Finding length of minute hand of clock

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The problem involves finding the length of the minute hand of a clock, given that it travels 6.28 inches in 15 minutes. The initial approach incorrectly used a linear speed formula, leading to confusion about the correct setup. The correct method involves recognizing that 6.28 inches represents a quarter of the circumference, allowing for the calculation of the full circumference using the formula for circumference (C = 2πr). By multiplying 6.28 by 4 and dividing by 2π, the length of the minute hand is determined to be approximately 8 inches. Understanding the relationship between arc length and radius is crucial for solving this type of problem.
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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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