A grinding wheel is in the form of a uniform solid disk of radius 6.94 cm and mass 1.94 kg. It starts from rest and accelerates uniformly under the action of the constant torque of 0.594 N·m that the motor exerts on the wheel.
(a) How long does the wheel take to reach its final rotational speed of 1100 rev/min?
(b) Through how many revolutions does it turn while accelerating?
convert revs/min to rad/s:
1100rev/min * 2π/60
115.19rad/s (final angular velocity)
find the accleration:
torque = inertia * acceleration
t = (mr^2) * a
0.594 = (1.94 * 0.0694^2)a
0.594 = 0.009a
a = 63.57rad/s^2
use this equation of rotational motion:
Wf = Wi + at
115.19 = 0 + 63.57t
115.19 = 63.57t
t = 1.81secs
rad = Wi(t) + 1/2at^2
rad = 0(1.81) + 1/2(63.57)(1.81)^2
convert rad to revs:
104.36 ÷ 2π
hope you get it^_^
Pinoy YFC’s answer is correct but the equation for inertia is .5mr^2, not just mr^2, so the seconds he supplied are actually twice as high as is correct. This comes seven years late, so I hope you have attained success in your life.
I believe that **PiNoY YFC** acceleration should be in m/s^2 while his is in rad/sec…
I don’t think that’s right