artemisxrpg
  • artemisxrpg
Earth revolves on its axis once every 24 hr. Assuming that Earth's radius is 6400 km, find the following. a. angular speed of earth in radians per day and radian per hours b. linear speed at the north pole or south pole c. linear speed at quito,ecuador, a city on the equator d. linear speed at salem, oregon (halfway from the equator to the north pole)
Trigonometry
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SOLVED
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chestercat
  • chestercat
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ganeshie8
  • ganeshie8
\(\large \omega = \frac{2 \pi}{T}\)
ganeshie8
  • ganeshie8
earth is taking 1 day to to rotate around itself. \(\large \omega = \frac{2 \pi~radians}{1~ day}\)
artemisxrpg
  • artemisxrpg
it's 2 pi because that how much one revolution is?

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More answers

ganeshie8
  • ganeshie8
Yup ! to go around circle once, it takes 2pi radians
ganeshie8
  • ganeshie8
|dw:1380369147272:dw|
artemisxrpg
  • artemisxrpg
so angular speed in 24 hrs would be pi/12
ganeshie8
  • ganeshie8
Yes, pi/12 radians per hour
ganeshie8
  • ganeshie8
2 pi radians per day us same as saying, pi/12 radians per hour
ganeshie8
  • ganeshie8
its just the unit conversion
ganeshie8
  • ganeshie8
*is
artemisxrpg
  • artemisxrpg
v = r*theta/t, so in this case it would be 6400*2pi/1 day or 24 hr?
ganeshie8
  • ganeshie8
theta/t is the angular speed \(\omega \)
ganeshie8
  • ganeshie8
linear speed v = r\(\omega \)
ganeshie8
  • ganeshie8
simply multiply radius wid \(\omega\)
ganeshie8
  • ganeshie8
whats the radius of earth at north pole ?
artemisxrpg
  • artemisxrpg
6400?
ganeshie8
  • ganeshie8
let me put it this way :- whats the distance from axis of rotation to you, when you are at north pole ?
artemisxrpg
  • artemisxrpg
I'm still confused|dw:1380370272714:dw|
artemisxrpg
  • artemisxrpg
this is where the north pole would be?
ganeshie8
  • ganeshie8
|dw:1380370352008:dw|
ganeshie8
  • ganeshie8
|dw:1380370410225:dw|
ganeshie8
  • ganeshie8
thats the axis of rotation, straight line from North pole to Sounth pole wats the distance ? :)
artemisxrpg
  • artemisxrpg
12800?
ganeshie8
  • ganeshie8
'point on North pole' , is right 'next to axis of rotation'
ganeshie8
  • ganeshie8
so, distance between them = r = 0
ganeshie8
  • ganeshie8
for a person on North / South poles :- linear velocity \(v = r \omega = 0\times \omega = 0 \)
ganeshie8
  • ganeshie8
see if that makes more/less sense
ganeshie8
  • ganeshie8
|dw:1380370749167:dw|
ganeshie8
  • ganeshie8
starting from equator, as you move towards the poles, the linear velocity decreases and becomes 0 at poles
artemisxrpg
  • artemisxrpg
so this is where the equator is?|dw:1380371306773:dw|
artemisxrpg
  • artemisxrpg
not at the center of the graph?
ganeshie8
  • ganeshie8
yup ! thats the equator, and since distance from axis is maximim, equals radius of earth, the speed wud be maximum at equator
artemisxrpg
  • artemisxrpg
so then radius would be 6400 at the equator?
ganeshie8
  • ganeshie8
lets call it, distance from axis. at equator, distance from axis = radius of earth = 6400
ganeshie8
  • ganeshie8
in ur speed formula, \(v = r \omega\) , \(r\) is distance form axis
artemisxrpg
  • artemisxrpg
then if the linear speed is halfway from the equator to north pole, radius would be 3200?
ganeshie8
  • ganeshie8
nopes, use trig to find r, the distance from axis
ganeshie8
  • ganeshie8
|dw:1380372161083:dw|
ganeshie8
  • ganeshie8
|dw:1380372288001:dw|
ganeshie8
  • ganeshie8
|dw:1380372490038:dw|
ganeshie8
  • ganeshie8
thats the distance from axis, \(r\) you can find it using proper trig ratio
artemisxrpg
  • artemisxrpg
180-45= 135 - 90= 45. since it's a special triangle, both side are 6400 so use pythagorean theorem and the answer should be 9050.97?
ganeshie8
  • ganeshie8
Nope, careful 6400 is the hypotenuse
ganeshie8
  • ganeshie8
|dw:1380442777970:dw|
artemisxrpg
  • artemisxrpg
sin45 = x/6400 = 4525.48?
ganeshie8
  • ganeshie8
Yup ! thats the \(r\)

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