Coin Slipping on a Spinning Wheel

In summary: And also because the centripetal component of the force increases as the coin speeds up.OHH I SEE!! So basically, it is a combination of both of those things I stated above.
  • #1
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Homework Statement
A coin is set on a large turntable a distance of 4 m from the center of the table. Two trials are performed with the table starting at rest. On Trial 1, the turntable increases its motion very gradually, and the coin slips on the turntable surface when the coin moves any faster than 6 m/s. On Trial 2, the turntable increases its motion so that the coin's speed increases at a rate of 8 m/s^2, and the coin slips when its speed reaches 4 m/s. Explain why the speed to make the coin slip in Trial 2 was less than in Trial 1.
Relevant Equations
##F = ma##, ##F_c = \frac {mv^2} r##
The coin slipped in Trial 2 at a lower speed because the tangetial acceleration was higher than it was in Trial 1. The coin slips whenever a force overcome static friction. There is a force of static friction acting tangetially to the circle upon the coin causing it to accelerate. If the tangential acceleration is too great, the tangential force from the turn table would be greater than the tangential max static friction, meaning the coin would slip tangentially. There are 2 possible ways it could slip, either from a centripetal force overcoming the centripetal static friction like in Trial 1, or a tangential force overcoming the tangential static friction like in Trial 2.

##F_{static} = \mu_{static} * F_N##
##F_N = F_g = mg##
##F_{static} = \mu_{static} * mg##

If positive x is defined to be tangetial to the circle, then:

##F_{netx} = ma_x = F_{static}##

So ##ma_x## has to be less than ##\mu_{static} * mg## for it to not slip because of tangential static friction.

I think that my explanation above is correct (I could be wrong though), however the one thing that I cannot seem to get around is why it doesn't slip immediately. If my explanation was correct in that the tangential force was greater than the tangential static friction causing to to slip, then why wouldn't it slip as soon as the table began accelerating?
 
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  • #2
uSee2 said:
The coin slipped in Trial 2 at a lower speed because the tangetial acceleration was higher than it was in Trial 1.
That's a given. It's not the full explanation.
uSee2 said:
The coin slips whenever a force overcome static friction. There is a force of static friction acting tangetially to the circle upon the coin causing it to accelerate.
Force is a vector. In this case, motion is two-dimensional.
uSee2 said:
If the tangential acceleration is too great, the tangential force from the turn table would be greater than the tangential max static friction, meaning the coin would slip tangentially.
True, but not the whole story.
uSee2 said:
There are 2 possible ways it could slip, either from a centripetal force overcoming the centripetal static friction like in Trial 1, or a tangential force overcoming the tangential static friction like in Trial 2.
This is not quite right. Force is a vector.
uSee2 said:
I think that my explanation above is correct (I could be wrong though), however the one thing that I cannot seem to get around is why it doesn't slip immediately.
Precisely. Therefore, your explanation cannot be correct.
uSee2 said:
If my explanation was correct in that the tangential force was greater than the tangential static friction causing to to slip, then why wouldn't it slip as soon as the table began accelerating?
Is the required static frictional force constant in time?
 
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  • #3
PeroK said:
That's a given. It's not the full explanation.

Force is a vector. In this case, motion is two-dimensional.

True, but not the whole story.

This is not quite right. Force is a vector.

Precisely. Therefore, your explanation cannot be correct.

Is the required static frictional force constant in time?
I believe that it is, as the roughness of the turn table does not change.
 
  • #4
uSee2 said:
I believe that it is, as the roughness of the turn table does not change.
Think again. And note the word required in my question.
 
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  • #5
PeroK said:
Think again. And note the word required in my question.
In terms of the force required to keep it stuck to the wheel, that force is not constant. Since the static friction force matches the force pushing it.
 
  • #6
uSee2 said:
In terms of the force required to keep it stuck to the wheel, that force is not constant. Since the static friction force matches the force pushing it.
Precisely!
 
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  • #7
What changes as the coin speeds up?
 
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  • #8
PeroK said:
What changes as the coin speeds up?
The velocity? Or is that a given when it speeds up already?
 
  • #9
uSee2 said:
The velocity? Or is that a given when it speeds up already?
What changes relevant to force and acceleration?
 
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  • #10
PeroK said:
What changes relevant to force and acceleration?
As it speeds up the centripetal force increases, which is in turn the static friction force?
 
  • #11
uSee2 said:
As it speeds up the centripetal force increases, which is in turn the static friction force?
Force, ##\vec F##, is a vector. In this case it has tangential and centripetal components. The magnitude of the force, ##F \equiv |\vec F|##, depends on both components.
 
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  • #12
PeroK said:
Force, ##\vec F##, is a vector. In this case it has tangential and centripetal components. The magnitude of the force, ##F \equiv |\vec F|##, depends on both components.
Oh I see, so only the centripetal component of the static friction force increases?
 
  • #13
uSee2 said:
Oh I see, so only the centripetal component of the static friction force increases?
Yes, that's what's changing with time. And why you need more static friction with time.
 
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  • #14
PeroK said:
Yes, that's what's changing with time. And why you need more static friction with time.
OHH I SEE!! So basically, it is a combination of both of those things I stated above. Static friction is a vector like you said, so it is a combination of its centripetal and tangential components. The acceleration contributes to the tangential part of static friction, and the velocity contributes to the centripetal part of static friction. But as either tangential acceleration or velocity increases, static friction as a whole increases. And if it goes over a threshold, it slips.

And the reason that it didn't slip at first was because it didn't have enough of centripetal and tangential static friction components to slip. However it slipped quicker because there was already a large tangential component of static friction, and so even with just a small tangential velocity, it increases the centripetal component of static friction just enough such that is slipped. But less because there is a tangential component of static friction backing it up. Is this correct?
 
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  • #15
uSee2 said:
OHH I SEE!! So basically, it is a combination of both of those things I stated above. Static friction is a vector like you said, so it is a combination of its centripetal and tangential components. The acceleration contributes to the tangential part of static friction, and the velocity contributes to the centripetal part of static friction. But as either tangential acceleration or velocity increases, static friction as a whole increases. And if it goes over a threshold, it slips.

And the reason that it didn't slip at first was because it didn't have enough of centripetal and tangential static friction components to slip. However it slipped quicker because there was already a large tangential component of static friction, and so even with just a small tangential velocity, it increases the centripetal component of static friction just enough such that is slipped. But less because there is a tangential component of static friction backing it up. Is this correct?
To make myself clearerr, tangential component of static friction is increased by the tangential acceleration. And centripetal component of static friction is increased by tangential velocity?
 
  • #16
uSee2 said:
To make myself clearerr, tangential component of static friction is increased by the tangential acceleration. And centripetal component of static friction is increased by tangential velocity?
I would definitely use the phrase "required static friction" in this and other problems. Static friction is a reaction force, which has a maximum possible value, but otherwise adjusts to that required by the system.

In this case, note that not only the magnitude but also the direction of the static friction force is continuously adjusting to maintain the position of the coin on the turntable.
 
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  • #17
PeroK said:
I would definitely use the phrase "required static friction" in this and other problems. Static friction is a reaction force, which has a maximum possible value, but otherwise adjusts to that required by the system.

In this case, note that not only the magnitude but also the direction of the static friction force is continuously adjusting to maintain the position of the coin on the turntable.
I'll keep this in mind. Thank you so much for the explanation! 🙏
 
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  • #18
PS in this problem, I suggest you use the vector form of Newton's second law and show that you understand that the forces are acting in two dimensions.
 
  • #19
PeroK said:
PS in this problem, I suggest you use the vector form of Newton's second law and show that you understand that the forces are acting in two dimensions.
I've never heard of the vector form of Newton's second law. Is this just F = ma but the directions are considered?
 
  • #20
uSee2 said:
I've never heard of the vector form of Newton's second law. Is this just F = ma but the directions are considered?
Newton's second law of motion is a vector equation:
$$\vec F = m\vec a$$This is shorthand for three "scalar" equations, which in Cartesian coordinates are:
$$F_x = ma_x, F_y = ma_y, F_z = ma_z$$Alternatively, in plane polar coordinates, we have:
$$F_r = ma_r, F_{\theta} = ma_{\theta}$$Which are the centripetal and tangential components. If we consider the magnitude of the vectors we have:
$$F = |\vec F | = |m\vec a| = ma$$And, we have an equation between the magnitude of the force and the magnitude of the acceleration. The full vector equation, however, has more information.
 
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  • #21
 

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