Angular and linear speed

[Timcago]

The radius of the planet mars is approximately 3400 kilometeres. The time of 1 Mars "day" is 24.6 hours.
A) What is the angular speed of a person standing at Mars' equator?
B) What is the linear speed of a person standing at Mars' equator?

My attempt:

A) Angular speed, w=(pheta/t)

3400*2pie = 6800pie

w=(6800pie/24.6) = 868.4 kilometers/hour?


B) Linear speed, v=(s/t)

s=r*pheta

v=(3400*6800pie)/(24.6) = 2952586.3 kilometers/hour?

 

[Timcago]

My numbers seem a little to high to be accurate.

 

[tkhunny]

1) General Comment: "pie" is for eating. "pi" is for Greek mathematicians.
2) Angular Speed: Why is angular velocity in kph?
3) Linear Speed: Why did you multiply by the radius twice?

 

[soroban]

Hello, Timcago!

You're making it complicated . . . and incorrect.

The radius of the planet Mars is approximately 3400 km.
The time of 1 Mars "day" is 24.6 hours.
A) What is the angular speed of a person standing at Mars' equator?
B) What is the linear speed of a person standing at Mars' equator?

A) Angular speed = (radians or degrees) per (hour).

\(\displaystyle \text{Angular speed}\:=\:\frac{2\pi\,\text{radians}}{24.6\,\text{hours}} \;\approx\;0.2554\,\text{radians/hr}\;\approx\;14.63^o/hr\)


B) Linear speed = {distance) per (hour)

Distance = circumference = \(\displaystyle 2\pi(3400)\:=\:6800\pi\text{ km}\)

\(\displaystyle \text{Linear speed}\;=\;\frac{6800\pi\,\text{km}}{24.6\,\text{hours}} \;\approx\;868.4\;\text{km/hr}\)

 

[galactus]

For part a. If you would've divided your 868.41 by the radius of Mars you would've arrived at your answer.

This is the amount of arc(in km) swept out by 1 hour of Mars's rotation:

\(\displaystyle \L\\\frac{6800{\pi}}{\frac{123}{5}}=\frac{34000{\pi}}{123}=868.41\)

Since \(\displaystyle s=r{\theta}\), you have

\(\displaystyle \L\\\frac{\frac{34000{\pi}}{123}}{3400}=\frac{10{\pi}}{123}\approx{.255}\ \text{\frac{rad}{hr}}.\)