Further Mechanics

(Dimensional A) 1 means..

No dimensions

Dimensional [torque]=

ML²T^(-2)

Dimensional [Number of things in a period of time]/frequency

T^-1

Dimesnional [period]=

T

Order of diff and int

SVA

Diff down int up

Acceleration=

v dv/dx

dv/dt

Integrating e^(x²,3,4..) method is

Set u=x²,3,3…

Work done for driving force is

Force x time

Work done can equal

Ke-Ke

Force x distance

The integral of a variable force

Arc length can equal

Pi d (x/360)

r x theta

Power=

Force x velocity

Momentum=

Mass x velocity

Impulse can equal

I=m(v-u) (Ns)

Force x time if force is constant

(Collisions) Velocity=

E u

Dimensional [e]=

1

NELR

VB-VA= -e(UB-UA)

CONS

MA(UA) +MB (UB)=MA (VA) + MB (VB)

Impulse of A on B

I=m(v-u) with Bs velocities

Collisions, if A changes direction..

VA<0 will give the values of e

I (A|B)=

A² +B² = I²

The red angle is

The original direction of motion

Find impulse and the angle by

Using the cosine rule

collisions, when a sphere collides with a smooth surface,

The parallel velocity is unchanged

Find impulse for the collision of sphere and surface

Use only perp components

Two smooth spheres collide..

Perp velocity unchanged

Impulsive tension in strings, When string is taut

CONS vB=vA=v

To find impulse:

IB=m(v-u)

IA=-IB

Completing orbits T=

2pi/w

Min value of u is

F<=uR

Hole in table, vmin

T-F=ma

Hole in table, vmax

T+F=ma

Banked tracks method

R⬆

R⬅=ma

Full circle motion conditions

Theta=180, T>0

Modulus of a |a|=

sqrt aT² + Ar²

When string horizontal, AT=

-g

r=ro+ut+1/2at² in vector format:

(ri|rj)=(sinx|cosx)+(vcosx|-vsinx)+1/2(0|-9.8)t²

Where does it land (ball falls off sphere)

• Ei=Ef from start to leaving point

• EF=ma from leaving point, R=0

• r=ro+ut+1/2at²

• rj=0

• t=

• ri=

Ball falls off circle, max height when

V²j=U²j+2ajsj

VECTORS work done=

F. X

½ m (v²-u²)

VECTORS V² can equal..

V² =vj² +vi²

V² = u² + 2 (aisi + ajsj)

VECTORS resistive wd is

Negative

VECTORS Propulsive wd is

Positive

VECTORS EF=

m (aj|ai)

VECTORS (1|5) x (10|25)=

(10|125)= 10+125=135