Adisc-shaped grindstone with mass 50 kg and diameter 0.52m rotates on frictionless bearings at 850 rev/min. an ax is pushed against the rim (to sharpen it) with a normal force of 160 n. the grindstone subsequently comes to rest in 7.5 seconds. what is the co-efficient of friction between ax and stone?

0.48

Explanation:

m = mass of disc shaped grindstone = 50 kg

d = diameter of the grindstone = 0.52 m

r = radius of the grindstone = (0.5) d = (0.5) (0.52) = 0.26 m

w₀ = initial angular speed = 850 rev/min = 850 (0.10472) rad/s = 88.97 rad/s

t = time taken to stop = 7.5 sec

w = final angular speed  0 rad/s

Using the equation

w = w₀ + α t

0 = 88.97 + α (7.5)

α = - 11.86 rad/s²

F = normal force on the disc by the ax = 160 N

μ = Coefficient of friction

f = frictional force

frictional force is given as

f = μ F

f = 160 μ

Moment of inertia of grindstone is given as

I = (0.5) m r² = (0.5) (50) (0.26)² = 1.69 kgm²

Torque equation is given as

r f = I |α|

(0.26) (160 μ) = (1.69) (11.86)

μ = 0.48

6/12=1/2

step-by-step explanation: