Torque Causing Angular Acceleration on a Uniform Disk

[hdp12]

Homework Statement

/* Intro, personal: I am taking Studio Physics 1 at FSU this summer semester. We have an online homework assigned weekly via a program called ExpertTA. This question has been giving me the HARDEST time and I seriously just need some help on it. I have completed and submitted sections a-g and all were correct. Part h), however, I must be doing wrong or something because I believe that I have tried every possibility that I can come up with. Whoever attempts to give this a shot, I genuinely appreciate it. I'm not one to usually ask for help, but like I said... I am stumped. */

PROBLEM:
A uniform disk of mass m and radius R can rotate about an axle through its center. Four forces are acting on it as shown in the figure. These forces are labeled F₁, F₂, F₃, and F₄. F₂ and F₄ act a distance d from the center of mass while the other two act on the outer edge of the disk. These forces are all in the plane of the disk.

RANDOMIZED VARIABLES:
m= 3.6 kg
R= 24 cm = 0.24 m
F₁= 2.5 N
F₂= 2.5 N
F₃= 6.5 N
F₄= 4.5 N
d= 3.5 cm = 0.035 m

a) Write an expression for the magnitude τ₁ of the torque due to force F₁
√ANSWER: τ₁ = F₁R

b) Calculate the magnitude τ₁ of the torque due to force F₁, in N·m
√ANSWER: τ₁ = 0.6

[c) Write an expression for the magnitude τ₂ of the torque due to force F₂
√ANSWER: τ₂ = F₂d

d) Calculate the magnitude τ₂ of the torque due to force F₂, in N·m
√ANSWER: τ₂ = 0.0875

[e) Write an expression for the magnitude τ₃ of the torque due to force F₃
√ANSWER: τ₃ = 0

[f) Write an expression for the magnitude τ₄ of the torque due to force F₄
√ANSWER: τ₄ = dF₄sin(53)

g) Calculate the magnitude τ₄ of the torque due to force F₄, in N·m
√ANSWER: τ₄ = 0.1258

h) Calculate the angular acceleration α of the disk about its center of mass in rad/s². Let the counter-clockwise direction be positive.
PREVIOUS ANSWERS:
Xα=3.95
Xα=25.65
Xα=25.6519
Xα=25.66

Homework Equations

τ=rFsin(θ)
τ=Iα
α=τ_NET/I
\downarrow Moment of Inertia of a uniform disk about its center of mass
I=\frac{1}{2}mR²

The Attempt at a Solution

I don't recall how I got the first answer that I submitted, but I do for the last three.
Since the directions of the torques varies, the net torque will be calculated by adding the two ccw torques and subtracting the cw torque
τ_NET=τ₁-τ₂+τ₄
Next I set up the angular acceleration equation using the net torque just written and the moment of inertia formula
α=τ_NET/I = (τ₁-τ₂+τ₄ )/ (\frac{1}{2}mR²)
This equation produces the answer in terms of m/s², but the question clearly asks for radians so, using the equation where 1 rad = 0.24 m = R
α=(τ₁-τ₂+τ₄ )/ (\frac{1}{2}mR²)·\frac{1 rad}{R}
This SHOULD give me the correct answer. And, plugging all of the numbers in, I get 25.6519 but that is allegedly incorrect.
Please help me.

 

[jbriggs444]

Next I set up the angular acceleration equation using the net torque just written and the moment of inertia formula
α=τ_NET/I = (τ₁-τ₂+τ₃ )/ (\frac{1}{2}mR²)
This equation produces the answer in terms of m/s²,
[...]

What leads you to think that the result of this equation is in meters per second squared? What are the units of torque? What are the units of moment of inertia? What are the units of the quotient?

 

[hdp12]

Error

I made an error in translating my work, I'm going to go back and fix it now.
Basically, τ₃ is 0, so in my writing out the work that I did I accidentally wrote τ₃ instead of τ₄
Fixing that now

 

[verty]

With full precision I get 25.6512463... . Perhaps the rounding is the problem.

 

[hdp12]

What leads you to think that the result of this equation is in meters per second squared? What are the units of torque? What are the units of moment of inertia? What are the units of the quotient?

Hm, okay. Lemme work this out.

Units of torque?​

τ = (#) N·m = (#) (kg·m·m/s·s) · m = (#) kg·m³·s^-2
m = (#) kg
R = (#) m
R² = (#) m²

Units of moment of inertia?​

I = 0.5·m·R·R = (#) kg·m²
α = τ/I = kg·m³·s^-2/kg·m² = kg·kg^-1·m³·m^-2·s^-2 =(cancelling) [STRIKE]kg·kg^-1[/STRIKE]·m[STRIKE]³·m^-2[/STRIKE]·s^-2 = m·s^-2

So that's my reasoning for why I believe the equation results in m/s²​

 

[hdp12]

With full precision I get 25.6512463... . Perhaps the rounding is the problem.

Honestly I thought so as well and that's why I tried a few variations of 25.65... they all came back as incorrect though. This homework program usually accepts answers within at least a tenth of a decimal point of the correct answer. For example is 25.65 WAS the correct answer it was looking for and I entered 25.5 or 25.7 it would accept it. I guess that's why I believe I'm not actually close to the correct answer.

 

[jbriggs444]

Hm, okay. Lemme work this out.

Units of torque?​

τ = (#) N·m = (#) (kg·m·m/s·s) · m = (#) kg·m³·s^-2

Try that again. What are the units of the Newton?

 

[hdp12]

Try that again. What are the units of the Newton?

Oh my..
N = kg·m/s²
I confused Newtons with Joules.
Wow.

So if α = τ_NET/I
= (τ₁-τ₂+τ₃ )/ (0.5mR²)
= (0.6 J - 0.0872 J + 0.1258 J) / (0.5 · 3.6 kg · 0.24 m · 0.24 m) = 0.6383 J / 0.10368 kg·m²
= 6.16 (kg·kg^-1·m²·m^-2·s^-2
= 6.16 ([STRIKE]kg·kg^-1[/STRIKE]·[STRIKE]m²·m^-2[/STRIKE]·s^-2
= 6.16 /s²

Which I'm assuming is in radians. Woooow. Kickin myself haha.

And that is correct I just submitted it. Thank you SO much. So frustrating that it was literally such a little mistake but it confused me for HOURS.