A jewel smith wishing to buff a finished piece of jewelry attaches a buffing disk to his drill. The radius of the disk is 4.30 mm, and he operates it at 2.50 104 rad/s. (a) Determine the tangential speed, in m/s, of the rim of the disk. m/s (b) The jeweler increases the operating speed so that the tangential speed of the rim of the disk is now 280 m/s. What is the period of rotation, in seconds, of the disk now? s

Solution (a) Find the angular displacement after 2.00 s, in both radians and revolutions. 1 A0 = w;t + Use Equation 7.8, setting @; 1.60 rad/s, a = 3.50 rad/s2, and t = 2.00 s. 1 Að = (1.60 rad/s) (2.00 s) + (3.50 rad/s²) (2.00 s)² 2 A0 = 10.2 rad Convert radians to revolutions. A0 = (Am rad)(1.00 rev/2r rad) AO = 16.02 X rev Your response differs from the correct answer by more than 100%. (b) What is the angular speed of the wheel at t = 2.00 s? Substitute the same values into Equation 7.7. @ = @; + at = 1.60 rad/s + (3.50 rad/s²)(2.00 s) W = rad/s Remarks The result of part (b) could also be obtained from Equation 7.9 and the results from part (a). - IN

A merry-go-round has a mass of 1450 kgkg and a radius of 7.50 mm . How much net work is required to accelerate it from rest to a rotation rate of 1.00 revolution per 8.10 ss ? Assume it is a solid cylinder.

A wheel with rotational inertia 46 kg.m2 is mounted on a fixed, frictionless axle. The angular speed of the wheel is increased from zero to 43 rad/s in 19-s. What is the AVERAGE power input to the wheel during this time interval?