A grindstone in the shape of a solid disk with diameter 0.520 m and a mass of 50

Question

A grindstone in the shape of a solid disk with diameter 0.520 m and a mass of 50.0 kg is rotating at 900 rev/min . You press an axe against the rim with a normal force of 160 N (see the figure below), and the grindstone comes to rest in 7.50 s . (Figure 1)

Part A

Find the coefficient of kinetic friction between the axe and the grindstone. There is negligible friction in the bearings.

Express your answer to three significant figures.

Figure 1 of 1

A grindstone in the shape of a solid disk with diameter 0.520 m and a mass of 50.0 kg is rotating at 900 rev/min . You press an axe against the rim with a normal force of 160 N (see the figure below), and the grindstone comes to rest in 7.50 s . (Figure 1)

Part A

Find the coefficient of kinetic friction between the axe and the grindstone. There is negligible friction in the bearings.

Express your answer to three significant figures.

Figure 1 of 1

Explanation / Answer

For a solid disk, radius of gyration, k = R/ 2
In this case, R = 0.52 /2 = 0.26
k = 0.26/2 = 0.1838m

Inertia of disk = mk^2
= 50 x 0.1838^2 = 1.689 kg.m^2

Initial angular velocity, i = 900 x 2 x /60 = 94.2 rad/s
Deceleration : i - f /t = (94.2 - 0) / 7.5 = 12.56 rad/s^2

Braking Torque = I = 1.689 x 12.56 = 21.213 Nm
Torque = force x radius
21.213 = Friction force x 0.26
Friction force = 21.213/ 0.26 = 81.59 N
Coefficient of friction = F/ R = 81.59 /160 = 0.5099

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