A grinding wheel is in the form of a uniform solid disk of radius 6.99.cm and ma

Question

A grinding wheel is in the form of a uniform solid disk of radius 6.99.cm and mass 1.99 kg. It starts from rest and accelerates uniformly under the action of the constant torque of 0.602 N . m that the motor exerts on the wheel. How long does the wheel take to reach its final operating speed of 1 190 rev/min? If you know the change in velocity and the acceleration you can certainly find the time. Can you determine the (angular) acceleration? S Through how many revolutions does it turn while accelerating? rev

Explanation / Answer

(a)
Final Speed, = 1190 * (2*pi)/60 rad/s
= 124.6 rad/s

Torque = I *
0.602 = 1/2 * m*r^2 *
= (0.602 * 2)/(1.99 * (6.95*10^-2)^2) rad/s^2
= 125.25 rad/s^2

We know,
= + t
124.6 = 0 + 125.25 * t
t = 0.99 s

(b)

we know,

² = ² + 2
124.6^2 = 0 + 2*125.25 *
= 61.97 rad
= 61.97/2*3.14 rev
= 9.87 rev

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