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?
Ok, I thought I needed to use the equation V=sqrt(Msgr). Ms=static friction. I converted 33.3rpm into rad/s and solved for r. But my answer just isn't coming out right. Any tips??
Am I even using the correct equation??
Thank you so much in advance!
2007-02-22 12:43:17 · 1 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Find a formula for the centripetal force necessary to keep a mass moving around a curve of a certain radius. Then find the force that's necessary to move your coin (write the coin's mass as m) across the record, and substitute its expression for centripetal force in the first formula. You should find that the expressions for mass cancel out.
There are about six radians in a circle, and 33 1/3 rpm is about half a rev per second. So 33.3 rpm oughta be maybe three radians per second. Mental calculation.
2007-02-22 12:56:25 · answer #1 · answered by 2n2222 6 · 0⤊ 0⤋