A grinding wheel is in the form of a uniform solid disk of radius 0.07 m and mass 2.00 kg. It starts from rest?

Question

A grinding wheel is in the form of a uniform solid disk of radius 0.07 m and mass 2.00 kg. It starts from rest and accelerated uni9formly under the action of the constant torque of 0.600 Nm that the motor exerts on the wheel.
(a)	How long does the wheel take to reach its final operating speed of 1200 rev/min
(b)	Through how many revolutions does it turn while accelerating


thanks!

Vince 505 · Accepted Answer

(a)   For a solid disk, radius of gyration, k = R/√2
 In this case, k = 0.07/√2 = 0.0495

 So, I    = mk^2
           = 2 x 0.0495^2
           = 2 x 2.45 x 10^-3 kg.m^2

 Torque = I α
     0.6 = 2 x 2.45 x 10^-3 x α
        α = 122.45 rad/sec^2

 Initial speed , ωi = 0
                     α = 122.45
                    ωf = 1200 rev/min i.e. 1200 x 2 π / 60 = 125.67 rad/sec
                    t   = ? secs
                    t   = (ωf - ωi) / α
                        = 125.67 / 122.45
Time taken is 1.026 seconds

(b)   θ  = ½  (ωf + ωi) x t
           =  0.5 x 125.67 x 1.026
           = 64.46 radians

 1 rev = 2 π rad.
No of revolutions = 10.26

? · Answer

Pinoy YFC's answer is actual however the equation for inertia is .5mr^2, no longer in elementary terms mr^2, so the seconds he provided are particularly two times as severe as is actual. This comes seven years overdue, so i'm hoping you have attained success on your existence.

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A grinding wheel is in the form of a uniform solid disk of radius 0.07 m and mass 2.00 kg. It starts from rest?

2 Answers