Circular Motion Flashcards
Period (T)
Time taken to complete one circle/revolution (unit: s).
Angular displacement (θ)
Angle in RADIANS through which a point or line has been rotated around a fixed point (unit: radians). Displacement is change in position, and b/c it’s a circle…
sin θ = s/r, where s is arc length (distance travelled by object) and r is radius, but we don’t bring the sin in b/c the angle is very small.
So, for a complete circle, θ = 2πr/r = 2π
Angular velocity (ω)
Change in angular displacement per unit time.
ω = Δθ/Δt
For a complete circle, ω = 2π/T, where T is the PERIOD.
Frequency (f)
Number of complete revolutions in one second (unit: Hz).
f = 1/T
ω = 2πf
Angular speed (v)
Recall,
speed (v) = distance/time
therefore, for one revolution:
v = 2πr/T
but angular velocity (ω) = 2π/T…
therefore,
v = ωr
Centripetal acceleration
Although an object is in uniform circular motion moves w/ a constant speed, its direction (and velocity, as velocity depends on direction) constantly changes. Therefore, the object is accelerating. This acceleration is known as centripetal acceleration, and it’s always acting towards the center b/c of direction for v (and, so, same for F).
centripetal acceleration = v^(2)/r
but v = 2πr/T
therefore,
centri. accel. = 4π^(2)r/T^(2)
Centripetal force
No acceleration w/o net force present, so…
The RESULTANT (you often have more than one force present) force that pulls objects towards the center of a circle. Always acts toward the center (hence the name). If not for this force, bob would’ve veered off, but keeps it in a circle. And it’s a general name, so the name varies w/ situation at any given time.
Types:
- For planets around a sun, gravitational
- W/ a car going round, friction (between tires, ground)
- Tension if string involved
Recall,
Fnet = ma
therefore,
Fc = mv^(2)/r
but v = ωr
therefore,
Fc = mω^(2)r
Mass on a string—horizontal (circular motion ex.)
T not plane (tilted up, so have to find horizontal component for it to be Fc). And ball has a mass, so have to consider gravity. No vertical force present, so weight (W) must be balanced by a vertical component of the tension (T).
*Motion only in horizontal direction (accel. in vertical 0, so net force is 0 [forces must be balanced]). There must be another force (upward arrow on ball).
θ is top angle…
For vertical component of T: Tcosθ = W
For horizontal component of T: Tsinθ = Fc = mv^(2)/r
Mass on a string—vertical (circular motion ex.)
When a mass is in a vertical circle, gravity can either act in same direction as its motion or against it (does it agree w/ direction of Fc).
Assuming perfect alignment (otherwise, have to find vertical component)…
At top:
mv^(2)/r = T + mg
Therefore,
T = mv^(2)/r - mg
At bottom:
mv^(2)/r = T - mg (longer - shorter)
Therefore,
T = mv^(2)/r + mg
NOTE:
- At the top, the mass loses KE to PE, resulting in the loss of speed.
- The MIN SPEED of the mass is achieved when its weight is enough to provide the centripetal force w/o the tension (T).
mv^(2)/r = mg or KE = PE
Gravitational field
Region where a test mass feels a force (every field has a force).
*When sphere involved, have to treat as point by going from center…
Newton’s Law of Universal Gravitation
States that the force of attraction between two point masses is directly proportional to the product of the two masses and inversely proportional to the square of their distance apart. Completely true when masses are tiny.
F ∝ m1m2/r^2
F = Gm1m2/r^2, where G is the universal gravitational constant (6.67 x 10^[-11] Nm^[2]kg^[-2]).
This, I = P/4πr^2 (intensity), and Coulomb’s Law, are called inverse square laws.
NOTE:
- Masses only attract (unlike in electricity).
- Law only completely true w/ point masses.
- If spherical masses involved, forces act from center.
Gravitational field strength (g)
Defined as force per unit test mass that enters a field.
g = F/m2, where m2 is our test mass.
Unit is Nkg^(-1)—same as ms^(-2) from F = ma (ma/m). So, acceleration is gravitational field strength. Since acceleration due to gravity (g) = F/m, gravitational field strength is the same as acceleration due to gravity.
Recall,
g = F/m
but F = Gm1m2/r^2
Therefore,
g = Gm1/r^2
g is a vector and can thus be represented using field lines.
For a test mass, just straight lines going toward the point.
For a sphere, lines become dashed when enter sphere and go towards center point (arrows on outside). Forces have to act from center, and inside the circle, they HAVE to be dashed.
For near the surface of the earth, just straight lines going down into a horizontal surface (b/c earth so big, can’t see curvature).
NOTE:
The closer the lines, the stronger the field (applies to every field line diagram).
*And only inwardly, as masses only attract…
Overall gravitational field strength
In order to calculate the above, vector addition is used. There’s usually a point between two masses where the overall g is 0. There’ll be force that they feel b/c of each other. Find the strengths due to other, and one will have effect in opposite direction (+, -).
g = Gm1/r^2
When 0, gm1 + (-gm2) = 0
gm1 = gm2
*if g at x = 0…
Orbital motion
For an object/mass going around in an orbit…
Recall,
Fc = m2v^2/r (m2 test mass)
Fg = Gm1m2/r^2
m2v^2/r = Gm1m2/r^2
v = (Gm1/r)^(1/2), where v is the speed in a circular orbit, and r is the orbital radius
But v = 2πr/T
v = (Gm1/r)^(1/2) = 2πr/T
4π^(2)r^(2)/T^2 = Gm1/r
T^2 = 4π^(2)r^(3)/Gm1
T^2 = r^3, where T is the period of the orbiting object
All constants (mass constant b/c referring to mass of
*Direction makes all the difference in circular motion…
