Problem Set 13: | 9 | 17 | 19 | 23 | 31 | 33 | 35 | 37 | 10:1 | Go up
 - by David Harms, 2002

9. A concrete highway is built of slabs 14 m long (20Cº).  How wide should the expansion cracks be (at 20Cº) between the slabs to prevent buckling if the temperature range is -30Cº to +50Cº?

I know I use the formula:
DL  = aL_oDT
DL  = ? (Change in length of the slabs. We are solving for this.)
a    =  12e-6 (Coefficient of expansion. I looked it up on page 388.)
L_o   =  14 m  (Initial length of slabs.)
DT  =  30 Cº (50Cº - 20Cº)  (Change in temperaure.  You only care about the hottest number since you are dealing with expansion.)

I then had the formula:
DL = 12e-6 *14 m * 30Cº
DL = .00504 m
DL = .50 cm

The space between the slabs must be the same as the expansion of the the slabs.  Therefore .50 cm is the correct answer.
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17. A quartz sphere is 14.5 cm in diameter.  What will its change in volume if it is heated from 30Cº to 200Cº? 

I had the formula:
DV = gV_oDT
DV = ? (Volumetric change in the sphere.  We are solving for this.)
g     =  1e -6 (Cº)^-1 (Coefficient of volumetric expansion.  It is on page 388 listed as b.)
V_o   =  (4/3) * p(.145 m/2)³ (Initial volume.  You must take the diameter and divide by to to get the radius.  Then you must stick it in the formula for volume of a sphere: (4/3)pr³
V_o   =  .001569 m³
DT  =  170Cº (Change in temperature.  200Cº -  30Cº.)

I then had the formula:
DV = 1e-6*.001569m³*170Cº
DV = 2.7 e-7 m³ *1e6 (Conversion from m³ to cm³.)
DV = .27 cm³

The answer is therefore .27 cm³.

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19. If the fluid is contained in a long, narrow vessel so it can expand in essentially one direction only, show that the effective coefficient of linear expansion a is approximately equal to the coefficient of volume expansion b (g in the IB packet.)? 

Since it can only expand in one direction, is is really linear expansion, not volumetric expansion. Therefore the two coefficients would be the same.

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23. A 23.4 kg solid aluminum wheel of radius 0.52 m is rotating about its axel in frictionless bearings with an angular velocity w = 32.8 rad/s.  If its temperature is now raised from 20.0Cº to 75.0Cº, what is the fractional change in w?

To solve this problem we must find I (Moment of Inertia) before it heats up and after it heats up.

I have the formula:
I₁ = Sm_ir_i²
I₁ = ? (Moment of Inertia.  We are solving for this.)
S  = 1/2 (looked it up on page 22).
m = 23.4 kg (mass)
r  = .52 m (radius)

I then had the formula:
I₁ = (1/2)*23.4kg * .52² 
I₁ = 3.16368 kg m² 

I then had to find the expansion of the radius using:
DL  = aL_oDT
DL  = ?
L_o   = .52 m
a    =  25e-6Cº^-1 
DT  = 55Cº

I then had:
DL = 25e-6Cº^-1 * .52 m * 55Cº 
DL = 7.14 e -4 m

I then had to find I₂:
I₂ = Sm_ir_i²
I₂ = ?
S  = 1/2
m = 23.4 kg
r = .520715

Then I had:
I₂ = .5 * 23.4 *.520715²
I₂ = 3.172386 kg m² 

I also know I have the formula:
I₁w₁ = I₂w₂ 
w₁ = 32.8 rad/s (Angular velocity.)
I₁   = 3.16368 (Found at beginning of problem.)
w₂ = ? (We are solving for this.)
I₂  =  3.172886 (Just solved for.)

I  then had:
w₂ =  32.7048 r/s

To find the percentage difference I took
(w_1-w₂)/w₁ 
( 32.8-32.7028)/32.8

Which equals the answer:
.3%

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31. If 3.0 m³ of a gas initially as STP is placed under a pressure of 4 atm, the temperature of the gas rises to 38.0Cº.  What is the volume? ?

I know I had to use the equation:
(P₁V₁)/T₁ = (P₂V₂)/T₂ 

P₁ = 1.013e5 Pa (1 atm* 1.013e5 Pa/atm)
V₁ = 3 m³  
T₁ = 273 k (Number of Kelvins at STP)
P₂ = 405200 Pa (4 atm * 1.013e5 Pa/atm)
V₂ =  ? (We are solving for this)
T₃ = 311 k (273 k +38)

I then solved this as got:
V₂ =  .85 cm³

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33. <Question goes here>? 

<solution goes here>
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35. <Question goes here>? 

<solution goes here>
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37. <Question goes here>? 

<solution goes here>
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10:1. <Question goes here>? 

<solution goes here>
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