Question

You have a grindstone (a disk) that is 80.0 kg, has a 0.370-m radius, and is turning at 80.0 rpm, and you press a steel axe against it with a radial force of 19.0 N.
(a)
Assuming the kinetic coefficient of friction between steel and stone is 0.30, calculate the angular acceleration (in rad/s2) of the grindstone. (Indicate the direction with the sign of your answer.)
(b)How many turns (in rev) will the stone make before coming to rest?

Answer 1

Given:
- m = 80.0 kg, R = 0.370 m, ω0 = 80.0 rpm = (80/60)·2π = 8π/3 = 8.3776 rad/s,
- N = 19.0 N, μk = 0.30.

Friction force: Ff = μk N = 0.30(19.0) = 5.70 N.
Friction torque (opposes rotation): τ = Ff R = 5.70(0.370) = 2.109 N·m.

Moment of inertia of disk: I = (1/2) m R^2 = 0.5(80.0)(0.370^2) = 5.476 kg·m^2.

(a) Angular acceleration: α = −τ / I = −2.109 / 5.476 = −0.385 rad/s^2 (negative sign = slowing down).

(b) Number of turns before stopping:
Use ωf^2 = ω0^2 + 2αθ with ωf = 0 → θ = −ω0^2/(2α) = ω0^2/(2|α|).
ω0^2 = (8π/3)^2 ≈ 70.18, so θ ≈ 70.18/(2·0.385) ≈ 91.12 rad.
Revolutions = θ/(2π) ≈ 91.12 / (2π) ≈ 14.5 rev.

Answers:
(a) α = −0.385 rad/s^2
(b) ≈ 14.5 revolutions before stopping.