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 33.0° north of the equator.

Explanation:

Step1: Find the angular speed of Earth

The Earth rotates once in 24 hours. One - full rotation is $\theta = 2\pi$ radians and $t=24\ h=24\times3600\ s = 86400\ s$. The angular speed $\omega=\frac{\theta}{t}$. So, $\omega=\frac{2\pi}{86400}\ rad/s\approx7.27\times 10^{-5}\ rad/s$.

Step2: Calculate speed and centripetal acceleration at the equator

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

Step3: Calculate the radius of the circular path at latitude $\theta = 33.0^{\circ}$

The radius of the circular path of a person at latitude $\theta$ is $r = R\cos\theta$, where $R = 6.38\times 10^{6}\ m$ and $\theta = 33.0^{\circ}$. So, $r=(6.38\times 10^{6}\ m)\cos33.0^{\circ}\approx5.35\times 10^{6}\ m$.

Step4: Calculate speed and centripetal acceleration at latitude $\theta = 33.0^{\circ}$

The linear speed $v=\omega r=(7.27\times 10^{-5}\ rad/s)\times(5.35\times 10^{6}\ m)\approx389\ m/s$. The centripetal acceleration $a_{c}=\omega^{2}r=(7.27\times 10^{-5}\ rad/s)^{2}\times(5.35\times 10^{6}\ m)\approx0.0282\ m/s^{2}$.

Answer:

(a) $v = 464\ m/s$, $a_{c}=0.0337\ m/s^{2}$
(b) $v = 389\ m/s$, $a_{c}=0.0282\ m/s^{2}$