The Glade 4.0
https://gladerebooted.net/

Simple physics problem (angular acceleration)
https://gladerebooted.net/viewtopic.php?f=2&t=10749		 Page 1 of 1
Author:  Stathol [ Sun Feb 09, 2014 6:51 pm ]
Post subject:  Simple physics problem (angular acceleration)
Quote:
A flywheel with a radius of 0.454m starts from rest and accelerates with a constant angular acceleration of 0.500 rad/s2

Compute the magnitude of the tangential acceleration, the radial acceleration, and the resultant acceleration of a point on its rim after it has turned through 55.3∘


Seems easy enough.

az(t) = 0.500 s-2
vz(t) = 0.500t + v0z = 0.500t (starting from rest, so v0z = 0)
sz(t) = (1/2)0.500t2 + s0z

Where az(t), vz(t), and sz(t) are angular acceleration, angular velocity, and angular position respectively. We don't know s0z, but it isn't relevant to the problem since we're dealing with a definite integral.

First we need to find tn, the time when the flywheel has rotated through 55.3 degrees. So we have:

The integral from t0 to tn of vz(t) = ~0.965 radians
0.250tn2 - 0.250t02 = 0.965

Since t0 = 0, this reduces to:
0.250tn2 = 0.965
tn = 1.965s

Angular velocity at t=1.965 is thus:
vz(1.965) = 0.500 * 1.965 = 0.982 s-1

Radial acceleration is just:
ar = vz2/r

Which gives us:
ar = (0.982 s-1)2 / 0.454m = 2.126 ms-2

Tangential acceleration is:

at = az*r = 0.500 s-2 * 0.454m = 0.227 ms-2

Total acceleration is:

a = at + ar = 2.353 ms-2

According to MasteringPhysics, this answer is wrong.

Our textbook supplies a similarly derived equation:

vz2=v0z2+2az*Δθ

Plugging in the values from the problem gives me exactly the same value for vr.

Also, the first part of this problem asked for at, ar, and a at t0. I answered 0.227, 0, and 0.227 respectively, which it marked as correct. Since angular acceleration is constant, tangential acceleration should also be. So I'm 99% sure that the problem is in my calculation of a{sub]r[/sub], but I can't figure out how it's wrong.

Edit: derped an equation. ar=vz2/r, not ar=vz2*r. It still doesn't like my answer, though.
Author:  Corolinth [ Sun Feb 09, 2014 8:06 pm ]
Post subject: 
The Mastering(Subject) line is kind of hit or miss. I've had a few students in Fields & Waves and Circuit Analysis gripe about their physics courses that relied heavily on MasteringPhysics.

Rotational motion problems can all be solved using the same equations as translational motion problems, and changing the variables to their appropriate rotational forms. That is to say, replacing x with theta, v with omega, and a with alpha. Honestly, they're having you jump through way too many intermediary steps to solve one equation and one unknown.

Off the top of my head, there is one thing I can think of that you're doing which MasteringPhysics may not like.

Your equation for total acceleration, while not wrong, is incomplete. Total acceleration is the vector sum of tangential and radial acceleration. Remember, tangential and radial motion are perpendicular to one another just like horizontal and vertical motion. Mid and upper level engineering courses actually treat tangential and radial components as their own coordinate system (you could think of them similar to i, j, and k components).

Given questions you were posting two months ago, I assume you're taking Physics I, presumably calculus-based? You would probably not be expected to learn tangential and radial coordinates, yet, so I would not think MasteringPhysics wants you to write your answer as a vector. Instead, I think it wants the magnitude of your total acceleration. Take your tangential and radial components, and combine them as legs of a right triangle.
Author:  Stathol [ Sun Feb 09, 2014 9:15 pm ]
Post subject:  Re:
Corolinth wrote:
Rotational motion problems can all be solved using the same equations as translational motion problems, and changing the variables to their appropriate rotational forms. That is to say, replacing x with theta, v with omega, and a with alpha. Honestly, they're having you jump through way too many intermediary steps to solve one equation and one unknown.

Well, the questions are custom written by the prof in this case, but yeah. She probably just expects us to plug in values to:
ω2 = ω02 + 2αΔθ
Solve for ω, and then use that in ω2/r = ar
I only went back and derived it the long way because the short approach wasn't working, and I like to actually understand where the equations come from in the first place. My memory is ****, so I try to lean on reasoning rather than rote whenever possible.

Corolinth wrote:
Your equation for total acceleration, while not wrong, is incomplete. Total acceleration is the vector sum of tangential and radial acceleration.
[...]
I think it wants the magnitude of your total acceleration. Take your tangential and radial components, and combine them as legs of a right triangle.

You know, I wasn't even thinking about that. It still doesn't like my answer, though, even with a = sqrt(ar2 + at2).

Would you mind sanity-checking me, though? Do you get at = 0.227m/s^2, ar = 2.126 m/s^2? Right now I'm not sure if the problem is with my answer or with MasteringPhysics being shitty.
Author:  Stathol [ Sun Feb 09, 2014 10:24 pm ]
Post subject: 
Well, ****. Nevermind. I finally got it.

α(t) = 0.500 /s^2
ω(t) = 0.500t /s
δ(t) = 0.250t2 + δ0

The flywheel has rotated 55.3 degrees at tn, so:

δ(tn)-δ(0) = 55.3π/180

0.250tn^2 = 55.3π/180

Solving gives tn = 1.965s, as before.

Now, at = 0.500 /s^2 * 0.454m = 0.227 m/s^2

And therefore: vt = at * 1.965s = 0.4460 m/s

Then instead of using ar = ω2r, I used:

ar = vt2/r = 0.4382 m/s^2

So:

at = 0.227 m/s^2
ar = 0.438 m/s^2
a = sqrt(at2 + ar2) = 0.493 m/s^2

But here's what I don't get...

ω2r = vt2/r
ω2 = (vt/r)2
ω=vt/r

Solving for ω from vt, I get 0.965 /s^2. But when I solve it using ω(1.965), I get 0.982 /s^2. The kicker is that I only found t = 1.965 in the first place by integrating that exact equation for ω(t), so I know it's got to be correct. What am I missing?
Author:  Rorinthas [ Sun Feb 09, 2014 11:26 pm ]
Post subject:  Re: Simple physics problem (angular acceleration)
I used to think I was smart, then I started reading threads like this.
Author:  Aethien [ Mon Feb 10, 2014 12:24 am ]
Post subject: 
Seriously. Well, I still think I'm smart, just not in this way.
Author:  Midgen [ Mon Feb 10, 2014 1:27 am ]
Post subject: 
I have no idea what this thread is about, but I've found some awesome new ascii characters to make unique MMO character names with!!

WOOHOO!!
Author:  Arathain Kelvar [ Mon Feb 10, 2014 7:56 am ]
Post subject: 
When I read threads like these, it's usually "oh, yeah, I remember that" - which is really just a reminder of how much I've forgotten.

Kind of sad. I tutored this **** back in the day.
Author:  Corolinth [ Mon Feb 10, 2014 8:39 am ]
Post subject:  Re: Simple physics problem (angular acceleration)
I'm getting the same value for time. I'd have to sit down and actually work the problem out, which I'll have to get back to later on.
Author:  Elmarnieh [ Mon Feb 10, 2014 8:41 am ]
Post subject: 
This reminds me of the endless "penny on a record" problems that involed this and coefficents of friction all to determine if the penny would spin off the record. The simplist answer is of course "turn the record player on and see".
Author:  Stathol [ Mon Feb 10, 2014 12:23 pm ]
Post subject:  Re:
Arathain Kelvar wrote:
When I read threads like these, it's usually "oh, yeah, I remember that" - which is really just a reminder of how much I've forgotten.

Kind of sad. I tutored this **** back in the day.

Welcome to my life for the last 2 years. I used to know all of this stuff in high school, but "use or it lose it".
Author:  Kaffis Mark V [ Mon Feb 10, 2014 12:33 pm ]
Post subject: 
Yeah. It makes me sad how much math is rusted tight.
Author:  Corolinth [ Mon Feb 10, 2014 12:43 pm ]
Post subject:  Re:
Stathol wrote:
Solving for ω from vt, I get 0.965 /s^2. But when I solve it using ω(1.965), I get 0.982 /s^2. The kicker is that I only found t = 1.965 in the first place by integrating that exact equation for ω(t), so I know it's got to be correct. What am I missing?

I'm looking over your calculations now. The first thing I want to point out is that your answers are within less than 2% of one another. What you are looking at is likely rounding errors. I would advise you carry out the same calculations while retaining your numbers out to eight decimal places instead of three, and see if they don't get closer together.

Alternatively, you could stop beating your head against a wall and move on. Whichever you prefer.
Author:  Stathol [ Mon Feb 10, 2014 1:18 pm ]
Post subject: 
Yeah, I did all of the calculations with full precision. I just rounded the output for the sake of legibility.
Corolinth wrote:
Alternatively, you could stop beating your head against a wall and move on. Whichever you prefer.

Where's the fun in that?

Maybe if I work it out symbolically the problem will become apparent.

α(t) = 0.500
ω(t) = 0.500t
δ(t) = 0.250t2 + δ0

integral from 0 to tn of ω(t) = δ(tn) - δ(0) = 55.3π/180
0.250t2 - 0.250t[sub]n2 = 55.3π/180
0.250tn2 + δ0 - [0.250*0 + δ0] = 55.3π/180
tn = sqrt(55.3π/45)

Method 1:

ω(tn) = 0.500 * sqrt(55.3π/45)

Method 2:

at = α*r = 0.500 * 0.454 = 0.227
vt(tn) = attn = 0.227 * sqrt(55.3π/45)

And per the previous post if:
ω2r = vt2/r
then:
ω=vt/r

So:
ω=0.227 * sqrt(55.3π/45) / 0.454
ω=1/2 * sqrt(55.3π/45)

Conclusion:

Now I'm getting exactly the same value either way. I obviously made a calculation error somewhere along the way yesterday. The thing is, I re-ran the equations and calculations at least 5 or 6 times, and somehow made the exact same mistake every time. WT actual F :psyduck:

Edit: TL;DR:

My original solution was wrong because I somehow got angular velocity mixed up with tangential velocity and was trying to plug the former into ar = v2/r, which is totally wrong. As for why I got a discrepancy between two different methods of solving for ω last night, I still have no clue. I'm going to chock that one up to getting tired and punchy.

Here's another fun fact: Quote =/= Edit
Author:  Corolinth [ Mon Feb 10, 2014 2:33 pm ]
Post subject:  Re: Simple physics problem (angular acceleration)
Typically, physics and engineering professors are not overly critical of answers within 5% of the accepted value. We've got a guy who has fired graders for marking such answers incorrect. There is a certain amount of error and uncertainty in the numbers provided in the problem statement itself that we can't really say which of the two values you posted is correct, given that they are so close.

To illustrate, try computing your answer with varying approximations of pi: 22/7, 3.14, 3.1416, and 355/113 instead of WolframAlpha's pi value.
Author:  Stathol [ Mon Feb 10, 2014 2:44 pm ]
Post subject: 
I wouldn't concern myself with it except that:

a) MasteringPhysics (I believe) checks for answers that are within 1%
b) I wanted to be sure that the problem was simply the result of calculation error rather than a problem with my understanding of the physics. Pragmatically, the variance was too small to probably matter for most applications, but it was too large to be accounted for by rounding error or truncation of pi given the calculator I was using.
Author:  Corolinth [ Mon Feb 10, 2014 4:01 pm ]
Post subject:  Re: Simple physics problem (angular acceleration)
I understand your concern with the MasteringPhysics package. As I've indicated, I'm not terribly keen on it. That would be something I'd ask your professor about, as I believe the margin of error should be something she has control over. It's worth verifying what that margin is. Moreover, I would pose the same question to her that you presented here. Even though you've worked out the solution for yourself, the more she sees your name/face in conjunction with her class, the better things go for you overall. Speaking from experience on both sides, there is a difference between a student in my class and one of my students. The former I give a grade to and never see again. The latter I will do things to assist them later on. They don't care that you're asking questions you already know the answer to, they care that you're interested.
Author:  Khross [ Mon Feb 10, 2014 4:07 pm ]
Post subject:  Re: Simple physics problem (angular acceleration)
Corolinth wrote:
Speaking from experience on both sides, there is a difference between a student in my class and one of my students. The former I give a grade to and never see again. The latter I will do things to assist them later on. They don't care that you're asking questions you already know the answer to, they care that you're interested.
This all sorts of win.
Author:  Diamondeye [ Tue Feb 11, 2014 12:08 pm ]
Post subject:  Re: Re:
Stathol wrote:
Arathain Kelvar wrote:
When I read threads like these, it's usually "oh, yeah, I remember that" - which is really just a reminder of how much I've forgotten.

Kind of sad. I tutored this **** back in the day.

Welcome to my life for the last 2 years. I used to know all of this stuff in high school, but "use or it lose it".

Same. I used to know how to do this, now I'm totally mystified.
Author:  Yzen [ Fri Feb 28, 2014 3:46 am ]
Post subject:  Re: Simple physics problem (angular acceleration)
Disappointing is being able to remember in detail the episode when Cousin Oliver joined the Brady Bunch better than everything from AP physics and calculus onward.
Author:  Vladimirr [ Fri Feb 28, 2014 7:57 am ]
Post subject:  Re: Simple physics problem (angular acceleration)
Yzen wrote:
Disappointing is being able to remember in detail the episode when Cousin Oliver joined the Brady Bunch better than everything from AP physics and calculus onward.


Hey man, long time no see.
Page 1 of 1 All times are UTC - 6 hours [ DST ]
Powered by phpBB® Forum Software © phpBB Group
https://www.phpbb.com/