Tension in string, vertical plane

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Homework Help Overview

The discussion revolves around a problem involving the tension in a string when swinging keys in a vertical circle. The keys have a mass of 0.100 kg and are attached to a 0.25 m long string. Participants are tasked with determining the tension at the bottom of the circle, given the velocity at the top of the circle is known to be 1.56 m/s.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need to find the velocity at the bottom of the circle and question whether to assume constant speed. Some suggest using conservation of energy to relate potential and kinetic energy changes. Others express uncertainty about the velocity at the bottom and the implications of the forces acting on the keys.

Discussion Status

The discussion is active, with various participants exploring different approaches to the problem. Some have provided insights into using conservation of energy, while others are questioning the assumptions made about speed and forces. There is no explicit consensus on the method to be used, but several productive lines of reasoning are being explored.

Contextual Notes

Participants note that the problem does not explicitly state to assume constant speed, leading to discussions about the implications of gravitational effects on the keys' speed as they move through the vertical circle. There is also mention of the potential for misunderstanding the relationship between speed and tension in the context of circular motion.

phee
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Homework Statement



Keys with a combined mass of 0.100 kg are attached to a 0.25 m long string and swung in a circle in the vertical plane.

What is the tension in the string at the bottom of the circle?

Known:
Mass = 0.100 kg
Radius = 0.25 m

Homework Equations



Fc = mv^2/r


The Attempt at a Solution



I know I need to find the velocity at the bottom to put it into the equation.

I just don't know how to figure out the velocity at the bottom, I know the velocity at the top of the circle is 1.56 m/s ^ 2


Any hints or tips on to what I should be doing?
 
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phee said:

Homework Statement



Keys with a combined mass of 0.100 kg are attached to a 0.25 m long string and swung in a circle in the vertical plane.

What is the tension in the string at the bottom of the circle?

Known:
Mass = 0.100 kg
Radius = 0.25 m

Homework Equations



Fc = mv^2/r


The Attempt at a Solution



I know I need to find the velocity at the bottom to put it into the equation.

I just don't know how to figure out the velocity at the bottom, I know the velocity at the top of the circle is 1.56 m/s ^ 2


Any hints or tips on to what I should be doing?

If this is a very simple introductory question you might be expected to assume constant speed, but more likely, the keys have a lower Potential energy at the bottom of the circle, so will have the equivalent increase in kinetic energy - and thus speed.

Handy to note that Fc and Kinetic energy both have mv2 as part of their "formula"
 
How would I go about finding the velocity at the bottom, I don't think I assume constant velocity.

104hhzn.png


A FBD I drew up, the dot at the bottom is the keys.
 
phee said:
How would I go about finding the velocity at the bottom, I don't think I assume constant velocity.

I'd do what PeterO suggests: Conservation of energy.

If you're not supposed to assume constant speed, then maybe you're supposed to assume constant energy in the system.

KEi + PEi = KEf + PEf

Sure, you could instead use kinematics, but conservation of energy is easier.

Btw, how do you know the velocity at the top of the circle is 1.56 m/s? That infomation wasn't in the problem statement. And is the square in the units (you typed 1.56 m/s ^2) a typo/mistake, or is that the acceleration?
 
The work done all the way from top to bottom is due to the Weight of keys, since Tension is centripetal and at every instant perpendicular to displacement.
 
stallionx said:
The work done all the way from top to bottom is due to the Weight of keys, since Tension is centripetal and at every instant perpendicular to displacement.
Okay, that's good. If there are no other forces or torques involved then you can use conservation of energy.

And btw, 1.56 m/s is ever so slightly not quite enough speed to keep the 0.25 m string from going slack. It's very close but not quite. Is this how you determined the "1.56" figure?

[Edit: I just realized that I wasn't replying to the original poster. (I'm doing this on a cell phone and it's tough to see everything.) But anyway, Phee, I pose the same statements/questions to you]
 
Last edited:
m*g + T(top) = m v**2 /r

T(bottom)- m*g = m V**2 /r

we should keep speed but not velocity constant for circular motion.
 
Last edited:
stallionx said:
m*g - T(top) = m v**2 /r

T(bottom)- m*g = m V**2 /r

we should keep speed but not velocity constant for circular motion.

Speed does not have to be constant for circular motion!

On a Ferris wheel, speed is approximately constant. for the "hot-wheels" toy set many children once used, the speed was certainly not constant as it did a vertical loop-the-loop.
 
phee said:
How would I go about finding the velocity at the bottom, I don't think I assume constant velocity.

104hhzn.png


A FBD I drew up, the dot at the bottom is the keys.

I am not all that convinced by the diagram. I would prefer FT to appear much longer than Fg as a way of emphasising that you knew it was larger.

I would also never show it finishing close to the centre of the circle, lest someone thought you meant it to be the same length as the radius.
 
  • #10
Should not the answer be : ( Taking g about 10 m/s**2 =

T( bottom ) = m *V**2/r + mg

which is :

0.97344 N


?
 
  • #11
And I found it is about 4 N

if we vary the speed ?
 
  • #12
stallionx said:
Should not the answer be : ( Taking g about 10 m/s**2 =

T( bottom ) = m *V**2/r + mg

which is :

0.97344 N ?

Is this a problem posted by stallionx or Phee ?
 
  • #13
stallionx said:
Should not the answer be : ( Taking g about 10 m/s**2 =

T( bottom ) = m *V**2/r + mg

which is :

0.97344 N
?
Please do not suggest answers to original poster's problem. It is against the forum rules. We're not here to do other people's homework for them.

That said, I will comment. No, that's not quite right. Conservation of energy, when involving gravity, generally implies that things speed up as they fall down. The keys' speed at the bottom is not the same as at the top.
 
Last edited:
  • #14
stallionx said:
The work done all the way from top to bottom is due to the Weight of keys, since Tension is centripetal and at every instant perpendicular to displacement.

And the significance of that is ?
 
  • #15
stallionx said:
And I found it is about 4 N

if we vary the speed ?

It is certainly going to be bigger than 1N. If the tension was less than 1N it would not even be stopped from going down - but it actually comes back up in a circular arc.
 
  • #16
PeterO said:
And the significance of that is ?

Ek(up) + m*g*L = Ek(down )
 
  • #17
stallionx said:
Ek(up) + m*g*L = Ek(down )

Firstly,

I assume that by (up) you mean when at the top, not while traveling up, and that by (down) you mean at the bottom, not while traveling down?

Secondly,

Where is Phee
 
  • #18
PeterO said:
Firstly,

I assume that by (up) you mean when at the top, not while traveling up, and that by (down) you mean at the bottom, not while traveling down?

Secondly,

Where is Phee

Yes Sir,

Your assumption is right.
 
  • #19
Ok first off, I got the velocity at the top 1.56 m/s from part A of the question.

TO find the 1.56 m/s I did v = [itex]\sqrt{}gr[/itex]

It doesn`t say to assume speed is constant throughout, and I don`t think it is as the keys traveling past the top point will end up going faster because of gravity.

Originally I had though to find the velocity at the bottom of the circle and plug the numbers into Fc=mv^2/r

I was stumped as to how to find the velocity in the first place at the bottom which led me to here.

This is my first unit of my course and I was not introduced to the conservation of energy. I will read up on it and see where it takes me, thank you.
 
Last edited:

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