A 55.0-kg grindstone is a solid disk 0.540 m in diameter. You press an ax down o

Question

A 55.0-kg grindstone is a solid disk 0.540 m in diameter. You press an ax down on the rim with a normal force of 150 N (Figure 1) . The coefficient of kinetic friction between the blade and the stone is 0.60, and there is a constant friction torque of 6.50Nm between the axle of the stone and its bearings.

a). How much force must be applied tangentially at the end of a crank handle 0.500 m long to bring the stone from rest to 120 rev/min in 9.00 s ?

b). After the grindstone attains an angular speed of 120 rev/min, what tangential force at the end of the handle is needed to maintain a constant angular speed of 120 rev/min?

c). How much time does it take the grindstone to come from 120 rev/min to rest if it is acted on by the axle friction alone?

Explanation / Answer

Given: mass of the grindstone m= 55 Kg; radius of grindstone r = 0.27 m ; opposite torque on grindstone due to axe = 150*0.6*0.27 = 24.3 N-m; and the opposite torque on grindstone due to bearing = 6.5 N-m ; initial angular velocity = 2f = 2*3.14*120/60 = 12.56 rad/s ;

(a) Now the force applied be F then the effective torque acting on the grindstone = 0.5*F - 24.3 - 6.5 = 0.5F - 30.8

Moreover applying third equation of motion in angular terms we get: 0 = - *t or    = 12.56/9 = 1.395 rad/s2

also = I* where the moment of inertia I = 1/2*m*r2 (for the disc)

so 0.5*F - 30.8 = 1/2*55*(0.27)2*1.395 or 0.5*F = 2.796 + 30.8 or F = 67.2 N

(b) Let the tangential force required be f then equating three torques we get

67.2*0.5 = f*0.6*0.27 + 6.5 or 0.162*f = 27.1 or f = 167.3 N

(c) for the axle friction alone the net angular acceleration = 6.5/I = 6.5/(0.5*55*0.272) = -3.242 rad/s2 (-ve sign is to remind that this angular acceleration would be in opposite direction to that of the motion)

now applying the first equation of motion in angular terms

0 = 120 - 3.242*t or time required t = 37 s

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