# Centripetal acc and tension

None_of_the
Hi,
please see the drawing I ' m not good with words.

If both string have the same lenght, will they have the same tension ?

I guess the answer is yes, they share the same centripetal acceleration, and they have the same lenght. The tension will be the same because they share too the force ( mg).

Its strange because when the stick dont turn only the upper string hold the weight.

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## Answers and Replies

Mentor
Don't guess: Figure it out. Label the forces acting on the bead and apply Newton's 2nd law.

cupid.callin
Hi,
please see the drawing I ' m not good with words.

If both string have the same lenght, will they have the same tension ?

I guess the answer is yes, they share the same centripetal acceleration, and they have the same lenght. The tension will be the same because they share too the force ( mg).

Its strange because when the stick dont turn only the upper string hold the weight.

Is the ball going in circular motion around stick? ... or something else?

Mentor
believe me I want to figure it out
Want i don't know is that it is or not possible that the ball on a one string system goes higher that the stick.
Is this a different problem? What about the first problem?

None_of_the
Sorry,
I took some time off to have a fresh start.
I dont think that both string have the same tension since one have to support the mass of the ball.
We know the verticals forces have to be equal since it make a equilateral triangle.
The ball is at 60 degree

So
T=tension upper string
t= tension lower string
Tcos(60)-mgcos(60)= tcos(60) cos(60)=0.5
T=(t/2+mgcos(60))/0.5
By this equation we know that the upper string tension will be more than the lower string tension.

thank you

Last edited:
Mentor
We know the verticals forces have to be equal since it make a equilateral triangle.
The vertical component of the net force must equal zero since there's no vertical acceleration.

So
T=tension upper string
t= tension lower string
Tcos(60)-mgcos(60)= tcos(60)
The weight is already vertical, so there's no cosine factor needed in that term.

Otherwise, OK!

To actually solve for the tensions, you'll need to consider the horizontal components.

Staff Emeritus