Need help with Tangential Acceleration

AI Thread Summary
The discussion focuses on calculating the coefficient of static friction for a coin on a rotating record. The coin revolves with the record at 33.33 rpm, and the user has already determined the velocity and radial acceleration. It is clarified that tangential acceleration can be considered zero since the record reaches speed gradually. The normal force acting on the coin is identified as the vertical force, mg, rather than the force opposing radial acceleration. Understanding these concepts is essential for accurately calculating the coefficient of static friction.
eyvhgi557r
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1. A coin placed on a 12" record will revolve with the record when it is brought up to a speed of 33.33 rpm. Provided the coin is not more than 4" from the axis...


2. a) What is the coefficient of static friction between the coin and the record?


3. I have calculated the velocity of the coin to be 27.92 in/s and the radial acceleration to be 194.92 in/s^2. Don't I need to calculate the tangential acceleration now to calculate the total acceleration? How do I calculate the tangential acceleration? I know the formula for ta is at = dv/dt.
 
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eyvhgi557r said:
… when it is brought up to a speed of 33.33 rpm.

Don't I need to calculate the tangential acceleration now to calculate the total acceleration? How do I calculate the tangential acceleration? I know the formula for ta is at = dv/dt.

Hi eyvhgi557r! :smile:

Yes, aT = dv/dt.

But in exam questions like this, you can assume that the record is "brought up" to the final speed so gradually that the tangential acceleration can be taken to be zero.

So just forget it! :wink:
 
Great! Thanks. So how do I calculate the coefficient of static friction from the radial acceleration/force? Is the force acting on the coin in the opposite direction of the radial acceleration considered "n", the normal force?
 
eyvhgi557r said:
Great! Thanks. So how do I calculate the coefficient of static friction from the radial acceleration/force? Is the force acting on the coin in the opposite direction of the radial acceleration considered "n", the normal force?

No … the normal force is always normal to the surface … in this case, the surface is horizontal, so it's the vertical force, mg. :smile:
 
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