Is My Calculation for a Microwave's Turntable Spin Correct?

[unique_pavadrin]

Homework Statement

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 \frac{{60}}{3} = 20 seconds
tangential velocity \frac{{2\pi r}}{T} = \frac{{2\left( \pi \right)\left( {15} \right)}}{{20}} = 4.7123\,m\,s^{ - 1}
centripetal force \frac{{mv^2 }}{r} = \frac{{0.25\left( {\frac{{2\left( {15} \right)\left( \pi \right)}}{{20}}} \right)}}{{15}} = 0.0785\,N 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.
unique_pavadrin

 

[robb_]

I didnt check the numbers but the rest looks okay. What is providing the centripetal force here?

 

[unique_pavadrin]

the centripetal force is provided by the spinning turntable of the microwave if I'm not mistaken.

 

[sidrox]

i presumed it is friction that is providing the centripetal force...

 

[unique_pavadrin]

friction of...