Question

what is the linear velocity of a person at equator of the earth due to its spinning motion ?(radius of the earth = 6400km)​

Answer 1

Given :

radius of the earth, R = 6400km.

To find :

The linear velocity of a person at equator

Solution :

The earth completes one revolution in 24 hrs.

When a body makes 'N' revolution in 't' sec then it's average angular velocity is

${ \leadsto{ \boxed{\sf{ \omega = \dfrac{2\pi N}{t} }}}}$

here,

ω denotes average angular velocity

N denotes no of revolution

t denotes time

According to the question ,

N = 1 and t = 24hr = 24 × 60 × 60sec

${ \leadsto{\sf{ \omega = \dfrac{2\pi \times 1}{24 \times 60 \times 60} }}}$

${ \leadsto{\sf{ \omega = \dfrac{\pi }{43200} \: rad/s }}}$

Now,

relation between linear and angular velocities is given by

• v = ωr

here,

v denotes linear velocity

ω denotes angular velocity

r denotes radius

by substituting all the given values

${ \leadsto{\sf{ v = \omega r }}}$

${ \leadsto{\sf{ v = 6.4 \times {10}^{6} \times \dfrac{\pi}{43200} }}}$

${ \leadsto{\sf{ v = 465.5 \: m/s}}}$

thus, the linear velocity of a person at equator is 465.5m/s.

Answer 2

At the equator, a person's linear velocity is 465.5m/s.

Find Linear Velocity

Given :

The Radius of Earth =

Step By Step Solution :

Step 1: Find Average angular velocity

• When an object moves along a straight route, the linear velocity is defined as "the rate of change of displacement with respect to time.

⇒No of revolution (N) = 1  

⇒Time (t) = 24 x 60 x 60

⇒ ω =  $\frac{2\pi X 1} {24 X 60 X 60}$

⇒ ω = $\frac{\pi }{43200} Rad / s$

Step 2: Find Linear Velocity

• The change in the angular coordinate, represented in radians, divided by the change in time is the average angular velocity. The angular velocity is a vector pointing in the rotational axis's direction.

⇒V = ωr

       ∴ r - Radius

       ∴ ω - Angular velocity

⇒V = $6.4 X 10^{6} X \frac{\pi }{43200}$

⇒ V = 465.5 m/s

At the equator, a person's linear velocity is 465.5m/s.