# Chapter 5: Circular Motion and Gravitation

Centripetal on the FBD - force is directly to the center of the circle

• Mass is never a factor of any uniform circular motion

### Uniform Circular Motion

Motion of an object traveling at a constant speed on a circular path Centripetal force is the ﻿$\sum F$﻿ that keeps an object in the circular path

﻿$F_{rad} = ma_{ac} = mv^{2}/r$﻿

﻿$v=2\pi r/T$﻿ where T represent the period of the motion

﻿$\underline{+ a_{c} = v^{2}/r \Rightarrow a_{c}}$﻿ represents the centripetal acceleration

﻿$a_[c] = 4\pi^{2}R/T^{2} \Rightarrow R$﻿ represents the radius

• An object will travel tangential to its original circular motion once it loses the centripetal force. (**Inertia**)
• Cannot be taken to image on the FBD

### Circular motion in reality

Friction - responsible for the centripetal force of a car making a curve with a specific ﻿$\mu$﻿ .

﻿$F_{c}= \mu _{s}F_{N}=\mu _{s}mg=mv^{\wedge }2/r\Rightarrow v= \gamma \mu _{s}gr$﻿

### Banking

On a tilt, the motion of an object in a circular direction

﻿$\underline{F_c = F_N Sin\Theta = mv^2 / r}$﻿ ﻿$\underline{F_{N}\ Cos\Theta = mg }$﻿

﻿$\underline{Tan\Theta = v^2 /rg}$﻿

#### On a Ferris wheel

﻿$N_B - mg = mv^2 / R \Rightarrow N_B = mg + mv^2 /R = m(g+v^2/R)$﻿

﻿$Mg - N_T = mv^2 / R \Rightarrow N_T = mg - mv^2 /R = m(g-v^2/R)$﻿

• The Normal force is greater @ bottom giving a greater pressurized force on a person.

﻿$@ v^2 / R= g$﻿ normal force ﻿$0=\rightarrow 0$﻿ weight / weightless (at top)

• Walking is considered uniform circular motion ﻿$F_{cent} = mg = mv^2 /g\Rightarrow \gamma gr = v$﻿

### Newton’s Law of Gravitation

﻿$G = 6.674 \cdot 10^{-11} \frac{N \cdot m^2}{kg^2}$﻿ - Gravitational Constant

﻿$F_G = Gm_1m_2 / r^2$﻿ - calculates attraction force between 2 masses

﻿$M_{\frac{1}{2}}=$﻿ the 2 masses, given r = distance between the 2 masses where: ﻿$Fg_{12}=-Fg_{21}$﻿

Weight = [m]g = G[mass] ﻿$(M_{earth}) / r_{earth}^2$﻿

#### Circular motion around the earth

﻿$F_c = GmM_E / r^2 = mv^2/r\Rightarrow v = \gamma(GM_E/r)$﻿ - Radius from center of earth to object

#### Keplar’s law

﻿$V = \gamma (GM/r) = 2\pi r/T \Rightarrow T =2\pi r^3 /2$﻿ ﻿$\sqrt{GM}$﻿ and ﻿$M = 4\pi ^2r^3 / GT^2$﻿

Escape Velocity - the speed required to leave earth’s gravitational force

### Apparent Weight

• The weight that we feel with other applied force on us.
• It is different from ﻿$F_{ga}$﻿