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#1

Vectors in mechanics:
o Solution of simple vector differential equations:
It’s the same as in previous modules (M3 and M4), where you had to solve scalar differential equations, but here you have vectors: all what you have to do is substitute v = ui + wj, where u an w are scalars, so dv/dt = du/dt i + dw/dt j, and d2v/dt² = d2u/dt² i + d2w/dt² j. Now equate coefficients of i and j, solve a if it is a scalar equation, hence you can find u and w.

o Work done by a constant force:
Work done = F.d
All what you have to do is to find the scalar product of the Force vector and the Displacement vector. Of course to find the distance: position vector of final point - position vector of initial point.

o Vector moment of a force:
Vector moment of a force = r × F
Where r is the position vector of any point on the line of action of F relative to the point where the moment is to be taken about.

o Resultant Force and Couples:
Resultant force = i = 1Σn Fi
Couples of moment G = i = 1Σn ri × Fi

If a system of forces can be reduced to a resultant force, then G = 0
If a system of forces can be reduced into a couple of moment, then FR = 0

A system s I equilibrium when both G and FR are equal to zero.
It doesn’t matter about what point the resultant force is about. It’s the same about any point.

Moments of inertia:
o Calculating moments of inertia:
M.I.of a point = mr²
M.I.total = iΣ mii
To find moments of inertia:

Take a small part of length δx which is at distance x from axis, the mass of that small part is: M/[length or area of the body] . [length or area of the small part].
Substitute the mass of the small part in the equation of M.I., r will be x,
Σ mii will contain δx,
So you let δx --> 0 then :
M.I. = lim δx --> 0 Σ mii , so you integrate to get M.I.
eg.:
Show that the moments of inertia of a uniform rod f mass M and length 2a about an axis through its centre perp. to its length to be 1/3 Ma².

Take a small part of length δx, which is at distance x from the axis.
Mass of rod per length = M/2a
So mass of the small piece which is of length δx = M/2a . δx
So M.I. small piece = mr² = M/2a . δx . x²
So total M.I. = lim δx --> 0 Σx=-ax=a M/2a . δx . x²
Which means : M.I. = ∫-aa M/2a . x² δx
= M/2a ∫-aa x² δx
= M/2a [x³/3]-aa
= M/2a [ a³/3 – (-a)³/3 ] = M/2a (2a³/3) = 1/3 M a².

o Standard moments of inertia that are given in the formula sheet:
For uniform bodies of mass m:
Thin rod, length 2l, about perpendicular axis through centre: 1/3 ml²
Rectangular lamina about axis in plane bisecting edges of length 2l: 1/3 ml²
Thin rod, length 2l, about perpendicular axis through end: 4/3 ml²
Rectangular lamina about edge perp. to edges of length 2l: 4/3 ml²
Rectangular lamina, sides 2a and 2b, about perpendicular axis through centre: 1/3 m(a²+b²)
Hoop or cylindrical shell of radius r about axis through centre: mr²
Disc or solid cylinder of radius r about axis through centre: ½ mr²

If two bodies have moments of inertia I1 and I2 about the same axis, then the moments of inertia of the composite body about that axis is I1 + I2.

eg. A uniform rod of mass 2m and length a, has a particle of mass m at distance 2a/3 from one of its ends. Find M.I. of the system.
Irod = 1/3 (2m) (a/2)² = 1/6 . ma²
Iparticle = m (1/3 a)² = 1/9 . ma²
Isystem = ma² ( 1/6 + 1/9) = 5/18 ma²

o Stretching rule:
If one body can be obtained from another body by stretching parallel to the axis without altering the distribution of mass relative to the axis, then the moments of inertia of the two bodies about the axis is the same.

For example: a solid cylinder is a disc stretched parallel to the axis through its centre perp. to it.
A uniform rectangular lamina is a rod stretched parallel to the axis through its centre perp. to it.

I = mk²
Or k = √[I/m]
Where k is the radius of gyration.

o Parallel axis theorem:
If a body of mass m has moments of inertia I about the axis through the cantre of mass, then the moments of inertia of that body about an axis PARALLEL to and is at distance d from the first axis is ICM+md².

o Perpendicular axes theorem:
If a lamina lies on the plane xy, where Ox and Oy are perp., and has moments of inertia Ix and Iy about Ox and Oy respectively, and Oz is an axis perp to Ox, Oy and the lamina, then moments of inertia about Oz is Iz, where Iz = Ix + I y.

Some more notes will be posted soon, I hope that those help, and I hope that I don’t have mistakes cheers. PLEASE add notes if u have some.
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#2
here are some more:
Rotation of a rigid body:

• K.E. of a body = ½ Iω²
P.E. of a body = M . g . hof centre of mass

You should know that conservation of energy should be used with such things.

• Moment of the resultant force about an axis L = Iabout the axis . θ” (proof to that is in Heinemann M5 book, page 85, if you don’t have the book, and still need the proof, please contact me). θ” is the angular acceleration.

• The components of acceleration of centre of mass (G) which is rotating about axis through O are:
r θ’² along GO
r θ” perpendicular to OG

• To find the force that the body is doing on the axis: put force X at the axis in the direction GO, and force Y at O perpendicular to OG. Then resolve any other forces on the body in the directions OY and OX, then use:
radial forceresultant = m r θ’²
transverse forceresultant = m r θ”
hence find X and Y, and then find their resultant force (if required to find the magnitude of the force acting on the axis).

• Angular momentum:
Angular momentum = moment of momentum = I θ’

(How to derive that is in the same book page 94, also contact me if needed.)

As angular momentum = moment of momentum, then angular momentum for a body moving in a straight line is: its momentum × the distance from the axis.

• Conservation of angular momentum:
Same as conservation of momentum, angular momentumbefore = angular momentumafter

Example: a rod ( mass m, length 4a ) is free to rotate about a vertical axis through its centre, it rests on a smooth horizontal table, a particle of mass 3m moving in a straight line perpendicular to the rod with speed u ms-1 hits the rod at distance a from one of its ends. The particle sticks to the rod. Find the angular speed of the body after collision.

Initial angular momentumrod = 0
Initial angular momentumparticle = (3m . u ) . a

Final angular momentumrod = Irodω
Final angular momentumparticle = Iparticleω

Irod = 1/3 . m . (2a)² = 4/3 ma²
Iparticle = m a²

Initial Angular momentumtotal = Final angular momentumtotal
3mua = 4/3 ma² ω + ma² ω
3u = 7/3 a ω
ω = 9u / 7a

• Effect of an Impulse on a rigid body that is free to rotate about an axis:

You know that:
L = I θ”
Integrate with respect to t
t1t2∫L dt = ω1ω2∫I θ” dt

t1t2∫L dt = [I θ’]ω1ω2

t1t2∫L dt = I Δ ω
usually L = F . r so:
t1t2∫F . r dt = I Δ ω
F . r (Δt) = I Δ ω
F.(Δt) = I Δ ω / r
Impulse = I Δ ω /r

• Pendulums:
Simple pendulum:
Period = 2π √(l/g)

Compound pendulum:
Period = 2π √[I/(mgh)] where h is the distance between axis and C.M. and I is moments of inertia about the axis.
Or it might be written in another form:
Period = 2π √[(k²+h²) / (gh) ] where k is the radius of gyration.

To find the simple pendulum with the same period of a certain compound pendulum: use T = T,
i.e. 2π √(l/g) = 2π √[I/(mgh)] which is simplified to l = I/mh
Or
2π √(l/g) = 2π √[(k²+h²) / (gh) ] which is simplified to l = [k² + h²] / h.
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15 years ago
#3
u lot are ****ing next level , cant even do M4 fully , and u lot are TALKING ABOUT M5,
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#4
well it's obvious that this thread is for those taking M5, so what's your problem?!?
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15 years ago
#5
Motion with variable mass

You only need to know the impulse-momentum principle:
Impulse = Change in momentum = Force * Time.

From this you can set up a differential equation that describes the motion of the system.

e.g. A particle is moving upward against gravity and is losing mass at a rate k mass/sec. The lost mass is ejected vertically downwards with speed u relative to the body.

To start off, the particle is moving against gravity. So the force acting on it is its weight. Thus:
I = Ft = -(m+δm)g δt, where δt is the time interval through which this impulse occurs and δm is the increase in mass (see below).

Now we want to find the change in momentum.
(i) particle
(m+δm)(v+δv) - mv, supposing that the body gains δm of mass and δv of speed (the fact that it loses mass will be accounted for later)
(ii) ejected mass
-δm(v+δv-u) - 0 = -δm(v-u), we always ignore δmδv since it's small.

So the change in momentum of the whole system is:
(m+δm)(v+δv) - mv - δm(v+δv-u)

Hence by the impulse-momentum principle:
(m+δm)(v+δv) - mv - δm(v-u) = -(m+δm)g δt

Some rearrangement (remember, we ignore δmδv):
mδv + uδm = -(m+δm)g δt

Divide by δt and take limits:
m(dv/dt) + u(dm/dt) = -mg

This is the differential equation that describes this motion. Now we take into account one last piece of information:
The body loses mass at constant rate, i.e. dm/dt = -k.

This transforms the differential equation into:
m(dv/dt) -ku = -mg, which you can now solve easily.

Some things you should know when solving questions related to this chapter:
- Limiting speed is achieved when (dv/dt)=0
- When dealing with rocket problems, sometimes the time of burnout helps. e.g. suppose that a rocket has variable mass m and that the mass of the fuel is Mf and of the rocket is Mr. The time of burnout T is when m=Mr. This fact is simple (and obvious), but easily over-looked.
- When dealing with rain-drop type problems, use mass=density*volume, i.e. m=pv. This is helpful because sometimes you're told that the rain-drop is spherical, and this helps you get enough information to find an expression for dm/dt. That's because you know that p is constant, v=(4/3)pi.r³, and you're usually given that dr/dt is constant. In other words: dm/dt = 4*p*pi*r²*(dr/dt).
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#6
cheers dvs. thank you.
0
#7
to prove that an oscillating body is doing SHM with small oscillation angle:
s = lθ
s' = lθ'
s'' = lθ''

now F = ma = ms''
and F = -mgsinθ
mgsinθ = -ms''
gsinθ = -lθ''

as θ is small and in radians==> sinθ = θ
then
θ'' = -g/l θ
which is if compared to SHM equation : x'' = -ω² x
then ω = rt[g/l]

where l is the distance of C.M. from the Axis. that can be used for simple and compound pendulum.
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#8
is there anyone doing M5 this Friday? are you prepared?
0
15 years ago
#9
I'm doing it on Friday. My teacher gave me a pack of questions to do from past papers since like 1990 so I'm gonna plough through them. Thanks for those btw.
0
13 years ago
#10
ok these are some pretty good notes but for some reason
i cant get around the
• The components of acceleration of centre of mass (G) which is rotating about axis through O are:
r θ’² along GO
r θ” perpendicular to OG
i just dont get it... anyone? exam tomorrow ...
0
6 years ago
#11
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