Centripetal Acceleration - Two people situated at different latitudes on Earth

[crono_]

For starters, I feel sorry for those who have to sift through all these questions and answer them. It seems like you've got a very busy job! Hopefully there are many of you to share the workload.

This is an assignment question that I get 3 attempts on. I've made two, both were incorrect. I'd like to try and salvage at least a couple of marks, if possible...

Thankfully, I managed to get part (a) correct, it's (b) is causing the stress.

Homework Statement

The Earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the plane of the equator. Assuming the Earth is a sphere with a radius of 6.38 x 10^6 m, determine the centripetal acceleration of a person situated (a) at the equator and (b) at a latitude of 17.0 ° north of the equator.

r_E = 6.38 x 10^6m

T = 86400s (1 day)

Homework Equations

a_c = v² / r

v = 2 \pir / T

The Attempt at a Solution

Part (a)

v = 2 \pir / T

v = 2 \pi6.38 x 10^6m / 86400s

v = 463.9666 m/s (I kept a few extra digits for now...that's okay, right?)

With v I went to calculate a_c = v² / r

a_c = v² / r

a_c = (463.9666 m/s)² / 6.38 x 10^6m

a_c = 0.0337 m/s²

Now on to (b)

I tried this two different ways. The first was trying to determine the new radius with trig and visualizing a couple of right angel triangles in the picture that came with the question. But this was a waste of time as the sides wouldn't add up properly due to the Earth's spherical shape.

After that, I tried to google the question to see if there was any advice online. Found a Yahoo answer stating that the new radius would be r_E cos17. Seemed easy enough...

r = 6.38 x 10^6m cos17 = 6.10 x 10^6m

a_c = v² / r

a_c = (463.9666 m/s)² / 6.10 x 10^6m

a_c = 0.0353 m/s²

But this came back wrong as well...

Any help would be appreciated. I've already lost the marks for the question, but would really like to know how to do it for next time!

Thanks!

 

[crono_]

Weight...

Would this be considered a "banked curve" problem? I may be stretching here...

tan \vartheta = v² / rg

Solve for r...

Then use that new r in

a_c = v² / r

?

Err...maybe not...

 

[rl.bhat]

You have found the velocity at the equator.
But the velocity at the given point is not the same.
So find angular velocity ω, which remains the same everywhere.
Then αc = (ω^2)Rcosθ