Tension in string for a bob pendulum

  • Thread starter Nicholaz99
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Homework Statement


-------------------------------------
|.\
|...\
|.....\
|.......\
Q.......O P

O=the bob
Teta=60 degree
The bob of a simple pendulum is released from rest at P. The mass of the bob is m and the lenght of the pendulum is L

What is the tension in the string when the pendulum bob is at position Q (where the string is vertical)?

I don't have any idea about this.. please guys help me to solve this problem... Thanks
 
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Answers and Replies

  • #2
Suraj M
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You could give it a try you know
Start off by identifying what forces act in the Bob as it moves to Q from P
 
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  • #3
At first, I think that the tension string will be the same as the Weight, so i answer T=mg, but the answer key is T=2mg.
So, I start to think about the centripetal and centrifugal force... but still I don't have any idea to solve it..
Is my idea about the centripetal and centrifugal is correct?
 
  • #4
Suraj M
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Perfect! Now that centrifugal force is arising due to the motion of the Bob right? How do you relate the force and motion?
 
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  • #5
haruspex
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At first, I think that the tension string will be the same as the Weight, so i answer T=mg, but the answer key is T=2mg.
So, I start to think about the centripetal and centrifugal force... but still I don't have any idea to solve it..
Is my idea about the centripetal and centrifugal is correct?
Yes, it's about centripetal force. What do you need to figure out in order to find the centripetal force at Q?
 
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  • #6
That's my problem.. i don't understand well about the centripetal force.. am i right if i draw the centripetal force in the same direction as T? So my equation will be
Fy=0
T=W-Fs

Also when does we use the centrifugal and when does we use the centripetal one? Thanks for all
 
  • #7
Suraj M
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Think of it practically Nichol
If you take a bob with a thread and hold it in your hand you feel a certainty pull now if you use it as a pendulum does that pull increase or decrease? Depending on that judge if your equation
T= W-Fs is right or not
Your Fs is what again?
Please define all the symbols you use :-)
 
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  • #8
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you have to get your frame of reference right.
if you are working in the frame of reference of the bob you have to take centrifugal force into consideration(it acts outwards)
but if you are working in the frame of the earth (or ground or the point of suspension of the bob) you have to take into consideration the centripetal force.
(since it is necessary for rotational motion,in this case the tension provides the necessary centripetal force)
if you don't know what that means i suggest first reading up on that.
 
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  • #9
Ah i see...Now I understand that my equation is wrong...
It must be T=W+Fcentrifugal
T=mg+mv^2/r
T=mg+m(omega)^2 L

Now, how do i change the omega ???
Thanks guys for your helps...
 
  • #10
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Ah i see...Now I understand that my equation is wrong...
It must be T=W+Fcentrifugal
T=mg+mv^2/r
T=mg+m(omega)^2 L

Now, how do i change the omega ???
Thanks guys for your helps...
you can use the law of conservation of mechanical energy it will give you the velocity at Q
then find omega and solve for T.
 
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  • #11
Thanks for the help guys... I appreciate it so much :D
Finally I solve this problem lol
 

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