PHY 111: Introduction to Physics

The content of this course is as shown below:

Circular Motion

  • Definition of circular motion
  • Angular velocity, acceleration and period
  • Centrifugal and centripetal forces
  • Conical pendulum

Gravitation

  • Gravitational field
  • Kepler's laws
  • Acceleration due to gravity
  • Newton's law of universal gravitation
  • Gravitational potential and escape velocity

Elasticity

  • Definition of elasticity
  • Hooke's law of elasticity and terms associated with elasticity
  • Young's modulus, strain energy and strain energy per unit value
  • Ductile and britle materials
  • Shear and bulk modulus
  • Workdone in stretching a spring

Circular Motion

The motion of an object in a curve about a fixed point is reffered to as a circular motion. Examples of circular motion include: an artificial satellite orbiting the Earth at a constant height, a ceiling fan's blades rotating around a hub, a stone which is tied to a rope and is being swung in circles, a car turning through a curve in a race track, an electron moving perpendicular to a uniform magnetic field, a gear turning inside a mechanism and the likes.

Consider the circular motion of a particle about the circular path ABC, in the figure above. The speed of the particle about the center O, is called the angular speed ω, of the particle and it is defined as the change in angle θ, per unit change in time t, i.e.,

$${ \omega = \frac{\Delta \theta}{\Delta t}} $$
when the Δt becomes very small, then we can write $${ \omega = \lim_{\Delta t \to 0}\frac{\Delta \theta}{\Delta t}=\frac{d\theta}{dt} ~~~-------------(1)} $$

The unit of angular speed is radian per second (rad s-1).
Similar to displacement = (uniform velocity) x (time), we can have

$${d\theta = \omega \cdot dt}$$

The time to complete a whole circle is known as the period T, and is given as

$${ T = \frac{\theta}{\omega} } $$

but θ = 2π, therefore (since πc = 180o, then 2πc = 360o; 360o is a complete circle!) $${ T = \frac{2\pi}{\omega} -----------------------(2)} $$ The SI unit of period T, is second(s).
The number of circles made in a second is defined as frequency f, and from this definition it could be seen that frequency is just the period inverted, i.e., $${ f = \frac{1}{T}. }$$ From the above equation, the unit of frequency could be deduced to be per second (i.e., 1/s), in SI unit 1 cycle per second is usually defined as 1 Hertz (Hz), hence the SI unit of frequency is Hz. Therefore, we could rewrite Eq. (2) as $${ \frac{1}{f} = \frac{2\pi}{\omega} } $$ that is $${ \omega = 2\pi f ~~~--------------------(3)} $$

The angle θ, in radian, is defined as the ration of the arc s to the radius r, i.e., $${ \theta = \frac{s}{r} } $$ therefore, $${ s = \theta \cdot r ~~~--------------------(4)} $$ if we divide both sides by time t, then we will have $${ \frac{s}{t} = r\frac{\theta}{t} } $$ Now, s/t is the linear velocity v, while θ/t is the angular speed ω, therefore $${ v = \omega r ~~~--------------------(5)} $$
The angular acceleration α, of the object as it moves round the circular path is given by $${\alpha = \frac{d\omega}{dt} ~~~--------------------(6)} $$
This angular acceleration α, for a uniform circular motion is always equal to zero (α = 0), since there will be nochange to the angula speed ω, of the object around the circle.
As the object moves round the circular path, the linear velocity of the object changes all the time. In Figure 1, linear velocity at point A is not the same as that at point B (the two velocities do not point in the same direction, even though they may have the same magnitude!). This therefore, implies that there is a change in linear velocity per unit change in time and this is defined as the linear acceleration a, i.e.,
$${a = \frac{dv}{dt} = r\frac{d}{dt}\frac{d\theta}{dt} = r\frac{d^{2}\theta}{dt^{2}} = r\cdot\omega^{2} } $$
therefore,
$${a = \omega^{2}\cdot r ~~~--------------------(7)} $$

As could be seen in Figure 2, acceleration a is always acting perpendicular to linear velocity v and most importantly, it is always directed towards the center; therefore, we say that the acceleration a is center seeking, in other word, centripetal.
According to Newton's second law of motion, associated with a body of mass m undergoing an acceleration a is a force F which is in the direction of the acceleration a, i.e., $${F = m\cdot a,} $$
this force F is also called centripetal force. If the the body of mass m is always at the circular path, that is, it is at equilibrium at the circular path then the Newton's third law is obeyed and there must be equal in magnitude but opposite in direction force to F, this opposite force is called centrifugal force. Now from Eq. (2), we can write F as; $${F = m\omega^{2}r,~~~-------------(8)} $$
From Eq.(5), we know that ω = v/r, therefore, $${F = m\cdot (\frac{v}{r})^{2}\cdot r,} $$
that is,
$${F = \frac{mv^{2}}{r}~~~--------------(9)} $$

So far, so good, we have succesfully learn some things in this section and it will be good if we recap on those things we have learn so far. To do this we are going to be answering some questions which are basically things learnt here. Click on the button below to start the quiz: