Mass on a rotating turntable problem.

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Homework Help Overview

The discussion revolves around a problem involving a mass on a rotating turntable, focusing on the forces acting on the mass and the conditions for circular motion. Participants are exploring concepts related to forces, friction, and motion in a circular path.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to analyze the free body diagram (FBD) of the block, questioning the forces acting in both the x and y coordinates. Some participants discuss the relationship between centripetal force and friction, while others inquire about the speed of the turntable and its relationship to the mass.

Discussion Status

The discussion is active, with participants providing insights into the forces involved and the relationship between speed and friction. There is an exploration of different interpretations regarding the forces acting on the mass and the conditions for motion.

Contextual Notes

Participants are considering the implications of static friction and the conditions under which the mass begins to translate radially. There may be assumptions about the system's setup that are being questioned.

btbam91
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[PLAIN]http://img841.imageshack.us/img841/7913/314pw.jpg

I'm having a bit of trouble with this one.

So the FBD of the block.

In the y coordinates, the normal equals its weight

in the x coordinates, we have static friction to the left, and is there a force to the right? There as to be right, since it eventually begins translating radially?

Just looking for a little guidance, thanks!
 
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From the point of view of the outside observer, a centripetal force is required to hold the mass in circular motion. This is provided by the force of friction. So you begin with Fc = Ff, put in the details and calculate the speed for which they are just equal.
 
the speed of the turntable and the block or just equal?

omega*R?
 
Yes, the speed of the turntable omega*R.
Then you can go for the time to reach that speed.
 

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