November 2nd, 2010, 09:31 PM  #1 
Newbie Joined: Oct 2010 Posts: 12 Thanks: 0  Physics Question.
A coin is placed 13.0cm from the axis of a rotating turntable of variable speed. When the speed of the turntable is slowly increased, the coin remains fixed on the turntable until a rate of 34rpm is reached and the coin slides off. What is the coefficient of static friction between the coin and the turntable?

November 2nd, 2010, 10:30 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs  Re: Physics Question.
We have that the magnitude of centripetal acceleration for circular motion of radius r and speed v is given by: Also, we have that the magnitude of the velocity (speed) for circular motion is where t is the time it takes to make one revolution. We are given a frequency f, or revolutions per time, thus: We want the frequency in revolutions per second, since that is the unit of time that is used for the constant of gravitational acceleration . And since frequency is revolutions per time, we may express the rotational speed as: , and therefore, our centripetal acceleration is: Now if we equate the maximum for the force of static friction with the centripetal force (using Newton's 2nd law), we have: Divide through by mass m, and solve for the coefficient of static friction : Now, using r = 0.13 m., f(rps) = f(rpm)/60 = 34/60 = 17/30, g = 9.81 m/sē, we have 
November 3rd, 2015, 09:47 PM  #3 
Newbie Joined: Nov 2015 From: eugene Posts: 1 Thanks: 0 
The circular velocity part was where I got caught. Very strait forward now, thank you.

November 12th, 2015, 05:35 AM  #4 
Senior Member Joined: May 2015 From: Varanasi Posts: 110 Thanks: 5 Math Focus: Calculus 
34 rpm means 68pi radians/60 secs that makes angular speed as 1.13pi radians per second. since coin is placed at a distance of 13 cm so linear velocity at this point will become v=rw v= (13/100)1.13pi m/sec v= 0.46 m/sec now centrefugal force on it would be m*v^2/r= m*0.46*0.46*100/13=m1.62 N frictional force would be Cmg here, c is coefficient of friction as both forces must be equal for equilibrium so Cmg=1.62m this means C=1.62/9.8 here g=9.8 acceleration due to gravity sp C=0.16 

Tags 
physics, question 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Applied Math and physics question  hunnybee  Applied Math  0  September 20th, 2013 11:51 PM 
physics time question  Darpa.Chief  Physics  1  April 16th, 2010 12:57 PM 
Need help with Application to physics question  wonger357  Calculus  4  July 24th, 2009 06:20 AM 
Physics question  MyNameIsVu  Physics  3  July 14th, 2009 12:18 AM 
challenging physics question  frenchie  Physics  0  November 22nd, 2007 02:26 PM 