My Math Forum Physics Question.

 Physics Physics Forum

 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: $a_r=\frac{v^2}{r}$ Also, we have that the magnitude of the velocity (speed) for circular motion is $v=\frac{2\pi r}{t}$ where t is the time it takes to make one revolution. We are given a frequency f, or revolutions per time, thus: $f(\text{rps})= f(\text{rpm})\frac{1\text{min}}{60\text{sec.}}$ We want the frequency in revolutions per second, since that is the unit of time that is used for the constant of gravitational acceleration $g$. And since frequency is revolutions per time, we may express the rotational speed as: $v=2\pi rf$, and therefore, our centripetal acceleration is: $a_r=\frac{(2\pi rf)^2}{r}=4\pi^2rf^2$ Now if we equate the maximum for the force of static friction with the centripetal force (using Newton's 2nd law), we have: $f_s=\mu_smg=m4\pi^2rf^2$ Divide through by mass m, and solve for the coefficient of static friction $\mu_s$: $\mu_s=\frac{4\pi^2rf^2}{g}$ Now, using r = 0.13 m., f(rps) = f(rpm)/60 = 34/60 = 17/30, g = 9.81 m/sē, we have $\mu_s=\frac{4\pi^2(0.13\text{m.})(\frac{17\text{re v.}}{30\text{s.}})^2}{9.81\frac{\text{m.}}{\text{s .^2}}}\approx 0.168$
 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

,

,

,

,

,

,

,

,

,

,

,

,

,

,

# a coin is placed 13 cm from the axis of a rotating turntable

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

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

 Contact - Home - Forums - Cryptocurrency Forum - Top