Question

The earth's radius is 6.37×106m; it rotates once every 24 hours.

a.)What is the earth's angular speed?

b.)Viewed from a point above the north pole, is the angular velocity positive or negative?

c.)What is the speed of a point on the equator?

d.)What is the speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator? (Hint: what is the radius of the circle in which the point moves?)

Answer 1

(a) angular speed ω = ∆θ/∆t

here, θ=2π and t= 24 x 3600 s (because you have to convert to seconds for rad/s).

so, ω = 2π / (24 x 3600s)=7.27 x 10^(-5) rad/s

(b) ω is positive for counter-clockwise rotation and negative in the clockwise direction.

From the north pole, you see that the Earth rotates counter-clockwise, therefore ω is positive.

(c) For determining the speed of a point on equator, we have to determine the tangential velocity.

The requisite equation for this is v = r*ω.

As we know, r = 6.37×10^6 m. We also know ω from A. Plug those in to get v_t=(6.37×10^6 m)(7.27 x 10^(-5) rad/s) = 463 m/s

(d) The angular difference between the equator and the north pole is 90º. 3/4 of that would be 67.5º.

Using rules for trig, you can find that the radius at that point is r=(Earth's radius)cos67.5º

                                                                                                    = 2.44 x 10^6 m

Then you plug that into the v = r*ω equation again to get

   v = (2.44x10^6)(7.27x10^(-5) rad/s) = 177.4 m/s