Here is a revision thread for M5, please feel free to add any notes you have:
• 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
, 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
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
If a system of forces can be reduced into a couple of moment, then FR
A system s I equilibrium when both G
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
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
will contain δx
So you let δx
--> 0 then :
M.I. = lim δx --> 0
, so you integrate to get M.I.
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
So total M.I. = lim δx --> 0
M/2a . δx
Which means : M.I. = ∫-aa
M/2a . x
= M/2a ∫-aa x
= 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
Hoop of radius r about a diameter: ½ mr²
Disc or solid cylinder of radius r about axis through centre: ½ mr²
Disc of radius r about a diameter: ¼ mr²
Solid sphere, radius r, about diameter: 2/5 mr²
Spherical shell of radius r about a diameter: 2/3 mr²
o Additive rule:
If two bodies have moments of inertia I1
about the same axis, then the moments of inertia of the composite body about that axis is I1
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.
= 1/3 (2m) (a/2)² = 1/6 . ma²
= m (1/3 a)² = 1/9 . ma²
= 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.
o Radius of Gyration:
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
o Perpendicular axes theorem:
If a lamina lies on the plane xy, where Ox and Oy are perp., and has moments of inertia Ix
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
+ 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.