Yahoo Answers: Answers and Comments for Physics Tetherball Problem.? [Physics]
Copyright © Yahoo! Inc. All rights reserved.
https://au.answers.yahoo.com/question/index?qid=20200227080309AA0bfXs
From Anonymous
enAU
Thu, 27 Feb 2020 08:03:09 +0000
3
Yahoo Answers: Answers and Comments for Physics Tetherball Problem.? [Physics]
292
38
https://au.answers.yahoo.com/question/index?qid=20200227080309AA0bfXs
https://s.yimg.com/zz/combo?images/emaillogoau.png

From Knr: for the moving object, force acting will be ce...
https://au.answers.yahoo.com/question/index?qid=20200227080309AA0bfXs
https://au.answers.yahoo.com/question/index?qid=20200227080309AA0bfXs
Thu, 27 Feb 2020 15:36:43 +0000
for the moving object, force acting will be centripetal force which is given by f = mv’2 /r, where “r” is the radius of the circle and it will be equal to length of the string in this case. Further what is the angle you asked for,Iam not sure.

From Steve4Physics: If θ is string’s angle to vertical, the radius...
https://au.answers.yahoo.com/question/index?qid=20200227080309AA0bfXs
https://au.answers.yahoo.com/question/index?qid=20200227080309AA0bfXs
Thu, 27 Feb 2020 11:01:43 +0000
If θ is string’s angle to vertical, the radius (of ball’s horizontal circular path) is Lsinθ.
In vertical direction: Tcosθ = mg (equation 1)
In horizontal direction: Tsinθ = mv²/r = mv²/(Lsinθ)
Tsin²θ = mv²/L (equation 2)
Then do some messy algebra to solve for θ and T.

From Anonymous: (m v^2 / r) sin@ = m g cos@
where
r = L cos@
...
https://au.answers.yahoo.com/question/index?qid=20200227080309AA0bfXs
https://au.answers.yahoo.com/question/index?qid=20200227080309AA0bfXs
Thu, 27 Feb 2020 11:01:13 +0000
(m v^2 / r) sin@ = m g cos@
where
r = L cos@
then
(m v^2 / L) tan@ = m g cos@
or
A tan@ = B cos@
where
A = m v^2 / L
B = m g
then
(A/B)^2 tan@^2 = cos@^2
taking into account that
cos@^2 = 1 / (1+ tan@^2)
on obtain
(A/B)^2 tan@^2 = 1 / (1 + tan@^2)
tan@^2 + (A/B)^2 tan@^4  1 = 0
or
x + (A/B) x^2  1 = 0
where
x = tan@^2
solve for x and find tan@^2 and then tan@ and @
tension is
T = (m v^2 / r) cos@ = m g sin@
r = L cos@
then
T = m v^2 / L + m g sin@