Data Booklet

$$ v = \omega r\\ a = \frac{v^2}{r} = \frac{4\pi^2r}{T^2}\\ F = \frac{mv^2}{r} = m\omega^2r $$

Circular Motion

A particle is said to be un uniform circular motion if it travels in a circle with constant speed.

Let’s consider an object rotating in a circle of radius r in a counter-clockwise direction, with constant speed v.

$$ \omega = \frac{\Delta\theta} {\Delta t}\\ \text{In other words, it means the total distance traveled divided by the total time.} $$

Examples

Example 1

The radius of the Earth’s orbit is about $1.5*10^{11}$m. Calculate:

1. the angular speed of the Earth as it rotates around the Sun
2. the linear speed of the Earth.

$$ \text{a: Angular speed: }\frac{2\pi}{T}\\\\b:\text{Linear speed} = \omega r\\ a)T = 365 d = 31536000s \\\text{Speed}_{ang} = \frac{2\pi}{31536000} = \bold{1.9910^{-7}rads s^{-1}}\\ b) \frac{2\pi}{31536000}1.510^{11} = \bold{3.010^{4}ms^{-1}} $$

Example 2

A large clock on a building has a minute hand that is 4.2 m long.

Calculate:

1. the angular speed of the minute hand

   $\omega = \frac{2\pi}{3600}\\\omega \approx 0.00174rad*s^{-1}$

2. the angular displacement, in radians, in the time periods

   1. 12 noon to 12.20

      $\frac{2\pi}{3}, \text{since 20 minutes is} \frac{1}{3} \text{of a full rotation }(2\pi)$

   2. 12 noon to 14.30

      $\text{2 full rotations (hours) and half of one} = 2.5(2\pi) = 5\pi$

3. the linear speed of the tip of the minute hand

   $v = \omega r\\ v = \frac{2\pi}{3600}*4.2\\ v=0.0073 ms^{-1}$