Calculate the frequency of rotation when a coin flies off a turntable

In summary, to calculate the frequency of rotation when a coin flies off a turntable, one must determine the turntable's angular velocity and the radius at which the coin is placed. The centripetal force acting on the coin is provided by the friction between the coin and the turntable. As the turntable rotates, if the frictional force is insufficient to keep the coin in circular motion, it will leave the surface. The frequency of rotation can be calculated using the formula \( f = \frac{v}{2\pi r} \), where \( v \) is the linear speed at which the coin travels when it detaches, and \( r \) is the radius of the turntable.
  • #1
M1N
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0
Homework Statement
Hi. I have tried and tried and I just can't figure this one out. I'm assuming the outer coin flies off first, then the inner coin. I'm assuming for that to happen, the rotation frequency must be higher (or do both coins come off at the same time or the inner coin before, with the inner coin not needing to have as high a velocity?) I believe the relationship between the velocities is v_outer = 4v_inner. I am not sure how to get the frequency for the inner coin (I have working out for different ideas but it's quite a mess so I won't include it here, but my thought process is expressed above) Any help will be greatly appreciated. Thanks!
Relevant Equations
v = wr, w = 2pif, F = mv^2/r, v = 2pir/T, T = 1/f,
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  • #2
You need to share your attempt to receive help via forum rules.
 
  • #3
M1N said:
I'm assuming the outer coin flies off first, then the inner coin.
ok, but why?
M1N said:
I'm assuming for that to happen, the rotation frequency must be higher (or do both coins come off at the same time or the inner coin before, with the inner coin not needing to have as high a velocity?) I believe the relationship between the velocities is v_outer = 4v_inner.
Relevant Equations: v = wr, w = 2pif, F = mv^2/r, v = 2pir/T, T = 1/f,
Which of those equations have you tried to use?
 
  • #4
haruspex said:
ok, but why?
The last sentence in the posted screenshot of the statement of the problem says "When the frequency of the rotation reaches 3.00 Hz the outer coin (2) flies off the disk." Physical reasoning aside, it is fair to assume that the coin mentioned is the one that slides first because it provides the data for predicting when the second coin will slide.
 
  • #5
kuruman said:
The last sentence in the posted screenshot of the statement of the problem says "When the frequency of the rotation reaches 3.00 Hz the outer coin (2) flies off the disk." Physical reasoning aside, it is fair to assume that the coin mentioned is the one that slides first because it provides the data for predicting when the second coin will slide.
Providing the rotation rate at which the inner coin flies off would also work, so the text is not definitive.
But I was hoping to elicit some physical reasoning.
 
  • #6
erobz said:
You need to share your attempt to receive help via forum rules.
Hi. I have added my working out now. Sorry!
 
  • #7
kuruman said:
The last sentence in the posted screenshot of the statement of the problem says "When the frequency of the rotation reaches 3.00 Hz the outer coin (2) flies off the disk." Physical reasoning aside, it is fair to assume that the coin mentioned is the one that slides first because it provides the data for predicting when the second coin will slide.
Hi. The problem is, in my working it appears I am going back to the same 'moment' in time when the outer coin flies off. I don't know how to show when it is flying off. My working suggests the inner coin should fly off at the same frequency of rotation and thus at the same time, but I am not sure if this is right. Question c says to find the relationship between the velocities of the inner and outer coin when they are flying off. When I put the velocity of the inner coin flying off (1.32m/s) and the radius into the equation v = wr it just gives the same period of 6pi, and when I put that into w = 2pi/T or 2pif I get the same frequency of 3.00Hz. What am I missing?
 
  • #8
M1N said:
What am I missing?
For starters, I don’t see Newtons Second law anywhere in that handwritten mess( sorry for being frank). Please see latex guide. That is how we wish to interact mathematically.
 
  • #9
M1N said:
What am I missing?
You are manipulating equations that relate linear and rotational displacements and velocities hoping that the answer will somehow pop out. It will not because, as @erobz noted, you have not considered Newton's second law that governs the motion of the coins. There is a force that accelerates them and causes them to go around in a circle but only up to the point when they slide off.

So you need to consider in terms of forces (a) what must be true for a coin to go around in a circle and (b) what must be true for a coin to slide off.
 
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FAQ: Calculate the frequency of rotation when a coin flies off a turntable

What factors determine the frequency of rotation at which a coin flies off a turntable?

The frequency of rotation at which a coin flies off a turntable is primarily determined by the coefficient of friction between the coin and the turntable surface, the radius at which the coin is placed from the center, and the mass of the coin. The centripetal force required to keep the coin in circular motion must be less than or equal to the frictional force; otherwise, the coin will fly off.

How do you calculate the critical frequency at which the coin flies off?

The critical frequency can be calculated using the formula \( f = \frac{1}{2\pi} \sqrt{\frac{\mu g}{r}} \), where \( \mu \) is the coefficient of friction, \( g \) is the acceleration due to gravity (approximately 9.81 m/s²), and \( r \) is the radius from the center of the turntable to the coin. This formula derives from equating the centripetal force to the maximum static frictional force.

Does the mass of the coin affect the frequency at which it flies off?

No, the mass of the coin does not affect the frequency at which it flies off. This is because the mass cancels out when equating the centripetal force (which is mass-dependent) to the frictional force (which is also mass-dependent). Therefore, the critical frequency is independent of the coin's mass.

What role does the radius play in determining the critical frequency?

The radius plays a significant role in determining the critical frequency. As the radius increases, the critical frequency decreases. This is because a larger radius requires a higher centripetal force to maintain the circular motion, which surpasses the frictional force at a lower frequency, causing the coin to fly off sooner.

Can the type of coin affect the critical frequency at which it flies off?

The type of coin can affect the critical frequency only if it changes the coefficient of friction between the coin and the turntable. Different materials and surface textures can have different coefficients of friction. However, if the coefficient of friction remains the same, the type of coin does not affect the critical frequency due to the mass independence in the calculation.

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