## artemisxrpg Group Title 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) 11 months ago 11 months ago

1. ganeshie8 Group Title

$$\large \omega = \frac{2 \pi}{T}$$

2. ganeshie8 Group Title

earth is taking 1 day to to rotate around itself. $$\large \omega = \frac{2 \pi~radians}{1~ day}$$

3. artemisxrpg Group Title

it's 2 pi because that how much one revolution is?

4. ganeshie8 Group Title

Yup ! to go around circle once, it takes 2pi radians

5. ganeshie8 Group Title

|dw:1380369147272:dw|

6. artemisxrpg Group Title

so angular speed in 24 hrs would be pi/12

7. ganeshie8 Group Title

8. ganeshie8 Group Title

2 pi radians per day us same as saying, pi/12 radians per hour

9. ganeshie8 Group Title

its just the unit conversion

10. ganeshie8 Group Title

*is

11. artemisxrpg Group Title

v = r*theta/t, so in this case it would be 6400*2pi/1 day or 24 hr?

12. ganeshie8 Group Title

theta/t is the angular speed $$\omega$$

13. ganeshie8 Group Title

linear speed v = r$$\omega$$

14. ganeshie8 Group Title

simply multiply radius wid $$\omega$$

15. ganeshie8 Group Title

whats the radius of earth at north pole ?

16. artemisxrpg Group Title

6400?

17. ganeshie8 Group Title

let me put it this way :- whats the distance from axis of rotation to you, when you are at north pole ?

18. artemisxrpg Group Title

I'm still confused|dw:1380370272714:dw|

19. artemisxrpg Group Title

this is where the north pole would be?

20. ganeshie8 Group Title

|dw:1380370352008:dw|

21. ganeshie8 Group Title

|dw:1380370410225:dw|

22. ganeshie8 Group Title

thats the axis of rotation, straight line from North pole to Sounth pole wats the distance ? :)

23. artemisxrpg Group Title

12800?

24. ganeshie8 Group Title

'point on North pole' , is right 'next to axis of rotation'

25. ganeshie8 Group Title

so, distance between them = r = 0

26. ganeshie8 Group Title

for a person on North / South poles :- linear velocity $$v = r \omega = 0\times \omega = 0$$

27. ganeshie8 Group Title

see if that makes more/less sense

28. ganeshie8 Group Title

|dw:1380370749167:dw|

29. ganeshie8 Group Title

starting from equator, as you move towards the poles, the linear velocity decreases and becomes 0 at poles

30. artemisxrpg Group Title

so this is where the equator is?|dw:1380371306773:dw|

31. artemisxrpg Group Title

not at the center of the graph?

32. ganeshie8 Group Title

yup ! thats the equator, and since distance from axis is maximim, equals radius of earth, the speed wud be maximum at equator

33. artemisxrpg Group Title

so then radius would be 6400 at the equator?

34. ganeshie8 Group Title

lets call it, distance from axis. at equator, distance from axis = radius of earth = 6400

35. ganeshie8 Group Title

in ur speed formula, $$v = r \omega$$ , $$r$$ is distance form axis

36. artemisxrpg Group Title

then if the linear speed is halfway from the equator to north pole, radius would be 3200?

37. ganeshie8 Group Title

nopes, use trig to find r, the distance from axis

38. ganeshie8 Group Title

|dw:1380372161083:dw|

39. ganeshie8 Group Title

|dw:1380372288001:dw|

40. ganeshie8 Group Title

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41. ganeshie8 Group Title

thats the distance from axis, $$r$$ you can find it using proper trig ratio

42. artemisxrpg Group Title

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?

43. ganeshie8 Group Title

Nope, careful 6400 is the hypotenuse

44. ganeshie8 Group Title

|dw:1380442777970:dw|

45. artemisxrpg Group Title

sin45 = x/6400 = 4525.48?

46. ganeshie8 Group Title

Yup ! thats the $$r$$