Dynamics & Relativity 1 Flashcards
for uniform motion r(t)=
r(t0)+v(to)(t-t0)+1/2A(t-t0)^2
velocity
defined as rate of change of positions
v(t) = dr/dt = dxp/dt i + dyp/dt j + dzp/dt k
acceleration
defined as the rate of change of velocity
a(t)=dv/dt=d^2r/dt^2 (same idea in vector components as velocity but with second derivative)
simpler notation for time dependence
r0 = r(t0)
r = r(t)
hence r=ro+vo(t-t0)+1/2a(t-t0)^2
motion in 1 dimension
pick just a single direction eg x
r=xi, v=vxi, a=axi
vx=dx/dt, ax=d&2x/dt^2 = constant
derivation of vx=vx0+ax(t-t0) using integration
start with ax=dvx/dt and integration both sides wrt time
on lhs, a constant so get ax(t-t0)
rhs gives vx-vx0
put together a rearrange
derivation of x-x0=vx0(t-t0)+ax/2(t-t0)^2
use vx=dx/dt
integrate both sides wrt time
insert previous equation for velocity and rearrange
how to make x-x0=vx0(t-t0)+ax/2(t-t0)^2 more familiar?
setting initial position and time to zero gives
x=vx0t+1/2axt^2
derivation of vx^2=vx0^2+2ax(x-x0)
combining previous two equations and eliminating (t-t0)
derivation of x-x0=1/2(vx+vx0)(t-t0)
combining previous two equations and eliminating acceleration
assumptions for free falling bodies
- gravitational acceleration due to Earth’s gravity is constant
- ignore gravity from everything but Earth
- ignore rotation of Earth
- Pretend Earth is flat
- Ignore air resistance
free falling body setup
ay=-g
vy=vy0-g(t-t0)
y-y0=vy0(t-t0)-1/2g(t-t0)^2
free falling bodies - things to check
- units
- signs (heights +Ve or 0, object moving down so y-component of velocity is -ve
3.magnitudes (timescale a few secodns, distance few tens of metres etc)
motion in 2 dimensions
study the motion in each dimension seperately
acceleration in x direction
0
hence vx=vx0
separating velocity into components
vx0=v0costheta
vy0=v0sintheta
how to find y as a function of x
take equation for y and use equation for x to eliminate t
how far does an object travel (x direction)
set vertical position to zero and rearrange for Xr
how high does object travel?
equations of motion for vy to find time when vy=0
or find dy/dx and set to 0 to find x at highest point. Use this to find y at highest point
or use symmetry if highest point is half of journey.
maximum height only depends on
g and vy0
relative motion
r p/a = r b/a + r p/b (similar for relative velocity if find derivative)
“the velocity of the point 𝑃
in coordinate system 𝐴 is equal to the
velocity of 𝑃 in coordinate system 𝐵 plus
the velocity of the origin of system 𝐵 as
measured in 𝐴”
gradient of x-t plot
vx
polar coordinates
r, theta
r=(x^2+y^2)^1/2
theta=tan^-1(y/x)
w, angular spped
rate of change of theta
w=d theta/dt
(rad per second)
time taken to complete revolution
T=2 pi/w
r vector
<r cos theta, r sin theta>
derivative of r vector
<-r sin theta, r cos theta>
v is just r rotated by
90 degrees in the theta direction and multiplied by w
for uniform circular motion, the velocity vector of a point is…
perpendicular to the position vector relative to the axis of rotation
acceleration vector points…
in the opposite direction to the position vector relative to the axis of rotation
for uniform circular motion, a towards centre
normal force
exerted on the box by the surface on which it is resting
always perpendicular to surface
resultant force
sum of all forces acting on a body
newton’s first law
a body acted on by no net force moves with constant velocity
equilibrium
no net force
ie R=0
Inertia
tendency of a body to remain at rest or keep moving if in motion
inertial reference frame
if a reference frame accelerates, Newton’s first law does not hold.
inertial reference frame is one where Newton’s first law DOES hold
Newton’s second law
if a net force acts on a body of mass m, the body will accelerate according to
ΣF=ma=R
approach for forces questions
- add up all forces to find ΣF=R (free body diagram useful)
- use N2nd to find a
- use a in eqns of motion
if we plot the magnitude of frictional force applied to box against time
friction force matches pushing force, up to a maximum at which point the box begins to move and friction force reduces to a constant
static friction
acts on surfaces not moving relative to one another
has maximum magnitude related to magnitude of the normal force due to contact between surfaces
static friction equation
F < or = Fs
Fs=µsn
µs is the dimensionless coefficient of static friction
kinetic friction
value proportional to the magnitude of the normal force acting on the object due to contact between the surfaces
kinetic friction equation
Fk=µkn
µk is the dimensionless coefficient of kinetic friction
coefficients of friction
depend on nature of surfaces
typically between 0 and 1 though can be >1
cannot be negative
µs>µk
SHM
force acting on an object is proportional to its displacement but acts in the opposite direction
