Time When Pendulum Clock is Put on Moon?

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Homework Help Overview

The discussion revolves around the behavior of a pendulum clock when taken to the moon, where the acceleration due to gravity is different from that on Earth. Participants explore how this difference affects the clock's timekeeping over a 24-hour period as measured on Earth.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the relationship between the pendulum's period and gravity, questioning how the reduced gravitational force on the moon affects the clock's accuracy. Some suggest using ratios of gravitational acceleration to analyze the problem, while others express uncertainty about the relevance of the moon's rotation.

Discussion Status

The discussion is ongoing, with various interpretations being explored. Some participants have proposed calculations based on their understanding of the pendulum's behavior in different gravitational fields, while others question the assumptions underlying the problem. There is no explicit consensus on the correct approach or outcome.

Contextual Notes

Participants note the potential for assumptions about the clock's operation in relation to the moon's rotation and the nature of the problem, with some expressing concern that it may be a trick question. There is also mention of classroom context, indicating varying levels of familiarity with the topic.

petern
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A pendulum clock that works perfectly on Earth is taken to the moon where g = 1.63 m/s^2. If the clock is started at 12:00 A.M., what will it read on the moon 24 Earth hours?

The equation you would probably use is T = 2(pi)(sq. root of m/k)

Can someone get me started?
 
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The pendulum clock works on SHM...how is angular velocity affected by the value of 'g' with a pendulum?

Use [tex]T=2\pi\sqrt{\frac{l}{g}}[/tex]are you allowed to use the value of g on earth? if so consider how you can express the acc.due to gravity on the moon in terms of 'g' on earth
 
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I don't really understand but I know that since gravity is less on the moon, the pendulum doesn't swing farther out than on Earth and therefore the clock is slower than on earth.
 
I would find the ratio so on Earth but plugging g on Earth and on the moon into T = 2(pi)(sq. root of m/k). The ratio I got for moon to Earth was .783 to .319 or 1 to 2.45455.

So 24 h x 2.45455 = 58.9092 h. So the answer would be 10:55 A.M. on the third day. Did I do this problem correctly?
 
Is it possible that this is a trick question? Doesn't a pendulum clock tell time on Earth by way of the Earth's rotation? Wouldn't a pendulum clock on the Moon depend on the Moon's rotation? And isn't the swing speed completely irrelevant?
 
I don't think it is. My physics class isn't that advanced and we usually never have any trick questions. Thanks for the input anyways. Do you think I'm doing it right?
 
well the way I intended to do it was the method with finding the ratio...so i think it would be correct...not too sure if you have to take into account the rotation of the moon and such
 
I think I did it right because my teacher has gone over anything related to the rotation so thanks! Problem solved!
 

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