EDEXCEl Mechanics 2 revision notes

Chapter 1 - Kinematics of a particle moving in a straight or plane

1.1 - Projectiles
A particle thrown into the air moves freely under gravity. Such a particle is generally called a projectile. It moves through the air along a curved path.
- The projectile has an acceleration is g ≈ 9.8m/s² vertically downwards.
- Problems involving projectiles can be solved by considering the vertical and horizontal components of the motion separately.
- As the acceleration is entirely vertical, the projectile has no accelertion horizontally and so its horizontal velocity is unchanged throughout the motion.
- Use the four uniform acceleration equations to find values for s,u,v,a,t.

Horizontal motion
Initial speed = Uh
Final speed = Vh = Uh
Acceleration = 0
Time = T
Displacement = Sh

s = ut + ½at²
Sh = UhT + 0
Sh = UhT

v = u + at
Vh = Uh + 0
Vh = Uh

Vertical motion
Initial speed = Uv
Final speed = Vv
Acceleration = -g
Time = T
Displacement = Sv

s = ut + ½at²
Sv = UvT + ½(-g)T²
Sv = UvT - ½gT²

v = u + at
Vv = Uv + (-g)T
Vv = Uv - gT

Horizontal projection
If an object is thrown horizontally from a height h, the object will fall downwards but travel horizontally as well as vertically while falling.
The object begins its journey at the highest point of the projectile path. At this highest point the stone's vertical velocity is zero but it has a vertical acceleration, g.

Projectiles using vector notation
Information about a projectile can be conveyed in vector form with i and j being the unit vectors in the horizontal and upward vertical directions respectively. Separate the horizontal and vertical information and solve as before.

E.g. if the object is thrown with initial speed (4i + 5j)m/s from a height (2i - 5j)m

Horizontal motion
s = s
u = 4i
a = 0
t = T
v = u = 4i

Vertical motion
s = s
u = 5j
a = -g
t = T
v = v

Calculate distance travelled in vector form and add it onto the initial position vector to find the final position vector.

Projection at an angle to the horizontal
If a particle is projected with initial speed V at an angle θ to the horizontal then for initial velocity;
horizontal component = Vcosθ
vertical component = Vsinθ

1.2 - Velocity and acceleration when displacement is a function of time
When x is a function of t, that is;
x = f(t)

then;
v = dx/dt
a = dv/dt
a = d²x/dt²

Determining the velocity and displacement when the acceleration or velocity is given
a = dv/dt = f(t)
v = ∫f(t) dt + c

Similarly if v = g(t)
x = ∫g(t) dt + k

1.3 - Differentiating and integrating vectors

Differentiating vectors
.
r = dr/dt
..
r = d²r/dt²

(The number of dots denotes the number of differentiations to be carried out.)

d(r1 + r2)/dt = dr1/dt + dr2/dt

If r = xi + yj
then, v = x dot i + y dot j
and, a = x double dot i + y double dot j

Integrating vectors
If a = f(t)i + g(t)j
then; v = ∫a dt
= i∫f(t) dt + j∫g(t) dt + C1i + C2j

- When you integrate a vector you must add an arbitrary constant vector C1i + C2j, where C1i and C2j are constants.

Similarly, if: v = F(t)i + G(t)j
then: r = i∫F(t) dt + j∫G(t) dt + K1i + K2j
where K1i and K2j are constants.

Chapter 2 - Centres of mass

2.1 - Centre of mass of a discrete mass distribution
The weight of the system is the resultant of the weights of the particles and this resultant force acts through a point called the centre of mass of the system.

Centre of mass of a system of particles distributed in one dimension
Let the centre of mass of the system be a distance x-bar from O.

The total mass M of a system is;
M = m1 + m2 + m3 + .... + mn
Or M = Emi

Taking moments about O
Mx-bar = m1x1 + m2x2 + m3x3 +....+mnxn
Mx-bar = Emixi

Since, M = Emi

x-bar = Emixi/Emi

Centre of mass of a system of particles distributed in two dimensions
x-bar = Emixi/Emi
y-bar = x-bar = Emiyi/Emi

2.2 - Centre of mass of a uniform plane lamina
The symmetries of a plane lamina can be used to determine its centre of mass.

Standard results for centres of mass
Uniform rod - mid-point of rod
Uniform rectangular lamina - point of intersection of lines joining mid-points of opposite sides
Uniform circular disc - centre of circle
Uniform triangular lamina - points of intersection of medians - in other words 2/3 distance from any vertex to the mid-point of the opposite side
Circular arc, radius r, angle at centre 2@ - [rsin@]/@ from centre
Sector of circle, radius r, angle at centre 2@ - [2rsin@]/3@ from centre

2.3 - Equilibrium of a plane lamina
Suspension of the lamina from a fixed point
- For a suspended lamina to be in equilibrium its centre of mass, G, must be vertically below the point of suspension.

Equilibrium of a lamina on an inclined plane- For a lamina on an inclined plane to be in equilbrium the line of action of the weight of the lamina must fall within the side of the lamina which is in contact with the plane.