Question

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 speed and centripetal acceleration of a person situated (a) at the equator and (b) at a latitude of 11.0° north of the equator.
(a) v=
units
a_c=
units
(b) v=
units
a_c=
units
save for later

Explanation:

Step1: Calculate the angular speed $\omega$

The Earth rotates once in a day. One - day $T = 24\times3600\ s$. The angular speed $\omega=\frac{2\pi}{T}$.
$\omega=\frac{2\pi}{24\times3600}\ rad/s\approx7.27\times 10^{-5}\ rad/s$

Step2: Find speed and centripetal acceleration at the equator

At the equator, the radius of the circular path $r = R_{e}=6.38\times 10^{6}\ m$.
The speed $v=\omega r$.
$v=(7.27\times 10^{-5}\ rad/s)\times(6.38\times 10^{6}\ m)\approx464\ m/s$
The centripetal acceleration $a_{c}=\omega^{2}r$.
$a_{c}=(7.27\times 10^{-5}\ rad/s)^{2}\times(6.38\times 10^{6}\ m)\approx0.0337\ m/s^{2}$

Step3: Find the radius of the circular path at latitude $\theta = 11.0^{\circ}$

The radius of the circular path $r = R_{e}\cos\theta$, where $\theta = 11.0^{\circ}$ and $R_{e}=6.38\times 10^{6}\ m$.
$r=(6.38\times 10^{6}\ m)\cos(11.0^{\circ})\approx6.26\times 10^{6}\ m$

Step4: Find speed and centripetal acceleration at latitude $\theta = 11.0^{\circ}$

The speed $v=\omega r$.
$v=(7.27\times 10^{-5}\ rad/s)\times(6.26\times 10^{6}\ m)\approx455\ m/s$
The centripetal acceleration $a_{c}=\omega^{2}r$.
$a_{c}=(7.27\times 10^{-5}\ rad/s)^{2}\times(6.26\times 10^{6}\ m)\approx0.0327\ m/s^{2}$

Answer:

(a) $v = 464\ m/s$, $a_{c}=0.0337\ m/s^{2}$
(b) $v = 455\ m/s$, $a_{c}=0.0327\ m/s^{2}$