Centripetal forces | finding coeff of friction with only velocity

In summary, the conversation revolves around solving a physics problem involving a ladybug on a turntable and finding the coefficient of friction. The given information includes the ladybug's slipping velocity of 0.5 m/s and equations related to force, acceleration, and friction. The conversation also touches on a possible version of the problem involving the ladybug crawling towards the rim of the turntable and the turntable's rotation rate of 33 1/3 revolutions per minute. However, it is unclear if this information was given in the original quiz. The conversation ends with uncertainty about the exact wording and accuracy of the original question.
  • #1
almosthavoc
10
0

Homework Statement


I don't remember exactly what the question was because it was on a quiz today. It said something along the lines of "a ladybug is on a turntable and begins to slip at .5 m/s. Find the coefficient of friction.

Homework Equations


Fc = (mv^2)/r
a = v^2/r
Ff = uFn

I think that's all I should need really for this?? Please let me know if I should be using anything else

The Attempt at a Solution


There hardly was an attempt unfortunately. I first set up the sum of the forces in the x direction. From that I had Fc - Ff = 0. I was able to cancel out mass but still had an unknown radius. Then I reread the problem, and the context made it sound like the lady bug wasn't under constant velocity (the lady bug begins to slip at .5 m/s - (to me at least) inferring that the ladybug was getting up to that speed). Overall I'm very confused. The physics department at my school is horrible, and all I really have for resources are yahoo answers and online forums.
 
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  • #2
We have to reconstruct a more complete problem before we can help you solve it.

Suppose that the lady bug starts at the center of the turntable and crawls toward the rim. She crawls slowly. The turntable is spinning at 33 1/3 revolutions per minute. When the bug gets far enough from the center that her tangential velocity is 0.5 m/s, she begins to slip.

How far can you get solving that version?

Edit: Corrected to match rotation rate of a 60's era LP
 
Last edited:
  • #3
jbriggs444 said:
We have to reconstruct a more complete problem before we can help you solve it.

Suppose that the lady bug starts at the center of the turntable and crawls toward the rim. She crawls slowly. The turntable is spinning at 33 1/3 revolutions per second. When the bug gets far enough from the center that her tangential velocity is 0.5 m/s, she begins to slip.

How far can you get solving that version?
where did you get the 33 1/3 from?
 
  • #4
almosthavoc said:
where did you get the 33 1/3 from?
Phonograph records (LP's) turn at a rate of 33 1/3 rotations per minute. I've edited my post to indicate minutes instead of seconds.
 
  • #5
jbriggs444 said:
Phonograph records (LP's) turn at a rate of 33 1/3 rotations per minute. I've edited my post to indicate minutes instead of seconds.
Wouldn't I be able to solve for r now since you gave me rpm? T = 1.8 and plugging it into v=(2pir)/T would give a radius of .14 meters.
Now I can set umg = (mv^2) / r. M cancels out. After plugging in I get u = .18
Did I do this correct? Even if I did, it still doesn't help me because the rpm wasn't given on the quiz
 
  • #6
almosthavoc said:
Wouldn't I be able to solve for r now since you gave me period? T = 1.8 a and plugging it into v=(2pir)/T would give a radius of .14 meters.
Now I can set umg = (mv^2) / r. M cancels out. After plugging in I get u = .18
Did I do this correct?
Let's see if I can follow. T=1.8a. That looks like a typo. The time for one rotation is 1.8 seconds. So far so good.
0.14 meters radius is right (within rounding error anyway).
0.18 for ##\mu## matches what I get as well.

It would be a good exercise to work the problem again with pure algebra rather than plugging in numbers. See if you can solve for ##\mu## in terms of the rotation rate (##\omega##), the tangential velocity at which the bug begins to slip (v) and the acceleration of gravity (g).
Even if I did, it still doesn't help me because the rpm wasn't given on the quiz
Since we do not know what was on the quiz, there is not much we can do to help.

If you do the algebra suggested above, it should become clear that the answer depends on the given value for ##\omega##.
 
  • #7
jbriggs444 said:
Let's see if I can follow. T=1.8a. That looks like a typo. The time for one rotation is 1.8 seconds. So far so good.
0.14 meters radius is right (within rounding error anyway).
0.18 for ##\mu## matches what I get as well.

It would be a good exercise to work the problem again with pure algebra rather than plugging in numbers. See if you can solve for ##\mu## in terms of the rotation rate (##\omega##), the tangential velocity at which the bug begins to slip (v) and the acceleration of gravity (g).

Since we do not know what was on the quiz, there is not much we can do to help.

If you do the algebra suggested above, it should become clear that the answer depends on the given value for ##\omega##.
The quiz literally only gave velocity. I did not leave out any givens.
 
  • #8
The quiz literally only gave velocity. I did not leave out any givens. The core of the question was slipping at .5 m/s velocity and find the coeff of friction. Everything else was useless info like draw a free body diagram
 
  • #9
almosthavoc said:
The quiz literally only gave velocity. I did not leave out any givens. The core of the question was slipping at .5 m/s velocity and find the coeff of friction. Everything else was useless info like draw a free body diagram
Again, you are asking for help with a question that you cannot provide.
 
  • #10
jbriggs444 said:
Again, you are asking for help with a question that you cannot provide.
I’m seriously giving the whole question. My wording definitely isn’t correct, but I’ve provided everything. Maybe the quiz was messed up? I highly doubt it, but it could be possible
 
  • #11
almosthavoc said:
I’m seriously giving the whole question. My wording definitely isn’t correct, but I’ve provided everything. Maybe the quiz was messed up? I highly doubt it, but it could be possible
How can we know? All you are able to give us is your recollection of the question. You have admitted that the recollection is not perfectly accurate.
 
  • #12
jbriggs444 said:
How can we know? All you are able to give us is your recollection of the question. You have admitted that the recollection is not perfectly accurate.
The only thing that’s not accurate is the exact wording. I remember all the key points. The question first starts off by saying a ladybug on a turntable begins to slip at .5 m/s. The next sentace said 1 point is awarded for drawing a free body diagram. The next sentence said 4 points are rewarded for finding the coeff of friction
 
  • #13
almosthavoc said:
The only thing that’s not accurate is the exact wording. I remember all the key points. The question first starts off by saying a ladybug on a turntable begins to slip at .5 m/s. The next sentace said 1 point is awarded for drawing a free body diagram. The next sentence said 4 points are rewarded for finding the coeff of friction
Then perhaps the test writer assumed that it is common knowledge that turntables rotate at 33 1/3 rpm. [Unless, of course, it's a 45, a 78 or a 16 2/3. https://blog.electrohome.com/vinyl-record-speeds-33-45-78-mean/]
 
  • #14
jbriggs444 said:
Then perhaps the test writer assumed that it is common knowledge that turntables rotate at 33 1/3 rpm. [Unless, of course, it's a 45, a 78 or a 16 2/3. https://blog.electrohome.com/vinyl-record-speeds-33-45-78-mean/]
Well then that would just further prove my point how horrible the physics department here is. Anyways, thank you for helping out as much as you could. I appreciate it
 
  • #15
almosthavoc said:
Well then that would just further prove my point how horrible the physics department here is. Anyways, thank you for helping out as much as you could. I appreciate it
You can apply dimensional analysis here.
You are given a velocity, you know a relevant acceleration, g, and you are asked for a dimensionless quantity. There is no way to combine a velocity and an acceleration to obtain such. You need another piece of information, whether it is a rotation rate or a radius, or whatever.
 
  • #16
haruspex said:
You can apply dimensional analysis here.
You are given a velocity, you know a relevant acceleration, g, and you are asked for a dimensionless quantity. There is no way to combine a velocity and an acceleration to obtain such. You need another piece of information, whether it is a rotation rate or a radius, or whatever.
haruspex said:
You can apply dimensional analysis here.
You are given a velocity, you know a relevant acceleration, g, and you are asked for a dimensionless quantity. There is no way to combine a velocity and an acceleration to obtain such. You need another piece of information, whether it is a rotation rate or a radius, or whatever.
I emailed my professor, and there was an error after all
 

1. What is a centripetal force?

A centripetal force is a force that acts towards the center of a circular path, keeping an object moving along that path. This force is perpendicular to the velocity of the object and is necessary to maintain circular motion.

2. How is centripetal force related to friction?

Friction is the force that opposes motion between two surfaces in contact. In the case of a circular path, friction acts as the centripetal force, providing the necessary force to keep the object moving in a circular path. The coefficient of friction is a measure of how much friction is present between two surfaces.

3. Can I find the coefficient of friction with only the velocity?

Yes, it is possible to find the coefficient of friction with only the velocity if other forces acting on the object are known. The equation for centripetal force is Fc = mv^2/r, where m is the mass of the object, v is the velocity, and r is the radius of the circular path. By rearranging this equation, the coefficient of friction can be calculated as μ = Fc/mg, where g is the acceleration due to gravity.

4. What factors affect the coefficient of friction in a circular motion?

The coefficient of friction depends on the nature of the surfaces in contact, the force pressing the surfaces together, and the velocity of the object. The coefficient of friction generally decreases as velocity increases, and it also varies with temperature and surface roughness.

5. How is the coefficient of friction used in real-life applications?

The coefficient of friction is an important factor in many real-life applications, such as designing tires for cars and calculating the force needed to keep objects in circular motion. It is also used in sports, such as ice skating and skiing, to control the speed and direction of movement. In engineering, the coefficient of friction is crucial in designing structures and machines that can withstand various forces and movements.

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