Question

A grinding wheel is in the form of a uniform solid disk of radius 6.96 cm and mass 2.05 kg. It starts from rest and accelerates uniformly under the action of the constant torque of 0.606 N · m that the motor exerts on the wheel.

(a) How long does the wheel take to reach its final operating speed of 1 190 rev/min?
  s

(b) Through how many revolutions does it turn while accelerating?
rev

Answer 1

mass m = 2.05 kg

radius r = 6.96 cm = 0.0696 m

Moment of inertia of the wheel I = (1/2) mr ²

                                              = 4.965 x10 ^-3 kg m ²

Initial angular speed = 0

Torque = 0.606 N m

We know = I

From this angulat accleration = / I

                                               = 122 rad/s ²

(a). Final angular speed ' = 1190 rev / min

                                        = 1190 x2 rad / 60 s

                                         = 124.6 rad/s

Required time t = ( ' - ) /

                        = (124.6-0) /122

                        = 1.021 s

(b).Angular displacment in time t is = t + (1/2) t ²

                                                       = 0 + (0.5 x 122 x 1.021 ² )

                                                       = 63.64rad

                                                       = (63.64/2) rev    where = 22/7

                                                       = 10.12 rev