Question

Consider the radius of the Earth to be 6.38×106 m. What is the magnitude of the centripetal acceleration experienced by a
person (a) at the equator and (b) at the North Pole due to the Earth's rotation?

Answer 1

The angular velocity of the earth is calculated:

The angular velocity is given by:

Where:

• Angular displacement in radians:

$w=1rev \frac{1\pi rad}{1rev}=2\pi rad$

• The time in seconds:

$t=1Day*\frac{24h}{1Day}*\frac{3600s}{1h}=86400s$

Remember that the earth makes a turn (360 degrees) in approximately 24 hours

Numerically evaluating: $w=\frac{2\pi rad}{86400s}=7.27*10^{-5}\frac{rad}{s}$

Centripetal acceleration is given by:

$a_{c}=w^{2}*r$

Where:

• Angular velocity:

$w=7.27*10^{-5}\frac{rad}{s}$

• Radio:

r

For each part the angular velocity is the same, just change the radius.

Part a

The radius is:

$r=6.38*10^{6}m$

The angular velocity is:

$w=7.27*10^{-5}\frac{rad}{s}$

​

Calculating angular acceleration:

$a_{c}=(7.27*10^{-5}\frac{rad}{s})^{2}*6.38*10^{6}m$

Finally: a$a_{c}=3.37*10^{-2}\frac{m}{s^{2}}$

Part b

The radius is:$r=0m$

Angular velocity:w$w=7.27*10^{-5}\frac{rad}{s}$

Calculating angular acceleration:

$a_{c}=(7.27*10^{-5}\frac{rad}{s})^{2}*0m$

Finally $a_{c}=0\frac{m}{s^{2}}$

Note: The radius is equal to zero, because the axis of rotation is considered to pass through the poles, therefore the distance from the axis of rotation to the point where the person is zero, consequently the centripetal acceleration is zero.