Angular Acceleration Caused by Axehead on a Rotating Grindstone (Disk)

[hdp12]

Homework Statement

PROBLEM:
You have a horizontal grindstone (a disk) that is mass m, has a radius r, and is turning at f in the positive direction. You then press a steel axe against the edge with a force of F in the radial direction.

RANDOMIZED VARIABLES:
m= 95 kg
r= 0.33 m
f= 92 rpm
F= 25 N

a) Assuming the kinetic coefficient of friction between steel and stone is 0.20, calculate the angular acceleration of the grindstone in rad/s²
√ANSWER: α = -0.319

See section 3 for explanation​

b) What is the number of turns, N, that the stone will make before coming to rest?
SUBMISSION HISTORY:
XN = 0.003
XN = 0.017

Homework Equations

τ=rFsin(θ)
τ=Iα
α=τ_NET/I
↓ Moment of Inertia of a uniform disk about its center of mass
I=[itex]\frac{1}{2}[/itex]mR²
F_k= μ_kF
ω_f=ω_i+αΔt
θ_f=θ_i+ω_iΔt+[itex]\frac{1}{2}[/itex]αΔt²

The Attempt at a Solution

I solved part a) by doing the following

Kinetic Friction Force​

F_k= μ_kF

Torque by F_k​

τ=rFsin(θ) = rF_k = rμ_kF

Magnitude of Angular Acceleration (will be negative)​

α=τ_NET/I = rμ_kF/[itex]\frac{1}{2}[/itex]mr²
= (0.33 m)(0.2)(25 N) / [itex]\frac{1}{2}[/itex](95 kg)(0.33 m)²
= -0.319 rad / s²

OKAY. Now onto the next part. Part b).
First I took the initial velocity, which is in rotations per minute, and converted it into radians per second
[STRIKE]rot[/STRIKE]/[STRIKE]min[/STRIKE] ·2∏r/[STRIKE]rot[/STRIKE] · 1 [STRIKE]min[/STRIKE]/60 s
=∏/30 rad/s

so if ω_i = ∏/30 rad/s
& ω_f = 0 rad/s
I can solve for time by using the formula ω_f=ω_i+αΔt
0 rad/s = ∏/30 rad/s + -0.319 rad/s²Δt
-∏/30 rad/s = -0.319 rad/s²Δt
Δt = 0.3283 s

now using this time, I plug it and the rest of my variables into the formula θ_f=θ_i+ω_iΔt+[itex]\frac{1}{2}[/itex]αΔt²
=(∏/30 rad/s)(0.3283 s) + [itex]\frac{1}{2}[/itex](-0.319 rad/s²)(0.3283 s)²
=0.03438 rad + -0.0172 rad
=0.01719 rad
1rot/2∏rad · 0.01719 rad = 0.002736 rot

I tried entering that, and it was incorrect. Then I tried entering the answer I got in radians and that was incorrect too.. So I guess I'm just confused.

 

[hdp12]

Oh wow okay

I read back over my post and I already realized what I did... so I'm sorry.

for the initial speed I didn't convert it to radians per second... I converted one rotation to radians per second... smoothe.
Lemme try that again
f=92 [STRIKE]rot[/STRIKE]/[STRIKE]min[/STRIKE] ·2∏r/[STRIKE]rot[/STRIKE] · 1 [STRIKE]min[/STRIKE]/60 s
=92∏/30 rad/s
= 3.07∏ rad/s

so if ω_i = 3.07∏ rad/s
& ω_f = 0 rad/s
I can solve for time by using the formula ω_f=ω_i+αΔt
0 rad/s = 3.07∏ rad/s + -0.319 rad/s²Δt
-3.07∏ rad/s = -0.319 rad/s²Δt
Δt = 30.2 s

now using this time, I plug it and the rest of my variables into the formula θ_f=θ_i+ω_iΔt+[itex]\frac{1}{2}[/itex]αΔt²
=(3.07∏ rad/s)(30.2 s) + [itex]\frac{1}{2}[/itex](-0.319 rad/s²)(30.2 s)²
=291.27 rad + -145.47 rad
=145.8 rad
1rot/2∏rad · 145.8 rad = 23.205 rot

okay yes this is correct. CRISIS AVERTED SORRY