radius gets halved. Angular momentum conservation.\[mvr=mV*(0.5r)\]\[T=mV ^{2}/(0.5*r)=8mv ^{2}/r\] you need to provide the complete question...I have guessed the question in this case n hence not completely sure :)

  2. christracted
    • 3 years ago
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    when the angular velocity is conserved, then: \[F _{tension, 1} = m*\omega ^{2}*r _{1}\] \[F _{tension, 2} = m*\omega ^{2}*\frac{ 1 }{ 2 }r _{1}\] ==> \[\frac{ F _{tension, 2} }{ F _{tension, 1} } = \frac{ m*\omega ^{2}*\frac{ 1 }{ 2 }r _{1} }{ m*\omega ^{2}*r _{1} } = \frac{ 1 }{ 2 }\] half the radius, half the force

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