Question

A jewel smith wishing to buff a finished piece of jewelry attaches a buffing disk to his drill. The radius of the disk is 2.20 mm, and he operates it at 2.00 104 rad/s.
Determine the tangential speed, in m/s, of the rim of the disk.
The jeweler increases the operating speed so that the tangential speed of the rim of the disk is now 290 m/s. What is the period of rotation, in seconds, of the disk now?

Answer 1

v = ωr = (2.00×10^4 rad/s)(2.20×10^-3 m) = 44.0 m/s.

For v = 290 m/s, T = 2πr / v = 2π(2.20×10^-3 m) / (290 m/s) = 4.77×10^-5 s.