mechanics Flashcards
Hookes law equation
T = λx / l
assumptions for hookes law
- the string/ spring is light
- The system has no friction
- The string / spring obeys hookes law
hookes law on two springs:
- draw a diagram for lengths, draw a diagram for tensions
- find the forces + equate if theyre in equilibrium
- apply hookes law to the tensions in the strings
- sub these tensions into the equations made in step 2
- solve
work done by a variable force in a string/ spring
λx²/2l
in perfectly elastic strings:
energy is conserved:
Final KE, GPE, EPE = Initial KE, GPE, EPE
if a non conservative force does work on a string
work done = change in energy
Solving vertical motion involving elastic forces using energy
- calculate loss/ gain in GPE and KE
2. use conservation of energy to work out λ or x
Solving vertical motion involving elastic forces using calculus
- Use F = ma where a is a derivative
- solve differential equation
- input initial variables
- input question conditions
all quantities can be expressed in terms of:
Mass
Length
Time
tangential speed, v =
rω
radial acceleration is
acceleration towards the centre of the circle
radial acceleration, a =
rω² = v²/r
Force towards the centre of the circle, F =
mrω² = mv²/r
Examples of centripetal force
- Tension in a string
- friction
- A reaction force
In horizontal circles consider the forces:
- forces towards/ away from the centre of the circle
- forces perpendicular to the plane of motion
working out speed in vertical circles
conservation of energy
GPE = KE
tangential acceleration =
ra
a is the angular acceleration
Constant angular acceleration with variable speed:
Suvat equations where ω = v ωo = u angular acceleration, a = a t = t θ = s
If a particle just leaves its circular path then
R>0 T=0
solving a circular motion problem
- use forces (F=ma)
- energy analysis
- sub equations together and solve
impulse =
Change in momentum = force * time
Ft = m(v-u)
∫Fdt =
impulse
∫Fds =
work done
work done vector form
W = F.d
power vector form
P = F.v
Kinetic energy vector form
KE = ½mv.v
alternate form for acceleration
dv/dt
d²x/dt²
vdv/dx
a force may be dependent on:
- time F= f(t)
- displacement F = f(x)
- velocity F = f(v)
including the effects of air resistance
- F = ma Where F is the sum of forces and a is a derivative
- separate variables
- integrate and add arbitrary constant
- use initial conditions to find constant
- input conditions given in question
When no forces acting except friction then
KE lost = work done against friction
resultant acceleration in a circle =
√(radial acceleration²+tangential acceleration²)
work energy pricipal
work done by forces = change in KE/GPE
coefficient of restitution, e =
speed of separation / speed f approach
if e = 0
the two velocities after impact are the same (the particles stick together), inelastic
If e = 1
the loss in KE is zero, elastic
Component of velocity perpendicular to the line of centres
unaffected by collision, momentum is conserved
Component of velocity parallel to the line of centres
Newtons law of restitution, momentum is conserved
When an object hits a plane there is no impulse…
parallel to the plane
Component of velocity parallel to plane
unaffected by collision, momentum is conserved
Component of velocity perpendicular to plane
Newtons law of restitution, momentum is conserved
newtons law of restitution
(Ua - Ub ) e = Vb - Va
Impulse =
integral
∫Fdt
Work done =
integral
∫Fdx
Using the impulse/ work done integrals
find the integral and set in equal to mv - mu or Pt
component of velocity that changes wall vs two spheres
- wall: perpendicular component changes
- spheres: component parallel to line of centres changes
stiffness =
λ/l
x̄ =
(x1m1 + x2m2 + … + xnmn)/ (m1 + m2 + … + mn)
ȳ =
(y1m1 + y2m2 + … + ynmn)/ (m1 + m2 + … + mn)
finding CoM of compsite shaped laminas
in a table put each shapes area, CoM x co-ordinate, CoM y co-ordinate.
Consider x or y co-ordinates i.e. x*A … =x̄A
If a hanging object is in equlibrium then…
the forces weight and tension act along the same line and there is no resultant moment.
The CoM lies directly below the point of suspension.
volume of revolution about x-axis
V = ∫πy²dx
volume of revolution about y-axis
V = ∫πx²dy
Vx̄ =
∫πy²xdx
Vȳ =
∫πx²ydy
Finding the CoM of a solid of revolution about the x-axis:
V = ∫πy²dx Vx̄ = ∫πy²xdx
Finding the CoM of a solid of revolution about the y-axis:
V = ∫πx²dy Vȳ = ∫πx²ydy
Finding the angle between the point of suspension and the vertical axes
- CoM directly below point of suspension so find x̄ and ȳ
- tanθ = ȳ/x̄
CoM of a lamina using strips parallel to y-axis
Ax̄ = ∫xydx Aȳ = ∫½y²dx
CoM of a lamina using strips parallel to the x-axis
Ax̄ = ∫½x²dy Aȳ = ∫xydy
if an object is about to slide
F = μR
friction is limiting
when the particle is at rest does friction act
no, there is no tendancy to slide
an object will slide if
the line of action sits within the base
an object will topple if
the line of action lies outside of the base
volume of a circle
4/3πr^3
volume of cone
h/3πr^2
a in circular motion suvat
angular acceleration
