A community for students.
Here's the question you clicked on:
 0 viewing
jt0770
 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.
jt0770
 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.

This Question is Open

sumanth4phy
 3 years ago
Best ResponseYou've already chosen the best response.0Well 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 :)

christracted
 3 years ago
Best ResponseYou've already chosen the best response.1when 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
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.