What Is the Minimum Speed and Tension in a Swinging String with Keys?

In summary, the conversation discusses finding the minimum velocity at which keys, with a combined mass of 0.100 kg, can swing and maintain a circular path on a 0.25 m long string in the vertical plane. The solution involves using the equations Fc = mv^2 /r and Fg = mg, and the concept of conservation of energy to determine the tension in the string at the bottom of the circle. The problem statement in the conversation differs from the solution in the textbook, which assumes uniform circular motion. To solve the problem, the concept of gravitational potential energy is needed, which is proportional to the mass, height, and acceleration of gravity. The height at the top of the circle is 0.50 m
  • #1
Specter

Homework Statement


Question:
Keys combined with a combined mass of 0.100 kg are attached to a 0.25 m long string swung in the vertical plane.
a) What is the slowest speed that the keys can swing and still maintain a circular path?

b) What is the tension in the string at the bottom of a the circle?

We are given:
m = 0.100 kg
r = 0.25 m

We need:
vminimum

Homework Equations


Fc = mv2 /r
Fg = mg

The Attempt at a Solution



My solution for part a which I think is correct:

Fc=mv2 /r
mg = mv2 /r
g = v2 /r
v^2 = g*r
v^2 = sqrt(9.8)(0.25)
v = 1.564 rounded 1.57 m/s.

Part b attempts:
I was told on another website to use conservation of energy but I haven't been taught that yet.

Attempt #1:
Fc = mv2 /r
Ft - mg = mv2 /r
Ft - (0.100)(9.8) = (0.100)(1.57)2 /0.25
Ft = 0.98 + 0.628
Ft = 1.608

Attempt #2:
Fc = mv2 /r
Ft-Fg = mv2 /r
Ft-mg = mv2 /r
Ft-g = v2 /r
Ft = v2 /gr
Ft = 1.572 /(9.8)(0.25)
Ft = 1.0 N

Attempt #3:
Fc = mv2 /r
Ft -mg = mv2 /r
Ft = mv2 /r + mg
Ft = (0.100)(1.57)2 /(0.25) + (0.100)(9.8)
Ft = 1.965 rounded to 2.0 N
 
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  • #2
Specter said:
Ft - mg = mv2 /r
Ft - (0.100)(9.8) = (9.8)(1.57)2 /0.25
One problem is the substitution here. You put in 9.8 for m.

The other problem is that you are using 1.57 for the velocity at the bottom of the circle. But you just finished deriving 1.57 m/s as the velocity at the top of the circle.

Conservation of energy means that the kinetic energy at the bottom of the circle must be equal to the kinetic energy at the top plus the extra potential energy at the top.

So... how much more potential energy does a 0.1 kg mass have at the top of a 0.25 m radius circle than it has at the bottom?
 
  • #3
jbriggs444 said:
The other problem is that you are using 1.57 for the velocity at the bottom of the circle. But you just finished deriving 1.57 m/s as the velocity at the top of the circle.

I thought that the question asked for me to find the minimum velocity at which the keys can be swung and still maintain a circular path. Isn't that what I solved for?

Also this is what I am going off of, I am trying to do it the same way as my course is doing it. I guess the only difference is that the velocity is given in the question and in my question I had to solve for it.
Question in the textbook
Solution for tension at the top of the circle (not needed for this question but I will include it anyways)
Solutuion for tension at the bottom of the circle.

I haven't learned anything about potential energy yet either, which is probably I don't understand this question... It's an online high school course provided by the government which means that I don't have a teacher or anything. They want me to solve this question with what I learned from the lesson, and the only tension question in the lesson is the one that I linked above so I am trying to use that to help me.

Sorry if this is confusing but I have been on this question on and off for 2 days and still can't find the answer. If I solve it in a way that's not shown in the lesson the will most likely mark it wrong and I can't contact the people who mark the homework.
 
  • #4
Specter said:
I thought that the question asked for me to find the minimum velocity at which the keys can be swung and still maintain a circular path. Isn't that what I solved for?
You solved for the minimum velocity which the keys would have to have at the top of the circle. This is not necessarily uniform circular motion. Gravity accelerates the keys on the way down and slows them on the way up. The lowest velocity will naturally occur at the top of the arc.

Potential energy is the kind of energy you have from being at the top of a hill rather than at the bottom. That's "gravitational potential energy" and is probably the most common sort of potential energy you will encounter in your course work. [There are other types, such as the potential energy in a wound up spring or that stored in a battery, but those need not concern us for now]

Gravitational potential energy is proportional to the height to which an object is raised. If one hill is twice as high as another your potential energy when on top will be twice as great on the high hill as on the low one.

Gravitational potential energy is proportional to the mass which is raised. Two wagons full of bricks on top of a hill have twice as much total potential energy as either separately.

Gravitational potential energy is proportional to the local acceleration of gravity. If gravity pulled twice as hard, then your potential energy on top of a hill would be twice as great.

Putting that together into a formula:$$PE=mgh$$ where m is the mass, g is the acceleration of gravity and h is the height [difference].

Now can you compute how much potential energy the keys have at the top of the circle compared to the bottom?
 
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  • #5
Having arrived at a location where I can view the solutions, it is clear that the solutions in the book have next to nothing to do with the problem posted in this thread. Those solutions contemplate uniform circular motion. The problem statement at the top of this thread does not.
 
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  • #6
jbriggs444 said:
Putting that together into a formula:
PE=mghPE=mgh​
PE=mgh where m is the mass, g is the acceleration of gravity and h is the height [difference].

I haven't been in high school for 5 years now so some of the stuff is new to me because I have never used the what I learned in high school and forgot it. I have also never done any physics or anything above grade 12 college level math (in Canada).

Would the height of the circle at the top be 0.50 m because r = 0.25 m , and at the bottom be 0 m? Or do I have to use the radius to find the height. I should probably know this but I don't.

Thank you for answering my questions it is very helpful!
 
  • #7
Specter said:
Would the height of the circle at the top be 0.50 m because r = 0.25 m , and at the bottom be 0 m? Or do I have to use the radius to find the height. I should probably know this but I don't.
Yes, the height would be 0.5. For purposes of potential energy, it is only differences that matter. It is convenient to label the bottom of the circle as 0 m.

It would be equally viable to label the center of the circle as 0 m, the top at 0.25m and the bottom at -0.25 m. Regardless of how where you choose to put the zero point, it's still 0.5 m from top to bottom.
 
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  • #8
jbriggs444 said:
Yes, the height would be 0.5. For purposes of potential energy, it is only differences that matter. It is convenient to label the bottom of the circle as 0 m.

It would be equally viable to label the center of the circle as 0 m, the top at 0.25m and the bottom at -0.25 m. Regardless of how where you choose to put the zero point, it's still 0.5 m from top to bottom.
Ahh okay! That makes sense. I'll try to solve part b when I can and post my solution when I am done.
 
  • #9
jbriggs444 said:
Yes, the height would be 0.5. For purposes of potential energy, it is only differences that matter. It is convenient to label the bottom of the circle as 0 m.

It would be equally viable to label the center of the circle as 0 m, the top at 0.25m and the bottom at -0.25 m. Regardless of how where you choose to put the zero point, it's still 0.5 m from top to bottom.

Alright so I have the potential energy at the top and at the bottom of the circle.

top:
PE = mgh
PE = 0.100kg * 9.8m/s2*0.5 m
PE = 0.49 J

bottom:
PE = mgh
PE = 100kg * 9.8m/s2*0 m
PE = 0 J

Is GPE measured in Joules or kgm/s?

How do I find the velocity at the bottom now that I have the potential energy at the top/bottom of the circle?

Thanks.
 
  • #10
Specter said:
Is GPE measured in Joules or kgm/s?

How do I find the velocity at the bottom now that I have the potential energy at the top/bottom of the circle?
Potential energy is in Joules -- kg m/s2. So 0.49 Joules is correct.

How much kinetic energy do the keys have at the top of the circle?
 
  • #11
jbriggs444 said:
Potential energy is in Joules -- kg m/s2. So 0.49 Joules is correct.

How much kinetic energy do the keys have at the top of the circle?
Would I convert 0.49 J to Kinectic energy? Or would the kinetic energy just be the gravity acting on the keys at the top?
 
  • #12
Specter said:
Would I convert 0.49 J to Kinectic energy? Or would the kinetic energy just be the gravity acting on the keys at the top?
The formula for kinetic energy is ##KE=\frac{1}{2}mv^2##
 
  • #13
jbriggs444 said:
The formula for kinetic energy is KE=12mv2
KE = 1/2 mv2
KE = 1/2 (0.100)(1.57)2
KE = 0.12 J
(top)

So now I have the potenital energy at the top, the kinetic energy at the top, and the velocity at the top.
 
  • #14
Specter said:
KE = 1/2 mv2
So now I have the potenital energy at the top, the kinetic energy at the top, and the velocity at the top.
Good. Now the total energy at the top is simply the total -- potential energy plus kinetic energy.

There is (presumably) no friction and no forces operating other than gravity. So conservation of energy applies and the total energy at the bottom will be the same.

The potential energy at the bottom will be zero. So the kinetic energy at the bottom must be... what?
 
  • #15
jbriggs444 said:
Good. Now the total energy at the top is simply the total -- potential energy plus kinetic energy.

There is (presumably) no friction and no forces operating other than gravity. So conservation of energy applies and the total energy at the bottom will be the same.

The potential energy at the bottom will be zero. So the kinetic energy at the bottom must be... what?
Etotal = PE + KE
Etotal = 0.49 J + 0.12 J
Etotal = 0.61 J

So if the potential energy at the bottom is zero, and the total energy at the bottom must equal the total energy at the top, the kinetic energy at the bottom would be 0.61 J?
 
  • #16
Specter said:
So if the potential energy at the bottom is zero, and the total energy at the bottom must equal the total energy at the top, the kinetic energy at the bottom would be 0.61 J?
Yep. It's just that simple.
 
  • #17
jbriggs444 said:
Yep. It's just that simple.
Alright. Now that I have the total energy at the bottom of the circle, how can I use that to find the tension in the string at the bottom?

Can I use the kinetic energy to find the tension in the string?
 
  • #18
Specter said:
Can I use the kinetic energy to find the tension in the string?
Not directly. But if you know the kinetic energy, you can find the velocity...
 
  • #19
jbriggs444 said:
Not directly. But if you know the kinetic energy, you can find the velocity...

KE = 1/2 mv2
now rearrange to solve for v
v2=√2(0.61 J)/0.100 kg
v = 3.49 m/s

Can I now use this velocity in Fc = mv2 to find tension?
 
  • #20
Specter said:
Can I now use this velocity in Fc = mv2 to find tension?
You are doing well.

Yes, you can now use that velocity to determine the centripetal force. Of course there's still another step. That centripetal force includes both gravity and tension. You are specifically interested in the tension alone.
 
  • #21
jbriggs444 said:
You are doing well.

Yes, you can now use that velocity to determine the centripetal force. Of course there's still another step. That centripetal force includes both gravity and tension. You are specifically interested in the tension alone.
Okay so I have done it like this.

Fc = Ft-Fg
Ft-Fg = mv2/r
Ft = mv2/r + mg
Ft = (0.100kg)(3.49m/s)2/0.25 m + (0.100kg)(9.8m/s2)
Ft = 48.73 N

I have also done it like this but that only solved for the centripetal force.

Fc = mv2/r
Fc = (0.100)(3.49)2/0.25
Fc = 4.87
 
  • #22
Specter said:
Ft = (0.100kg)(3.49m/s)2/0.25 m + (0.100kg)(9.8m/s2)
Ft = 48.73 N
Seems like you slipped a digit there.
 
  • #23
jbriggs444 said:
Seems like you slipped a digit there.
Oops.
Ft = 5.852 N
 
  • #24
Specter said:
Oops.
Ft = 5.852 N
I've not carefully checked the arithmetic, but the algebra seems sound.
 
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  • #25
jbriggs444 said:
I've not carefully checked the arithmetic, but the algebra seems sound.
Thanks for all of your help! I finally understand the question :)
 
  • #26
jbriggs444 said:
Good. Now the total energy at the top is simply the total -- potential energy plus kinetic energy.

There is (presumably) no friction and no forces operating other than gravity. So conservation of energy applies and the total energy at the bottom will be the same.

The potential energy at the bottom will be zero. So the kinetic energy at the bottom must be... what?
I was reading through this again and I have a question. Why is the potential energy (gravitational potential energy?) zero at the bottom of the circle?
 
  • #27
Specter said:
I was reading through this again and I have a question. Why is the potential energy (gravitational potential energy?) zero at the bottom of the circle?
This question was touched upon briefly up-thread.

There is no reason that potential energy needs to be zero at the bottom. The zero point is an arbitrary reference. You can pick any point you like and say "potential energy is zero here". Then potential energy at any other point will depend on how far that other point is above or below the reference point.

Say, for instance that you were swinging these keys from a point h meters above ground. You would go through your calculations...

The potential energy at the top is ( h meters plus 0.5 meters ) times m times g...
The potential energy at the bottom is ( h meters ) times m times g...
So the keys must have 0.5 meters times m times g more kinetic energy at the bottom than at the top.

No matter where you put the zero point, the answer you get it the same. You can even put the zero point underground. [In some circumstances, one even puts the zero point "at infinity" -- but let's not worry about that just yet].

Only differences in potential energy are physically meaningful. The energy itself is irrelevant.

If you want something physical to hang this intuition on... It takes the pretty much the same amount of energy to climb a ten foot flight of stairs regardless of whether that stairs is in Death Valley or on top of Pike's Peak.
 
  • #28
jbriggs444 said:
This question was touched upon briefly up-thread.

There is no reason that potential energy needs to be zero at the bottom. The zero point is an arbitrary reference. You can pick any point you like and say "potential energy is zero here". Then potential energy at any other point will depend on how far that other point is above or below the reference point.

Say, for instance that you were swinging these keys from a point h meters above ground. You would go through your calculations...

The potential energy at the top is ( h meters plus 0.5 meters ) times m times g...
The potential energy at the bottom is ( h meters ) times m times g...
So the keys must have 0.5 meters times m times g more kinetic energy at the bottom than at the top.

No matter where you put the zero point, the answer you get it the same. You can even put the zero point underground. [In some circumstances, one even puts the zero point "at infinity" -- but let's not worry about that just yet].

Only differences in potential energy are physically meaningful. The energy itself is irrelevant.

If you want something physical to hang this intuition on... It takes the pretty much the same amount of energy to climb a ten foot flight of stairs regardless of whether that stairs is in Death Valley or on top of Pike's Peak.
Okay that makes sense. Thanks for clearing that up.
 

Related to What Is the Minimum Speed and Tension in a Swinging String with Keys?

1. What is circular motion?

Circular motion is the movement of an object along a circular path. This type of motion is characterized by a constant speed and a change in direction, causing the object to continuously move in a circular shape.

2. How is tension related to circular motion?

Tension is the force that is exerted by a string, rope, or any other type of flexible material that is keeping an object in circular motion. Without tension, the object would not be able to maintain its circular path and would instead move in a straight line.

3. What is the relationship between velocity and circular motion?

In circular motion, the velocity of an object is constantly changing due to the object's change in direction. The velocity is always tangent to the circular path and is directed towards the center of the circle at every point along the path.

4. How does centripetal force affect circular motion?

Centripetal force is the force that keeps an object moving in a circular path. It is directed towards the center of the circle and is equal to the mass of the object multiplied by its centripetal acceleration. Without this force, the object would move in a straight line tangent to the circle.

5. What are some real-life examples of circular motion and tension?

Some examples of circular motion and tension in everyday life include a car turning a corner, a roller coaster moving along its track, and a pendulum swinging back and forth. In each of these cases, the object is experiencing circular motion and tension is necessary to keep it moving along its circular path.

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