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You have a horizontal grindstone (a disk) that is 85 kg, has a 0.39 m radius, is turning at 85 rpm (in the positive direction), and you press a steel axe against the edge with a force of 19 n in the radial direction. 50% part (a) assuming that the kinetic coefficient of friction between steel and stone is 0.20, calculate the angular acceleration of the grindstone in rad/s2.

Respuesta :

m = mass of disk = 85 kg

r = radius of disk = 0.39 m

moment of inertia of the disk is given as

I = (0.5) m r²

I = (0.5) (85) (0.39)² = 6.5 kgm²

F = normal applied force at the edge = 19 N

[tex] \mu [/tex] = coefficient of friction = 0.20

frictional force is hence given as

f = [tex] \mu [/tex] F

f = 0.20 x 19 = 3.8 N

[tex] \alpha [/tex] = angular acceleration

Torque due to frictional force is given as

[tex] \tau [/tex] = f r

Torque is also given as

[tex] \tau [/tex] = I [tex] \alpha [/tex]

hence

I [tex] \alpha [/tex] = f r

inserting the values

(6.5) [tex] \alpha [/tex] = (3.8) (0.39)

[tex] \alpha [/tex] = 0.23 rad/s²

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