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Horizontal Circular Motion

  1. Mar 9, 2007 #1
    1. The problem statement, all variables and given/known data
    A microwave's turntable spins 3.0 revolutions per minute. It has a 250.0g cup of coffee placed 15.0cm from its centre. Find its period of revolution, its tangential velocity and the centripetal force on the cup.

    2. The attempt at a solution
    period of revolution [tex]\frac{{60}}{3} = 20[/tex] seconds
    tangential velocity [tex]\frac{{2\pi r}}{T} = \frac{{2\left( \pi \right)\left( {15} \right)}}{{20}} = 4.7123\,m\,s^{ - 1} [/tex]
    centripetal force [tex]\frac{{mv^2 }}{r} = \frac{{0.25\left( {\frac{{2\left( {15} \right)\left( \pi \right)}}{{20}}} \right)}}{{15}} = 0.0785\,N[/tex] towards the centre of the circle

    I was wondering if somebody could kindly take the time to look over my working and see if it is correct. Thank you greatly for your time and effort.
  2. jcsd
  3. Mar 9, 2007 #2
    I didnt check the numbers but the rest looks okay. What is providing the centripetal force here?
  4. Mar 9, 2007 #3
    the centripetal force is provided by the spinning turntable of the microwave if I'm not mistaken.
  5. Mar 10, 2007 #4
    i presumed it is friction that is providing the centripetal force.....
  6. Mar 10, 2007 #5
    friction of.....
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