Tension at the Top of the String

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To calculate the tension in the cord at the top of a vertical circle, it is essential to understand the forces acting on the ball. The centripetal force is derived from the equation T + mg = mv²/r, where T is the tension, m is the mass, g is the acceleration due to gravity, v is the velocity, and r is the radius of the circle. At the top of the swing, both the tension and gravitational force act in the same direction, leading to the equation T = mv²/r - mg. The confusion arises from misinterpreting centripetal force as a standalone force rather than an acceleration. Correctly applying the forces will yield the appropriate tension value at the top of the swing.
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Homework Statement



If a ball is being swung in a vertical circle at the end of a light cord at a constant speed of 3.35m/s when the the radius of the vertical circle is 0.70m and the mass of the ball is 0.160kg calculate the tension in the cord at the top of the swing.

Homework Equations


fc=T+fg


The Attempt at a Solution


m=0.160kg v=3.26m/s r=0.7m
T=fc-fg fc=T+fg T= ( mv2/r) - mg
now what confuses me is that this equation is used for tension in the cord at the bottom of the circle and it gives me the same answer what can i do to get a different answer for tension in the cord at the top of the cord
 
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Welcome to PF!

Hi Ahmad786! Welcome to PF! :smile:

(try using the X2 tag just above the Reply box :wink:)
Ahmad786 said:
fc=T+fg

I think you're confused because you're using fc, by which I assume you mean centripetal (or centrifugal?) force.

This is a bad idea.

This is an F = ma equation, with all the forces on one side, and the acceleration on the other side.

What you are calling "fc" is not a force, it's the centripetal acceleration …

just remember that both T and centripetal acceleration are always inward, but fg can be either inward or outward. :smile:
 


tiny-tim said:
Hi Ahmad786! Welcome to PF! :smile:

(try using the X2 tag just above the Reply box :wink:)


I think you're confused because you're using fc, by which I assume you mean centripetal (or centrifugal?) force.

This is a bad idea.

This is an F = ma equation, with all the forces on one side, and the acceleration on the other side.

What you are calling "fc" is not a force, it's the centripetal acceleration …

just remember that both T and centripetal acceleration are always inward, but fg can be either inward or outward. :smile:


fc is centripetal force and is given by centripetal force= mv2/r this question is supposed to be answered using T= mv2/r + mg but i am confused on how to execute it
 
Ahmad786 said:
fc is centripetal force and is given by centripetal force= mv2/r this question is supposed to be answered using T= mv2/r + mg but i am confused on how to execute it

No no no

v2/r is centripetal acceleration

multiply it by m, put it in F = ma, and it should be equal to the total forces (in the vertical direction), in this case T and ±mg.

(if your book says that it is always +mg, then your book is wrong)

You are confused because you are tyring to treat this as a problem with three forces … when you accept that there are only two forces, everything will become clearer. :smile:
 
tiny-tim said:
No no no

v2/r is centripetal acceleration

multiply it by m, put it in F = ma, and it should be equal to the total forces (in the vertical direction), in this case T and ±mg.

(if your book says that it is always +mg, then your book is wrong)

You are confused because you are tyring to treat this as a problem with three forces … when you accept that there are only two forces, everything will become clearer. :smile:

there are example problems for this but they are on the downswing of the string they use the equation that I am using but its supposed to be used differantly for the upswinng I am still very confused:shy:
 
On the downswing, T and mg are in opposite directions, but on the upswing, T and mg are in the same direction.

So the first has ma = T - mg, the second has ma = T + mg. :smile:
 
tiny-tim said:
On the downswing, T and mg are in opposite directions, but on the upswing, T and mg are in the same direction.

So the first has ma = T - mg, the second has ma = T + mg. :smile:

so for my question it would be ma=T+mg and the ma is mv2/r so T=mv2/r -mg so (0.160kg)(3.26m/s)2/(0.7m) - (0.160kg)(-9.81m/s2) or do i use positive 9.81 for this example and do (0.160kg)(3.26m/s)2/(0.7m) - (0.160kg)(+9.81m/s2)
 
Ahmad786 said:
so for my question it would be ma=T+mg and the ma is mv2/r so T=mv2/r -mg so (0.160kg)(3.26m/s)2/(0.7m) - (0.160kg)(-9.81m/s2) or do i use positive 9.81 for this example and do (0.160kg)(3.26m/s)2/(0.7m) - (0.160kg)(+9.81m/s2)

We always say g = 9.71 (positive), so the downswing has ma = T - mg, and the upswing has ma = T + mg.
 
Thx I got it :smile:
 
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