Introduction to Two-Dimensional Motion:
We resolve this type of motion into two separate cases of one-dimensional
motion by regarding the horizontal and vertical components of the motion
independently. For these two directions we use x and y, respectively. For
example, if an object is projected from the ground with a velocity u at an
angle of elevation θ , then we can use SohCahToa to find out how fast it is
moving in the x and y directions.
1. Any object launched from the ground at some angle θ is called a
projectile. The path it travels is an inverted parabola called its trajectory.
A classic example would be the motion of golf ball when struck with a
golf club. Can you think of a few more?
2. The initial velocity in the x direction is ux = u·cos(θ). The
acceleration in the x direction is ax = 0. The velocity of the object in the
y direction is uy = u·sin(θ). The acceleration is that of gravity which
acts only in the y direction. So, we can say that ay= -g = -9.81 m/s2.
3. We still have the five motion formulas from the study of kinematics
developed by Galileo (1564-1642). We know them as: Δs = vavg·Δt ,
vavg = (u+v)/2 , v = u + a·Δt , v2 = u2 + 2a·Δs ,
Δs = u ·Δt + ½a·Δt2 . The task now is to adjust these for the separate
x and y directions.
4. In the absence of air resistance a projectile has a constant horizontal
velocity and a constant downward free-fall acceleration which effects the
vertical velocity, subtracting 9.81m/s from it on the way up, and adding
9.81m/s to it on the way down.
Introduction to Circular Motion:
5. Newton's second law applied to a particle moving in uniform circular
motion states that the net force must be toward the center. This is the
Centripetal Force, Fc.
6. Uniform circular motion occurs when an acceleration of constant
magnitude is perpendicular to the tangential velocity and the object
maintains a constant speed but is accelerated toward the center of
7. This introduces the concept of centripetal acceleration, ac = v2/r,
and, by Newton's second law, centripetal force, Fc = mv2/r. We also
know that v = 2pr/T .
8. The central force acting on an object that provides the centripetal
acceleration could be have its origin in the following: (i) the force of
gravity (as in satellite motion), (ii) the force of friction (as in a car
rounding a curve), or (iii) a force exerted by a string (motion in a
9. In the case of motion in a vertical circle, the force of gravity
provides the tangential acceleration and part or all of the
centripetal acceleration. At the top of the circle, the net Force on the
object is zero when the string slackens, so that Fc = FG which implies
mv2/r = mg . Therefore the critical velocity vcrit = √(rg).
At the bottom FNET = mv2/r + mg .
10. In the case of a car rounding an unbanked curve, the force of static
friction is the central force. And we derive the equation v = √(μrg).
11. When the curved roadway is banked at an angle, then the horizontal
component of the normal force is centripetal. And we derive the equation
v = √(rgtan(θ)).
12. Considering Periodic Motion, which is back and forth over the same
path, we have the Pendulum, with period, T=2p√(l/g). We also have
the mass-spring system, with period T=2p√(m/k), with the spring
constant k = F/s.
Introduction to Gravitation:
Much of what we know about universal gravitation is due to the work
of the following astronomers and mathematicians.
(i) Nicolaus Copernicus (1473-1543), Poland, suggested that the Earth
and all other planets revolve in circular orbits around the Sun, a
heliocentric system, not the geocentric model that persisted for 1400
(ii) Tycho Brahe (1564-1601), Denmark, charted the positions of the
planets and 777 stars for 20 years.
(iii) Johannes Kepler (1571-1630), Germany, Brahe's assistant who
studied the data from the charts for 16 years and finally formulated 3
laws of planetary motion.
(iv) Galileo Galilei (1564-1642), Italy, who perfected the telescope and
later was placed under house arrest and force to recant for supporting
the heliocentric theory.
(v) Isaac Newton (1642-1727), England, developed the Law of Universal
Gravitation which states that all masses attract each other with a mutual
force that varies with the inverse-square of the distance.
13. Kepler's 3 laws of planetary motion state the following:
Law (1): All planets revolve in elliptical, nearly circular, orbits around the
Law (2): A straight line from a planet to the sun sweeps out equal areas in
equal time intervals.
Law (3): The cube of the orbital radius of any planet divided by the square
of its period is constant. r3/T2 = k
14. Newton's law of Universal Gravitation: "The force of attraction between
two bodies is directly proportional to the product of their masses but varies
inversely with the square of the distance between them." F = G·m1·m2/r2 .
15. The value of the Universal Gravitational constant, G, was also predicted
16. In 1798 the value of G was carefully measured with a torsion apparatus
by Henry Cavendish (1731-1810), England, confirming Newton's prediction.
G = 6.67x10-11 Nm2/kg2 (known as "Big G")
17. The mass of the Earth can be found by using Newton's Gravitation Law.
It is ME = 5.98x1024 kg. The mass of the Sun can be found from the
period and radius of a planet's orbit. The Sun's mass is computed to be
MS = 2.0x1030 kg.
18. The mass of a planet can be found only if it has a satellite orbiting it.
19. A satellite in a circular orbit, radius R, accelerates centripetally toward
Earth at a rate equal to the acceleration of gravity at its orbital radius.
20. The following properties of satellite motion can all be proven:
(i) the velocity is given by the equation v = (2πR)/T
(ii) acceleration due to gravity at the orbital radius, R, is g = (G·ME)/R2
(iii) the minimum or critical velocity for stable orbit is v = √(Rg)
21. All bodies have gravitational fields around them, which can be
represented by a collection of vectors representing the force per unit
mass at all locations.
22. The mass of an object can be determined in two ways, gravitationally
and inertially. Both result in equivalent determinations of mass.
And still, we need these steps to solve any problem in Physics:
(i) read the problem and identify the given variables
(ii) determine what you are asked to solve for
(iii) find the correct motion formula to use
(iv) use algebra to isolate the unknown
(v) substitute-in the given information and simplify.
For the C-Code and B-Code Gravitation problem set 1. Click HERE.
For the C-Code and B-Code Gravitation problem set 2. Click HERE.
Get the template to type-up your Lab Abstract. Click HERE.
For the Gravitation slides, click HERE.
For the Chapter end problem solutions, click HERE.