Physics: Energy, Circular Motion, Gravitational Field

EPE

1/2kx^2 or 1/2Fx, where F is spring (applied) force

work is done when

WD by a constant force on a body is defined as the product of the force and the displacement in the direction of the force. WD = Fscos𝚹

gravitational force

F = GMm/r^2. r is the distance bw the two cg of the masses

circular motion formulae

Fc = m*ac = m x (v2/r = rω^2 = vω)

for POCOE qsn

loss in energy a = gain in energy b

KE

1/2mv^2

POCOE formula

KEi + PEi + WD by ncnsf - wd against ncnsf = KEf + PEf

uniform circular motion

An object undergoing uniform circular motion must have constant magnitude of velocity and its acceleration should always point to the centre of the circle

circular motion qsn

centripetal force is provided by: usually Mg, N, T, f etc.

• for a drifting car, usually Fc is frictional force

• for rollercoaster, Fc top: N+W, where N = 0 for min velocity

• for spinning water bucket, its T instead of N

eqn connecting power and force

P = Fv

banked track qsn

∑Fy = 0

Ncos𝚹 = mg

∑Fx = m*ac

Nsin𝚹 = mv^2/r

tan𝚹 = v^2/rg

roller coaster qsn

∑Fy top = mg = mv(top)^2/r

∑Fy bottom = N - W = mv(bottom)^2/r

• thus, v top,min = (rg)^1/2

TEi = TEf (where i is bottom, f is top)
KEi = KEf + GPEf
0.5mv(bottom)^2 = 0.5mv(top)^2 + mg(2r)

• thus, v bottom = (5rg)^1/2

KE in gfield

GMm/2r. derive from GMm/r^2 = mv^2/r, then manipulate mv^2/r into 1/2mv^2

angular speed formula

ω = 2π/T = ∆𝚹/∆t

period of geostationary satellite

24 hours

is there work done when obj is underfoing uniform circular motion?

no. force is towards the centre and the displacement it perpendicular to the force, thus WD = Fcos90 = 0 joules

how to find WD from F-x graph

WD = area under F-x graph

work energy theorem

∑WD = ∆KE

Power

work done / time

when a spring is gently lowered until from its unstretched length to equilibrium, the loss is gpe is more than the gain in epe. why?

theres upward work done by your hand. since WD = F cos theta, where theta = 180 degrees (since the displacement is downwards), WD is -ve.

by POCOE, loss in gpe + WD = gain in EPE. since WD is -ve, loss in gpe - wd = gain in epe, thus gain in epe < loss in gpe

WD by gas

WD = pressure x ∆Volume

when gas expands, work is done by gas

when gas contracts, work is done on gas

efficiency

eff = useful output/ total input

formula for g in g-field

F = GMm/r^2 = mg
thus, g = GM/r^2

formula for U in g-field

U = mϕ = GMm/r

why does the moon orbit with a specific radius from the earth?

the G-force of the earth acting on the moon is sufficient to prove centripetal force

kepler's third law

T^2 is proportional to r^3. T is the orbital period, r is the orbital radius.
equate GMm/r^2 = mr(2pi/T)^2

angular velocity

rate of change of angular displacement with respect to time

gravitational field strength

gravitational force per unit mass acting on a small test mass placed at that point. g = GM/r^2

how to find escape velocity

By POCOE, Et at earth's surface = Et at infinity

Ep + Ek at earth = Ep + Ek at infinity

-GM(e)m/r^2 + 1/2mv^2 = 0 + 0

then u manipulate to find v. m = obj's mass. M = earth's mass

why are small planets devoid of an atmosphere

• usually ur to prove escape velocity

• when speed of gas particles > v(esc), particles will have sufficient KE to overcome the gravitational pull of the planet and escape to infinitys

escape velocity formula

(2GM/r)^1/2

if eqn is a differentiation of given values / graph

find the two closest coordinates to find most accurate gradient at that point. refer to gfield DQ 8b

when is there weighlessness?

when N = 0.
when is N = 0? when the surface supposed to provide contact has the same acceleration as the person. thus, the surface does not push against the person.

gravitational potential

The gravitational potential at a point in a field is defined as the work done per unit mass by an external agent in bringing a small test mass from infinity to that point.

why does gravitational potential have a negative value?

gravitational potential at infinity is taken to be zero.

since G-force is attractive, external force will be acting radially away from the source mass

F ext is opposite in direction to displacement of the mass

Thus WD by Ext mass to bring a unit mass from infinity to a point in the field will be -ve.

[WD = Fcos(180)x = -Fx]

how are g-force and g-potential related?

F = mg = m(-dϕ/dr). gravitational field strength = gradient of ϕ against radius graph.

variation of potential energy with distance due to earth and moon

its negative, key in 1/x graph in gc. for moon, mirror it. DQ12

how to find GPE when theres a spring compression involved. aka, lets say a ball falls down into a spring.

loss in GPE = h+x. x is the spring compression.