Rotational Motion

  • #1
I've been having difficulties solving this problem. I will say how I have been trying to solve it beneath the question.


A .005 kg nickel rests .100 m from the center of a record player at rest. The nickel and player have a coefficient of friction of .33.
a) If the player can accelerate at 2.50 rad/s^. How many revolutions will it go through before the nickel is thrown off?
b) The player is level .985 m from the ground. Find the nickel's displacement from leaving the player to hitting the ground.


To solve this, I figured I need to find out what force is required to overcome the friction, so I found the nickel's weight as .005kg*9.80m/s^2 = .049N and then .049N*.33 = .016 N required for the nickel to be thrown off. From here, I wasn't exactly sure what to do. I'm pretty sure I need to find out what velocity is required to get that force, so I used acceleration=force/mass to get 3.2 m/s^2 as the acceleration. To get the velocity, I used centripetal acceleration = velocity^2 / radius and got .57 m/s. I have no idea if this is correct, and if it is, I don't know how to use the 2.50 rad/s^2 acceleration of the record player to see how long it takes to reach that velocity. Once I have the time requred to reach that speed, I would just find out how many revolutions it would make in that time.

For part b) I need to know both distance and direction. I'm pretty sure I can get distance on my own, but I don't know about direction.


Thank you for the help.
 

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