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

1. Jul 17, 2014

### hdp12

1. The problem statement, all variables and given/known data
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
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

2. Relevant equations

τ=rFsin(θ)
τ=Iα
α=τNET/I
↓ Moment of Inertia of a uniform disk about its center of mass
I=$\frac{1}{2}$mR2
Fk= μkF
ωfi+αΔt
θfiiΔt+$\frac{1}{2}$αΔt2

3. The attempt at a solution

I solved part a) by doing the following
Kinetic Friction Force​
Fk= μkF
Torque by Fk
τ=rFsin(θ) = rFk = rμkF
Magnitude of Angular Acceleration (will be negative)​
α=τNET/I = rμkF/$\frac{1}{2}$mr2
= (0.33 m)(0.2)(25 N) / $\frac{1}{2}$(95 kg)(0.33 m)2

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

so if ωi = ∏/30 rad/s
I can solve for time by using the formula ωfi+αΔt
Δt = 0.3283 s

now using this time, I plug it and the rest of my variables into the formula θfiiΔt+$\frac{1}{2}$αΔt2
=(∏/30 rad/s)(0.3283 s) + $\frac{1}{2}$(-0.319 rad/s2)(0.3283 s)2

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.

2. Jul 17, 2014

### 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

so if ωi = 3.07∏ rad/s
I can solve for time by using the formula ωfi+αΔt
Δt = 30.2 s

now using this time, I plug it and the rest of my variables into the formula θfiiΔt+$\frac{1}{2}$αΔt2
=(3.07∏ rad/s)(30.2 s) + $\frac{1}{2}$(-0.319 rad/s2)(30.2 s)2