Question

Find the centripetal accelerations of (a) a point on the Equator (b) at North Pole and (c) at the location of
Ada (40° N lattitude), due to the rotation of Earth about its axis.

Answer 1

Hi

so, for all parts think about the information given (you have the Earth radius?) [its 6371km]. It wants the centripetal acceleration for circular motion, which has the formula
\[a = \frac{ v^2 }{ r }\] so we will need the velocity and radius.
a) - think about if the object there is even traveling in a circle.. is it?
b) you must calculate the velocity of an object at the equator, using speed = distance / time. distance is the circumference (in m), and time is a day (converted to sec). then use the formula.
c) this is the same as part b), but the circle is not the same size. At 40 deg latitude, the circle being traveled in is smaller. You calculate the new circumference by multiplying the circumference from part b) by cos40, (see http://mathforum.org/library/drmath/view/54158.html for details).

Hope that helps?

Answer 2

where do i get the time and velocity from?

Answer 3

read the post carefully - the time is the time taken to go around the circle once. this is one day on the earth. you should be able to work out how many seconds in a day - that is you time. you should also be able to work out the circumference of the Earth - that is how far the person 'travels' in a day as the Earth spins. divide circumference by the time taken (speed = distance / time) and you have your velocity to use in the equation at the top.

Answer 4

I don't get the A part of the answer @furnessj

Answer 5

try to imagine sitting on top of the Earth as it spins round. If you are sat right at the top, do you trace out a circle from above, or do you just rotate on the spot..? (if you don't make a circle, you cannot have an inward acceleration)

Answer 6

so you saying the answer is 0 right? @furnessj

Answer 7

yes :)

Answer 8

thanks.. How you find the circumference?

Answer 9

A steel cylinder of mass 0.325 kg is tied to a string and allowed to revolve in a circle of radius 1.15 m on
a frictionless horizontal table. The other end of the string passes through a hole in the center of the table, and a
mass of 1.00 kg is tied to it. The suspended mass remains in equilibrium while the cylinder on the tabletop
revolves. (a) What is the tension in the string? (b) What is the horizontal force acting on the steel cylinder? (c)
What is the speed of the steel cylinder?

Answer 10

@furnessj please see if you can also help me with this? thanks