A grindstone (a uniform disk of stone) has a diameter of 0.35 m and a mass of 80

Question

A grindstone (a uniform disk of stone) has a diameter of 0.35 m and a mass of 80 kg. The grindstone is initially rotating at 560 rev/min when an ax is sharpened by pressing the blade onto the rotating stone with a normal force of 160 N. The stone then comes to rest in 9.5 s. What is the rotational acceleration of the disk? rad/s^2| What magnitude of torque is required to produce this acceleration? N A ¢ m Assuming the force of friction from the ax is the only force slowing down the rotation of the stone, what is the force of friction the ax exerts on the stone? What is the coefficient of kinetic friction between the ax and the stone?

Explanation / Answer

Ans:- here d= 0.35m; m = 80kg wi = 560rev/min = 9.33rad/s ; Fn = 160 N t= 9.5s Wf = 0

A ] = Wf – Wi / t = 0- 9.33 / 9.5 =-0.98rad/s^2

B] torque =I* = ½*MR^2 * (-0.98) =1/2*80*(0.175)^2*(-0.98) =- 1.2Nm

    = 1.2Nm

c]by Newton’s law

F = Fk

M* = Fk

80*(-0.98) =-78.4N =Fk

D]Fk= µk*N = µk*160

µk= 78.4/160 = 0.49

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