Question

A disc spins at a rate of 2 rps. There is a 15 g object on the disc located 10 cm away from the center. Find the object's tangential velocity and the frictional force that has to be exerted on the object to keep it from slipping off the disc.

My answer:
Tangential velocity = 10*2 = 20 cm/s
Frictional force = centripetal force = mv^2/r = 600 Newtons

Answer 1

If 'rps' here stands for revolutions per second, then you must convert it to radians per second first, otherwise 20cm/sec is correct

Centripetal force = mv^2/r
= 0.015*0.2*0.2/0.1
= 0.015*0.4
= 0.006 N

Gotta convert your units

Answer 2

Circumference = pi*2r = 3.14 * 20 = 62.8 cm.

V = 62.8m/rev. * 2rev/s = 125.7 m/s.

Answer 3

Correction: V = 62.8cm/rev * 2rev/s = 125.7 cm/s.