At What Temperature Will the Ring Just Fit Over the Rod?

[Saladsamurai]

A certain steel rod has a diameter of 3.00 cm at 25 celsius and brass ring has an inner diameter of 2.992 cm at 25 degrees celsius. At what common temperature will the ring just fit over the rod?

I am using rate of linear expansion: \Delta d=d_0\alpha*\Delta T

\Rightarrow d_f=d_0(\alpha \Delta T+1)

So, letting d= diameter of steel rod and d'= diameter of brass ring and setting the two equal to each other I have:

d_0(\alpha \Delta T+1)=d'_0(\alpha' \Delta T+1)

\Rightarrow d_0(\alpha \Delta T+1)-d'_0(\alpha' \Delta T+1)=0

\Rightarrow d_0\alpha \Delta T+d_0-d'_0\alpha' \Delta T+d'_0=0

\Rightarrow d_0\alpha \Delta T-d'_0\alpha' \Delta T=-d_0-d'_0

\Rightarrow T_f=\frac{-d_0-d'_0}{d_0\alpha-d'_0\alpha'}+T_0

Now d_0=3.0 cm and d'_0=2.992 cm
alpha=11*10^(-6) and alpha'= 19*10^(-6)
and T_0=25

I am getting a number like 280,000 degrees which is obviously ridiculous.

The answer is supposed to be 360 celsius. What am I screwing up here?

Thanks,
Casey

 

[mgb_phys]

Wouldn't it be simpler to say that the difference in expansivity is (19E-6 - 11E-6) = 8E-6
You need to make up a difference of 0.008/3 = 2.7E-3

So T difference = 2.7e-3/8E-6 = (2.7/8) * 1000 K = 333K

ps. Nice latex though

 

[Saladsamurai]

Wouldn't it be simpler to say that the difference in expansivity is (19E-6 - 11E-6) = 8E-6
You need to make up a difference of 0.008/3 = 2.7E-3

So T difference = 2.7e-3/8E-6 = (2.7/8) * 1000 K = 333K

ps. Nice latex though

I am sorry mgb, but I cannot follow your thought process. And the solution is 360 C. Moreover, I would like to know what exactly is incorrect about my method

Casey

 

[mgb_phys]

It's a differential expansion, the steel and brass are getting bigger at different rates.
This is equivalent to a unchanging rod and a ring expanding at the difference rate ie. 8E-6 /K

You need to make up 3.000 - 2.992 = 0.008mm in 3mm = 0.008/3 = 2.67E-3
Every degree increase it by a factor of 8E-6, you need a factor of 2.67E-3
So the temperature difference is = 334deg
Add the original 25deg and the answer is 360 deg (roughly)

 

[Saladsamurai]

Could someone please explain why my method in post #1 does not work?!

It seems like I have taken everything into account. But something goes wrong...

Casey

mgb, I can appreciate your intuitive method, but it unfortunately I cannot answer questions for my class in an intuitive manner. I need to be able to show mathematically why this is true.

 

[Saladsamurai]

Come on Dick. You're usually willing to slap some sense into me.
It seems like the more clearly I present a problem to PF, the less people like to respond to me.

Maybe I should start a new account with no 'gold letters' and present my problems like a tard...like this: Plz I NEED helllpppppppp! And then show no attempt at a solution.

Okay. I am done venting. Sorry.

Casey

 

[katchum]

How about you made a mistake in the algebra? Third equation.

 

[Saladsamurai]

How about great!? Problem solved. Is that why they call katchum?...'cause that's what you do?

Casey

 

[katchum]

Heheh, no katchum's just that guy in Pokemon! I just happened to keep my first nickname from 10 years ago.