You have a horizontal grindstone (a disk) that is 86 kg, has a 0.33 m radius, is turning at 92 rpm (in the positive direction), and you press a steel axe against the edge with a force of 24 n in the radial direction. a) assuming the kinetic coefficient of friction between steel and stone is .2, calculate the angular acceleration of the grindstone in rad/s^2b) how many turns will the stone make before coming to rest

a) α = 0.338 rad / s²  b)   θ = 21.9 rev

Explanation:

a) To solve this exercise we will use Newton's second law for rotational movement, that is, torque

    τ = I α

    fr r = I α

Now we write the translational Newton equation in the radial direction

    N- F = 0

    N = F

The friction force equation is

    fr = μ N

    fr = μ F

The moment of inertia of a saying is

    I = ½ m r²

Let's replace in the torque equation

    (μ F) r = (½ m r²) α

    α = 2 μ F / (m r)

    α = 2 0.2 24 / (86 0.33)

    α = 0.338 rad / s²

b) let's use the relationship of rotational kinematics

    w² = w₀² - 2 α θ

    0 = w₀² - 2 α θ

    θ = w₀² / 2 α

Let's reduce the angular velocity

     w₀ = 92 rpm (2π rad / 1 rev) (1 min / 60s) = 9.634 rad / s

    θ = 9.634 2 / (2 0.338)

     θ = 137.3 rad

Let's reduce radians to revolutions

    θ = 137.3 rad (1 rev / 2π rad)

    θ = 21.9 rev