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April 1, 2015

April 1, 2015

Posted by **jean** on Thursday, April 19, 2007 at 6:02am.

a. If only the ring is heated, what temperature must it reach so that it will just slip over the rod?

b. WHAT IF? If both are heated together, what temperature must they both reach so that the ring just slips over the rod? Would the latter process work?

sorry., this one i really couldn't figure out. although the last 2 i got it after awhile.

calculate the circumference of the rod.

Calculate the difference in the circumerance of the ring and the rod.

a) LEt that be delta L

deltaL= Ringcirc*coeffAl*deltaT

solve for delta T.

b) for each, the original material Circumer + delta L is equal to the sum of L and delta L for the other.

aluminumcirc=rodcirc

but the rod expands, it has a delta V. So the delta L for it has to be determined by changes in V.

Volume= 4/3 PI r^3

dV= 4 PI r^2 dr= V(coefficientvolume)deltaT

or

dr= V*coeffV*deltaTemp/4PIr^2 where r is the original r. And

newCircumference= 2PI (r + dr)

Now, set the aluminum length heated equal to the brass change in circumference.

Solve for deltaTemp.

so i figured out the first part

but im still lost about part b

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