What is the max speed the turntable (rpm) can go before the object 10cm from centre moves with static friction coefficient 0,6?

The object is 10cm from the centre, and the static friction is 0,6.

This is all the info available. I don't understand how to calculate this without the mass of the object.

Help would be greatly appreciated

1 Answer
A08
Feb 23, 2018

This is what I get

Explanation:

When the object having mass #m# is placed on a rotating turn table at a distance #r# from the center, the force of static friction provides the centripetal force for its circular motion.

  1. Centripetal force #=mromega^2#
    where #omega=2pif#
  2. Force of static friction#=mumg#

So long as the centripetal force is smaller than the force of static friction, the object will stay on the turn table. Once centripetal force becomes just greater than the force of static friction, the object will slip tangentially due to centrifugal force acting on it. Which is equal and opposite to the centripetal force - Newton's Third Law of Motion.

Equating (1) and (2) we get

#mromega_max^2=mumg#
#=>r(2pif_max)^2=mug#
#=>f_max=sqrt((mug)/(4pi^2r))#

Taking #g=9.81\ ms^-2# and inserting given values we get

#f_max=sqrt((0.6xx9.81)/(4pi^2xx0.1))#
#=>f_max=1.491\ "cps"#
#=>f_max=89.5\ "rpm"#

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