# Physics equations/Uniform circular motion

#### Newton's second law of motion

$\Sigma {\vec {F}}=m{\vec {a}}=m\,{\frac {\rm {{d}{\vec {v}}}}{\rm {{d}t}}}=m\,{\frac {\rm {{d^{2}}{\vec {r}}}}{\rm {{d}t}}},$ ##### *** Problem: Explain the units of ΣF=ma

Solution: Σ has no units (i.e. is dimensionless}; it is an instruction to sum all the forces acting on an object. The symbol Σ is a simple example of an operator. Another dimensionless operator is d, which is an instruction to take the difference beteen two values that are nearly equal to each other. In our SI system of units: m (mass) is measured in kg; a (acceleration) is measured in m s-2. The equation informs us that F (force) is measured in kg m s-2, which is also called a Newton (abbreviated as N). Note that F and a are boldfaced to indicate that they are vectors.

##### *** Problem: What was Newton's first law?

A 2kg concrete block is being dragged along by a rope with a tension of 4N. The block is moving west at a constant speed of 10m/s. What is the net force?

Solution: If v is constant then a is zero. Therefore the net force is zero (possibly because 4N of friction is acting in the easterly direction. Newton's first law states that if the net force (the vector sum of all forces acting on an object) is zero, then the velocity of the object is constant.

#### Newton's law of universal gravitation

Newton published this in 1687, his knowledge of the numerical value of the gravitational constant was a crude estimate. For our purposes, it can be conveniently state as follows :

$\mathbf {F} _{12}=-G{m_{1}m_{2} \over {\vert \mathbf {r} _{12}\vert }^{2}}\,\mathbf {\hat {r}} _{12}{\mbox{ where }}G\approx 6.674\times 10^{-11}\ {\mbox{m}}^{3}\ {\mbox{kg}}^{-1}\ {\mbox{s}}^{-2}$ ##### *** Problem: Explain each symbol in Newton's law of gravity

Solution:

$\mathbf {F} _{12}$ is the force applied on object 2 due to object 1
$G$ is the gravitational constant
$m_{1}$ and $m_{2}$ are respectively the masses of objects 1 and 2
$\vert \mathbf {r} _{12}\vert \ =\vert \mathbf {r} _{2}-\mathbf {r} _{1}\vert$ is the distance between objects 1 and 2
$\mathbf {\hat {r}} _{12}\ {\stackrel {\mathrm {def} }{=}}\ {\frac {\mathbf {r} _{2}-\mathbf {r} _{1}}{\vert \mathbf {r} _{2}-\mathbf {r} _{1}\vert }}$ is the unit vector from object 1 to 2

(Note: the minus sign is a complexity that is often ignored in simple calculations. Don't fuss with minus signs unless you have to.)

#### Weight and the acceleration of gravity

The force of gravity is called weight, ${\vec {w}}$ , If one of two masses greatly exceeds the other, it is convenient to refer to the smaller mass, (e.g.stone held held by person) as the test mass, $m_{0}$ . A vastly more massive body (e.g. Earth or Moon) can be referred to as the central body, with a mass equal to $m_{C}$ . It is convenient to express the magnitude of the weight ($w=|{\vec {w}}|$ ) as,

$w=F=G{\frac {m_{0}m_{C}}{r^{2}}}=m_{0}g$ ,

where $g=Gm_{C}/r^{2}$ is called the acceleration of gravity (or gravitational acceleration). Near Earth's surface, ${\vec {g}}=$ is nearly uniform and equal to 9.8 m/s2. In general the gravitational acceleration is a vector field, meaning that it depends on location, g = g(r) or even location and time, g = g(r,t).

#### Uniform circular motion Position, velocity and acceleration for uniform circular motion. All vectors rotate with the same angular frequency.

Consider a particle is moving in a circle or radius, R, at a uniform speed, v. The period of orbit is T, and is defined by

$vT=2\pi R$ .

It is convenient to define angular frequency, ω, by the relation,

$\omega T=2\pi$ ,

so that $v=\omega R$ . It can be shown that the acceleration is towards the center, and has magnitude given by,

$a={\frac {v^{2}}{R}}=\omega v=\omega ^{2}R$ ##### *** Problem: Prove that a = v2/R

Solution: Two solutions are offered:

Without Calculus:

The figure depicts a change in the position and velocity of a particle during a brief time interval $\Delta t$ . The distance traveled is $\Delta \ell =a\Delta t$ , and the change in velocity $|\Delta {\vec {v}}|=a\Delta t$ . The two triangles are similar, so that $a\Delta t/v=v\Delta t/r$ , which leads to $a/v=v/r$ With Calculus Let ${\vec {r}}=x{\hat {i}}+y{\hat {j}}=R\cos \omega t{\hat {i}}+R\sin \omega t{\hat {j}}$ Take the derivative twice to get first ${\vec {v}}$ then ${\vec {a}}$ :

${\vec {v}}=-\omega R\sin \omega t{\hat {i}}+\omega R\cos \omega t{\hat {j}}$ ${\vec {a}}=-\omega ^{2}R\cos \omega t{\hat {i}}-\omega ^{2}R\sin \omega t{\hat {j}}$ Otain the desired result by taking the magnitude of each vector (i.e. a = (ax2+ay2)1/2

##### ***Problem: Generalize a=v^2/R to include nonuniform motion on a curve

Solution: Components of acceleration for a curved motion. The tangential component at is due to the change in speed of traversal, and points along the curve in the direction of the velocity vector (or in the opposite direction). The normal component (also called centripetal component for circular motion) ac is due to the change in direction of the velocity vector and is normal to the trajectory, pointing toward the center of curvature of the path.
${\vec {a}}=\kappa \left({\frac {d\ell }{dt}}\right)^{2}{\hat {\eta }}+{\frac {d^{2}\ell }{dt^{2}}}{\hat {\tau }}$ ,

where $\ell$ is the path length along the direction of motion, ${\hat {\tau }}$ is a unit vector parallel to that path (so that velocity, ${\vec {v}}={\frac {d^{2}\ell }{dt^{2}}}{\hat {\tau }}$ , and ${\hat {\eta }}$ is a unit vector that is perpendicular to the direction of motion, and points "inwards" towards the direction of the turn. The curvature, $\kappa$ equals $1/R$ where $R$ is the radius of curvature of the turn.

Two solutions offered:

1. (from Physics with calculus Wikibook)
2. Wikipedia article(See tangential and centripetal acceleration, if it is still on the current page.

#### Newton and Kepler's third law for planetary motion

1.     $ma=m{\frac {v^{2}}{r}}={\frac {mMG}{r^{2}}}$ , where m is the mass of the orbiting object, and M>>m is the mass of the central body, and r is the radius (assuming a circular orbit).
2.     $vT=2\pi r$ , where m is the mass of the orbiting object, and M>>m is the mass of the central body, and r is the radius (assuming a circular orbit). After some algebra:
3.     $r^{3}={\frac {MG}{4\pi ^{2}}}T^{2}$ • $a^{3}={\frac {(M+m)G}{4\pi ^{2}}}T^{2}$ , is valid for objects of comparable mass, where T is the period, (m+M) is the sum of the masses, and a is the semimajor axis: a = ½(rmin+rmax) where rmin and rmax are the minimum and maximum separations between the moving bodies, respectively.

#### Newton's third law

The forces on a collection of N interacting particles can be described as either internal or external forces. Since forces are generally viewed as interactions between particles, the external force on the j-th particle, ${\vec {F}}_{j}^{\text{ext}}$ is an interaction with particles outside this collection of N particles. The sum of all forces on the jth particle is:

$m_{j}{\vec {a}}_{j}=\sum _{j=1}^{N}{\vec {F}}_{ji}+{\vec {F}}_{j}^{\text{ext}}$ Newton's third law states that the internal force on the jth particle by the ith particle is ${\vec {F}}_{ji}$ , and Newton's third law states that ${\vec {F}}_{ji}=-{\vec {F}}_{ij}$ . Since the only number that equals its additive inverse is zero, it follows that no particle can exert a force on itself, e.g. ${\vec {F}}_{ii}=0$ (for all i), and one wonders if the motion of physical objects could ever have been fathomed if particles could induce forces on themselves. When many particles are connected (either rigidly or by forces that allow relative motion), it can be shown that Newton's second law (F-ma) applies to the center-of-mass, defined as:

${\vec {R}}={\frac {\sum _{i}m_{i}{\vec {r}}_{i}}{\sum _{i}m_{i}}}={\frac {1}{M}}\sum _{i}m_{i}{\vec {r}}_{i}$ 