Gravity and Motion Flashcards
Projectile Motion, Inclined Planes, Circular Motion, Gravitational Forces and Field and Orbits
What component of SUVAT needs to be resolved in projectile motion?
Inital Velocity
Horizontal Motion
The velocity remains the same as the initital velocity throughout the motion
Maximum Height
When the projectile is at maximum height, vertical velocity is 0
Velocities
Horizontal and vertical velocities are at 90 degree angles to each other which is why they can be resolved to find initial velocities
Forces acting on an inclined planes
Normal force, weight, frictional force, applied force
What force needs to be resolved into components
The weight force, there is motion on the vertical component of the weight force (Fgcos) but none on the horizontal (Fgsin)
Normal Force Equation
mgcos-theta
Acceleration of Inclined Planes
gsin-theta
Fnet in inclined planes
All the forces acting on the object on the inclined plane
generally f=ma, f= fg + fn
Method of calculating forces on an inclined plane
Resolving all forces found into x and y components through a table or other
Uniform Circular Motion
Motion along a circular path in which there is no change in speed, only direction
constant velocity tangent to path, constant force towards centre
CM - Period
Time of one ‘cycle’
distance = rate x time
time = distance/rate
therefore, period = 2pir/v
Role of Tension in CM
Without tension, the object being flung around moves in a straight line. Tension changes the direction of the object
Centripetal Acceleration
Acceleration directs towards the centre of the circle
Ac = v (squared)/R
F = ma = mv(squared)/R
Conical Pendiulum
When a mass is moving in a horizontal circle of radius (R) at the end of the cord of length (L)
Tension is diagonal from the mass and can be resolved into horizontal and vertical components
Fnet is given by Tsintheta
Tsintheta = mv2/R
Tcostheta = mg
tan theta = v2/gR
Vertical Circular Motion
Top of circle - mg + T = mv2/r
Bottom of circle - T - mg = mv2/r
Top of circle, T = mv2/r - mg
Bottom of circle, T = mv2/r + mg
Gravitational Force - why do masses attract
All objects with mass are attracted to each other
Gravitational fields emit radially away from the centre of the mass
Gravitational Fields
Region around a mass where another mass experiences a force of gravitational attraction (more lines - more massive)
Gravitational Force Equation
F = GMm/r2
mg = GMm/r2
What does m (mass 1) feel from M (mass 2)
g force = Gm/ r squared
g force = f/m
Orbits - Keplars First Law
Planets move in one elliptical (eclipse shape) orbit with the sun at one focus
Aphelion - the point on the orbit which is furtherst from the Sun
Perihelion - the point on the orbit which is the closest to the Sun
Law of Universal Gravitation
Newton realised that the force of attraction between two massive objects:
- Increases as the mass of the objects increases
- Decreases as the distance between objects increases
Orbits - Keplars Second Law
The line joining the planet to the sun sweeps out equal areas in equal intervals of time
A1 = A2
T1 = T2
Orbits - Keplars Third Law
The ratio of the squares of the periods of any two planets revolving aorund the sun is equal to the ratio of the cubes of thier mean distances from the sun
Third Law Derivation
Gravity is equal to centripetal force in a orbit
(sun and earth example)
Fg = Fc
GMsme/ r2 = Mev2/r
GMs/R = v2
GMs/R = (2pir/t)2
GMs/R = 4pi2r2/T2
GMsT2= 4pi2r3
T2 = 4pi2r3/GMs
T12/T22 = 4pi2r3/GMs/4pi2r3/GMs
T12/T22 = R23/R13
When does F normal force = F weight force in inclined planes
when it’s 0 degrees to the horizontal
