Question

A grinding wheel is in the form of a uniform solid disk of radius 6.93 cm and mass 2.02 kg. It starts from rest and accelerates uniformly under the action of the constant torque of 0.597 N · m that the motor exerts on the wheel. How long does the wheel take to reach its final operating speed of 1300 rev/min? Through how many revolutions does it turn while accelerating?

Answer 1

The moment of inertia of the grinding wheel about its axis of rotation is given by, mr 2 (2.02 kg) (6.93x10 m) 4.85x10-3 kg m The torque acting on the grinding wheel is Rearrange the equation in terms of angular acceleration (α) 0.597 N 123 rad/s I 4.85 x10-3 kg m* USe the following formul to calculate the time 2π rad)( 1 min) (1300 rewmin) 1 rev人60 s 123 rad/s 1 t-1.106s The angular displacement of grinding wheel can be calculated as, -(0)(123 rad/s)(1.106 s) -75 rad The angular displacement for one complete revolution is 2π radians, so the number of revolutions to cover the angular displacement is, n 75 rad 2π rad -12 rev

Answer 2

T = Iα = mr²α/2

=> α = 2T/mr² = 2 * 0.597 / (2.02 * 0.0693²) = 123.08 rad/s²

Time taken, t = ω/α = mr²ω/2T = (1300 * 2π) / 123.08 = 66.4 s

Revolutions turned, θ = (1/2π)(αt²/2) = αt²/4π = 123.08 * 66.4² / 4π = 43183 revolutions