Finding length of minute hand of clock

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Homework Help Overview

The problem involves determining the length of the minute hand of a clock based on the distance it travels in a given time. The context is related to circular motion and arc length calculations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the setup of equations related to arc length and the relationship between distance traveled and the radius of the circle. Questions arise about the division by time and the use of linear speed formulas. Some participants suggest calculating the full circumference based on the given arc length.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and questioning the assumptions made in their calculations. Some guidance has been offered regarding the relationship between arc length and radius, but no consensus has been reached on the correct approach.

Contextual Notes

Participants are working under the constraints of a homework problem, which may limit the information available for solving the problem. There is also uncertainty about the correct setup of the equations and the relevance of the time factor in the calculations.

QuantumCurt
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Homework Statement



The minute hand of a clock travels 6.28 inches in 15 minutes. Find the length of the minute hand. Approximate to the nearest inch, if necessary.




The Attempt at a Solution



I figured the arc length to be \frac{\pi}{2} because it's traveled one quarter of the way around the clock.

I set the equation up like this-

\frac{\pi/2}{15}x=6.28

Then I solve for x by multiplying by the reciprocal, and I'm getting 59.97.

This was a question on a test I had earlier today, and when I did it then, I got approximately 15.001 or something like that. That answer didn't seem right to me, and the answer I'm getting now doesn't seem anymore right. I can't remember exactly how I set it up. What am I doing wrong? The rest of the test was simple, this was the only one that threw me off.
 
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QuantumCurt said:

Homework Statement



The minute hand of a clock travels 6.28 inches in 15 minutes. Find the length of the minute hand. Approximate to the nearest inch, if necessary.




The Attempt at a Solution



I figured the arc length to be \frac{\pi}{2} because it's traveled one quarter of the way around the clock.

I set the equation up like this-

\frac{\pi/2}{15}x=6.28

Then I solve for x by multiplying by the reciprocal, and I'm getting 59.97.

This was a question on a test I had earlier today, and when I did it then, I got approximately 15.001 or something like that. That answer didn't seem right to me, and the answer I'm getting now doesn't seem anymore right. I can't remember exactly how I set it up. What am I doing wrong? The rest of the test was simple, this was the only one that threw me off.

Why are you dividing by 15?
 
Mark44 said:
Why are you dividing by 15?

I had been trying to use a linear speed formula to find the radius, but as I've thought about it more it made less and less sense.

Should I have found the circumference by multiplying the 6.28 by 4 to get the entire circumference of the circle, then divide by 2pi?
 
QuantumCurt said:
I had been trying to use a linear speed formula to find the radius, but as I've thought about it more it made less and less sense.

Should I have found the circumference by multiplying the 6.28 by 4 to get the entire circumference of the circle, then divide by 2pi?

The pertinent fact is that circumference = 2π * radius; from this you can find the length of one quarter of the circumference. In general you want to find a formula including the unknown and the known. This is the algebraic way of thinking - get a formula that for conceptual reasons you know is correct, then plug in the numbers with less chance of making a mistake.
 
I agreed Verty.

You may start with the definitlion of \pi=\frac{\mathcal{C}}{2R}, where \mathcal{C} is the circumference lenght.

Then, since you have the length of a quarter of the circumference, can clear the radio.
 
All that is relevant is that (\pi/2)r is the length of a quarter circle arc. The time it takes to travel that is irrelevant. You only need the fifteen minutes to know that the minute hand traveled a quarter circle.
 
So then I would want to set it up like-6.28(4)/2pi? So...approximately 8 inches?
 
QuantumCurt said:
So then I would want to set it up like-6.28(4)/2pi? So...approximately 8 inches?

You're dividing a quarter of an arc between a full arc angle...
 
QuantumCurt said:
So then I would want to set it up like-6.28(4)/2pi? So...approximately 8 inches?

6.28 ≈ 2##\pi##, so the expression above simplifies to approximately 4.
 

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