Angular and Linear Speed....Part 2

[mathdad]

Assume that the Earth is a sphere with a radius of 3,960 miles and a rotation of 1 revolution per 24 hours.

A. Find the angular speed. Express your answer in units of radians/sec, and round to two significant digits.

B. Find the linear speed of a point on the equator. Express the answer in units of miles per hour, and round to the nearest 10 mph.

Part A

I decided to convert 24 hours to seconds and found that to be 86,400 seconds. The reason is, I thought, because the answer must be written in radians/sec.

I then needed to use w = θ/r but not before finding θ.

θ = (86,400 seconds)(2 pi radians).

θ = 172,800 pi radians

w = 172,800 pi radians/sec.

w is about 542,867.21054031 radians/sec.

I rounded to two significant digits.

My answer for Part A is w is approximately 540,000 radians/sec.

Book's answer is 0.000073 radians/sec.

Part B

Since the radius of Earth is given to be 3,960 miles,
I doubled the radius of Earth to get the equator.

Let E = equator length

Equator = 2r

E = 2(3,960 miles)

E = 7,920 miles.

In Part A above, I found θ to be 172,800 pi radians.

I next decided to find s, the arc length. At this point, I am convinced that s = d in the linear speed equation v = d/t.

s = (172,800 pi radians)(radius of earth)

s = 684,288,000 pi miles = d.

The numbers here look very out of touch with reality.

The linear speed formula is v = d/t.

So, v = 684,288,000 pi miles/hour.

v is about 2,149,754,153.7396 miles/hour.

Again, the numbers just did not look right but I decided to confidently proceed.

I then rounded to the nearest 10 mpr.

My answer for Part B is v is approximately 2,149,754,150 miles/hour.

Book's answer is 1040 mph.

 

[MarkFL]

Let's first look at part A. We are asked for $\omega$, given by:

\(\displaystyle \omega=\frac{\theta}{t}\)

Now, we know the Earth makes 1 revolution per day, so:

\(\displaystyle \omega=\frac{1\text{ rev}}{1\text{ day}}\cdot\frac{2\pi\text{ rad}}{1\text{ rev}}\cdot\frac{1\text{ day}}{86400\text{ s}}=\frac{\pi}{43200}\,\frac{\text{rad}}{\text{s}}\approx0.000073\,\frac{\text{rad}}{\text{s}}\)

Now, for part B., we want the linear speed at the equator, so we use:

\(\displaystyle v=\omega r\)

We have $\omega$, and $r$ is simply the radius of the Earth. What do you get?

 

[mathdad]

For Part A, you included 1 day/86,400 seconds. I understand why you did that but, can we also write the fraction as 24 hours/86,400 seconds and get the same result?

Part B

v = wr

v = (0.000073 rad/sec)(3,960 miles)

Is this the correct equation set up?

I noticed that you decided to use v = wr instead of v = d/t, which is also given in Section 6.1 as the linear speed formula. Why did you decide to use v = wr and not v = d/t?

 

[MarkFL]

For Part A, you included 1 day/86,400 seconds. I understand why you did that but, can we also write the fraction as 24 hours/86,400 seconds and get the same result?

Yes, since 1 day = 24 hours, that would be equivalent.

Part B

v = wr

v = (0.000073 rad/sec)(3,960 miles)

Is this the correct equation set up?

I noticed that you decided to use v = wr instead of v = d/t, which is also given in Section 6.1 as the linear speed formula. Why did you decide to use v = wr and not v = d/t?

Personally, I wouldn't use the rounded value for $\omega$, I would use the exact value, and then round the result.

Since we have already computed $\omega$ and are given $r$, it seemed straightforward to use $v=\omega r$. Of course, you could get the same result using the other formula.

 

[mathdad]

Mark FL:

Can you show me how to get the same answer using the formula v = d/t?

This is funny. I did all that work leading no where, and you got the answer in 2 or 3 steps. You are a mathematician. If I understood math at your level, I certainly would have a better job.

BTW, you are not wasting time with me. I actually do apply everything you teach me to other similar questions. I just love this course. I love David Cohen. Wish I could express to him how much this book means to me.

 

[MarkFL]

Mark FL:

Can you show me how to get the same answer using the formula v = d/t?

The distance traveled is the circumference of the equator, and the time taken to travel that distance is 24 hours (I use hours because the questions asks for the speed in miles per hour).

\(\displaystyle v=\frac{d}{t}=\frac{2\pi r}{t}=\frac{2\pi(3690\text{ mi})}{24\text{ hr}}=\frac{615\pi}{2}\text{ mph}\)

Now, let's look at:

\(\displaystyle v=\frac{2\pi r}{t}=\frac{2\pi}{t}r=\frac{\theta}{t}r=\omega r\)

 

[mathdad]

Very interesting. I will move on to Section 6.2. I will post questions and my solution from Section 6.2 as needed.