Centripetal acceleration and static friction

In summary, the conversation is discussing a physics problem involving a rotating record and a coin placed on it. The goal is to determine the maximum distance the coin can be placed from the center without slipping off, given the record's rotation speed and the coefficient of static friction between the coin and the record. The conversation includes equations for finding the period of the record's rotation and the velocity of the coin, as well as a reminder to use the correct units in the equations.
  • #1
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Homework Statement




A coin is placed on a record that is rotating at 33.3 rpm. If the coefficient of static friction between the coin and the record is 0.2, how far from the center of the record can the coin be placed without having it slip off?

Homework Equations



mv^2/r = mg(mu)

The Attempt at a Solution



I can't figure out what I'm doing wrong. I converted the 33.3 rpm to radians /sec and pluged that in for v then solved for R.

I keep getting 6.204254889


 
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  • #2
Welcome to PF!

Hi shaanyr! Welcome to PF! :smile:

(have a mu: µ and an omega: ω :wink:)
shaanyr said:
mv^2/r = mg(mu)

I can't figure out what I'm doing wrong. I converted the 33.3 rpm to radians /sec and pluged that in for v

Nooo … that's ω, not v isn't it? :cry:

Use ω2r. :smile:
 
  • #3
im sorry i don't understand

use

[tex]\omega^2/r[/tex] = g x [tex]\mu[/tex]
 
  • #4
High school physics to the rescue!
Make a proportion to find the period:
33 turns/60 s = T turns/1 second
Once you have the period you can use v = circumference/period.

This is an interesting problem because r is unknown, yet needed to find v. In this kind of problem, you have to work with the letters and put in numbers only at the last stage when you have an r = formula.

Show your equations if you would like more help or a check!
 
  • #5
Just got up :zzz: …
shaanyr said:
[tex]\omega^2/r[/tex] = g x [tex]\mu[/tex]

No …

look at the dimensions (or the units) …

v = ωr,

so centripetal acceleration = v2/r = ω2r. :smile:
 

1. What is centripetal acceleration?

Centripetal acceleration is the acceleration experienced by an object moving in a circular motion. It is always directed towards the center of the circle and its magnitude is dependent on the speed and radius of the circular motion.

2. How is centripetal acceleration related to static friction?

Static friction is the force that prevents two surfaces from sliding against each other when there is no relative motion between them. In the case of centripetal acceleration, static friction is the force that acts as the centripetal force, keeping the object moving in a circular path.

3. Can static friction be greater than centripetal acceleration?

Yes, static friction can be greater than centripetal acceleration. This can happen when the speed of the object is too high or when the radius of the circular motion is too small. In this case, the static friction force needs to be greater in order to provide the necessary centripetal force to keep the object moving in a circular path.

4. How does centripetal acceleration affect the magnitude of static friction?

The magnitude of static friction is directly proportional to the centripetal acceleration. This means that as the centripetal acceleration increases, the magnitude of static friction also increases. This is because a higher centripetal acceleration requires a greater static friction force to keep the object moving in a circular path.

5. What is the role of centripetal acceleration and static friction in circular motion?

Centripetal acceleration and static friction both play important roles in circular motion. Centripetal acceleration is responsible for keeping the object moving in a circular path, while static friction is responsible for providing the necessary centripetal force. Without these forces, the object would not be able to maintain its circular motion and would move in a straight line instead.

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