# Physics equations/Uniform circular motion

#### Newton's second law of motion

[edit | edit source]***** Problem:** Explain the units of Σ**F**=m**a**

[edit | edit source]**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*. The equation informs us that

^{-2}*(force) is measured in kg m s*

**F**^{-2}, which is also called a

*Newton*(abbreviated as

*N*). Note that

*and*

**F***are boldfaced to indicate that they are vectors.*

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

[edit | edit source]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

*is zero. Therefore the net force is zero (possibly because*

**a***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.

^{[1]}

#### Newton's law of universal gravitation

[edit | edit source]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
^{[2]}:

***** Problem:** Explain each symbol in Newton's law of gravity

[edit | edit source]**Solution:**

- is the force applied on object 2 due to object 1
- is the gravitational constant
- and are respectively the masses of objects 1 and 2
- is the distance between objects 1 and 2
- 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

[edit | edit source]The force of gravity is called **weight**, , 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**, . A vastly more massive body (e.g. Earth or Moon) can be referred to as the **central body**, with a mass equal to . It is convenient to express the magnitude of the weight () as,

,

where is called the acceleration of gravity (or gravitational acceleration). Near Earth's surface, is nearly uniform and equal to 9.8 m/s^{2}. 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

[edit | edit source]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

- .

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

- ,

so that . It can be shown that the acceleration is towards the center, and has magnitude given by,

***** Problem:** Prove that *a = v*^{2}/R

[edit | edit source]^{2}/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 . The distance traveled is , and the change in velocity . The two triangles are similar, so that , which leads to

**With Calculus**
Let

Take the derivative twice to get first then :

Otain the desired result by taking the magnitude of each vector (i.e. a = (a_{x}^{2}+a_{y}^{2})^{1/2}

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

[edit | edit source]**Solution:**

- ,

where is the path length along the direction of motion, is a unit vector parallel to that path (so that velocity, , and is a unit vector that is perpendicular to the direction of motion, and points "inwards" towards the direction of the turn. The **curvature**, equals where is the **radius of curvature** of the turn.

Two solutions offered:

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

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

[edit | edit source]- , 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).
- , 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:

- , 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(r_{min}+r_{max}) where*r*and_{min}*r*are the minimum and maximum separations between the moving bodies, respectively._{max}

#### Newton's third law

[edit | edit source]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, is an interaction with particles outside this collection of N particles. The sum of all forces on the jth particle is:

Newton's third law states that the internal force on the j^{th} particle by the i^{th} particle is , and Newton's third law states that . Since the only number that equals its additive inverse is zero, it follows that no particle can exert a force on itself, e.g. (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: