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Inaugural Post - Sep 29, 2009 at 2:31PM
Dave Avatar
Dave
San Marcos, Texas
361 Posts

Hi Everyone,

     We are starting a new thread in which we will post and discuss questions from sample physics GRE exams.  The idea is that we will spend a week discussing approaches to answering the question, and hopefully at the end of the week have the correct answer.  Please post any discussion you have in this thread.

Here is this week's question:

The coefficient of static friction between a small coin and the surface of a turntable is 0.30.  The turntable rotates at 33.3 revolutions per minute.  What is the maximum distance from the center of the turntable at which the coin will not slide?

A. 0.024 m
B. 0.048 m
C. 0.121 m
D. 0.242 m
E. 0.484 m

Let the discussion commence.  If we get the answer to this question before next week, I will post another.

Dave


Try not to become a man of success, but rather try to become a man of value -- Albert Einstein

Replies to Inaugural Post

Re: Inaugural Post - Oct 01 2009 1:23PM
James Thompson
2 Posts

This is why I need to study up a lot before taking the GRE.  The velocity of the coin is (pi)*r*11/10.  I know that centripetal force is provided by friction and there is an equation that is in one of my general physics books for solving this problem without the mass of the coin, as is the immidiate obstacle.  Maybe I will get back to this later, but for now its off to Digital Electronics.

peace and physics,
Zeph



Re: Inaugural Post - Oct 08 2009 10:01AM
Dave Avatar
Dave
San Marcos, Texas
361 Posts

Didn't get much discussion on this one.  Hopefully more folks will participate as time goes on.  Anyway, here's how this problem works.

At the maximum radius, the static frictional force will be providing the needed centripetal acceleration.  We can therefore say

mu*N=m*omega^2*r

Because the turntable is horizontal, the normal force is equal to the weight of the coin.  We can also calculate the angular velocity, omega, from the fact that the turntable rotates at 33.3 revolutions per minute.  When we do this we find that

omega=3.487 rad/s

Putting this all together gives

mu*m*g=m*omega^2*r

Solving for r we get

r=mu*g/omega^2

Plugging in numbers gives r=0.242 m


Try not to become a man of success, but rather try to become a man of value -- Albert Einstein


Re: Inaugural Post - Jul 14 2010 1:06AM
Jorosh john
1 Posts

I've been studying constantly this summer, all I can do is hope and pray I pass the GSE exams!! make money from home



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