Question

You have a grindstone (a disk) that is 91.0 kg, has a 0.370-m radius, and is turning at 68.0 rpm, and you press a steel axe against it with a radial force of 11.0 N.

(a)

Assuming the kinetic coefficient of friction between steel and stone is 0.50, calculate the angular acceleration (in rad/s²) of the grindstone. (Indicate the direction with the sign of your answer.)

_____rad/s²

(b)

How many turns (in rev) will the stone make before coming to rest?

_____rev

Answer 1

a) by doing the following Kinetic Friction Force​ Fk= μkF

Torque by Fk​ is τ=rFsin(θ) = rFk = rμkF

Magnitude of Angular Acceleration (will be negative)​ α=τ_NET/I = rμkF/0.5mr^2 = (0.37 m)(0.5)(11 N) / 0.5*(91)(0.37)^2

= -0.3267 rad / s^2

b)  ωi = 68*∏/30rad/s =2.267*∏ rad/s

& ωf = 0 rad/s

I can solve for time by using the formula ωf=ωi+αΔt

0 rad/s = 2.267*∏ rad/s + (-0.3267 rad/s2)Δt

-2.267*∏ rad/s = (-0.3267rad/s2)Δt

Δt = 21.72 s now using this time,

I plug it and the rest of my variables into the formula θf=θi+ωiΔt+ 0.5αΔt^2

=(2.267*∏ rad/s)(21.72 s) + 0.5(-0.3267rad/s2)(21.72 s)^2 =77.55 rad

(1rot/2∏rad) *77.55 rad =12.34 rotations