2 more thermal expansion questions

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SUMMARY

The discussion focuses on two thermal expansion problems involving a steel rod and a brass ring, and the effect of temperature on a grandfather clock's pendulum. For the first problem, participants confirm that the linear expansion equation, ΔL = α(Lo)(ΔT), is appropriate to determine the temperature at which the brass ring will fit over the steel rod. The second problem addresses how the pendulum's length changes with temperature, affecting the clock's timing. The consensus is that the clock will run fast at higher temperatures due to the increased length of the brass pendulum.

PREREQUISITES
  • Understanding of linear thermal expansion, specifically the equation ΔL = α(Lo)(ΔT)
  • Knowledge of the coefficients of linear expansion for materials like steel and brass
  • Familiarity with pendulum mechanics and how length affects period
  • Basic principles of temperature effects on physical dimensions
NEXT STEPS
  • Research the coefficients of linear expansion for steel and brass
  • Study the relationship between pendulum length and period in detail
  • Explore real-world applications of thermal expansion in engineering
  • Learn how to calculate time discrepancies in pendulum clocks due to temperature changes
USEFUL FOR

Students in physics, engineers dealing with thermal expansion, and anyone interested in the mechanics of pendulum clocks and material properties.

F|234K
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1.) a steel rod is 3 cm in diameter at 25 C. a brass ring has an interior diameter of 2.992 cm at 25 C. at what common temperture will the ring just slide into the rod?

for this Q, i am not too sure which equation to use. i think it's the volume change one, which is delta V= beta(Vo)(delta T)

but i don't know how the temperature and the diameter is going to fit in...

2.) a grandfather's clock is calibrated at a temperature of 20 C. the pendulum is a think brass rod with a heavy mass attached to the end. on a hot day, when the temperature is 30 C, does the clock run fast or slow? how much time does it gain or lose in a 24 hour period?

this Q...i just have no idea how to start...

thanks in advance, any suggestions/comments are greatly appreciated.
 
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F|234K said:
1.) a steel rod is 3 cm in diameter at 25 C. a brass ring has an interior diameter of 2.992 cm at 25 C. at what common temperture will the ring just slide into the rod?

for this Q, i am not too sure which equation to use. i think it's the volume change one, which is delta V= beta(Vo)(delta T)

but i don't know how the temperature and the diameter is going to fit in...
Forget about volume; the rod will just slide into the ring when their diameters are the same. You need to find the \Delta T that will have them end up with the same diameter.
2.) a grandfather's clock is calibrated at a temperature of 20 C. the pendulum is a think brass rod with a heavy mass attached to the end. on a hot day, when the temperature is 30 C, does the clock run fast or slow? how much time does it gain or lose in a 24 hour period?
Hint: How does the period of a pendulum depend on its length?
 
for number one, are u suggesting taht i should use the linear equation instead of the volume one?

linear equation being delta L=alpha(Lo)(delta T)
 
Last edited:
"Hint: How does the period of a pendulum depend on its length?"

taht the thing i don't no...so i can't do the Q...
 
F|234K said:
for number one, are u suggesting taht i should use the linear equation instead of the volume one?
Absolutely.

Regarding the pendulum: Look it up! :smile:
 
thanks a lot.
 

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