Dynamics of Rigid Bodies
A rigid body is one in which the distances between constituent particles is constant
throughout the motion of the body, i.e. it keeps its shape.
There are two kinds of rigid body motion:
1. Translational
Rectilinear forces acting.
Particles move on straightline paths.
2. Rotational
Rotational forces, i.e.
torques, acting.
Particles move on circular
paths
Kinematics of Rigid Bodies
dθ
dt
Let α be a constant.
s = Rθ
ω=
Then,
d 2θ
=α
dt 2
α=
Let ω = ω0 at t = 0, ω =
Integrating,
Let θ = 0 at t = 0,
dω d 2 θ
= 2
dt
dt
Integrating,
dθ
= αt + ω 0
dt
θ = ½ αt 2 + ω 0 t + c
θ = ½ αt 2 + ω0 t
Angle θ is in
radians
dθ
= αt + C
dt
COMPARE THESE
WITH THE
EQUATIONS FOR
LINEAR MOTION
Note: θ is in radians
Angular Momentum
Angular momentum is the rotational equivalent of linear momentum. It is a
conserved quantity
A rigid body can be thought
of as a large number of
y
r
particles i, at positions Ai,
ω
with masses mi, at position
vectors from an origin on the
axis of ri, at distances Ri from
Ai
the axis, and with velocities
vi.
x
z
The particle at the point Ai has linear momentum p = mivi and Angular Momentum
about the origin
L i = ri × p i
= mi ri × v i = mi ri vi nˆ
where nˆ is unit vector normal to both ri and pi.
Note that L is not generally parallel to the axis of rotation.
Moment of Inertia
The component of the angular momentum in the y-direction, i.e. along th axis of
rotation, is
Liy = mi ri vi sin θ i
However, ri sin θi= Ri, so
Li = mi Ri vi
The magnitude of ωi is the same for all points, so we drop the index i, and vi = Riω.
Then
Liy = mi Ri2 ω
Summing or integrating over all points,
L y = L1 y + L2 y + K = ∑ Liy = ω∑ mi Ri2
i
This sum, I = ∑
i
mi Ri2
i
is called the Moment of Inertia , and L = Iω (c.f. p = mv)
Moment of Inertia of Potato-Shapes
The Moment of Inertia depends on the axis of rotation.
The Angular Momentum is generally not parallel to the axis of rotation.
For a body of general shape (an asteroid, a potato . . .) there are three mutually
perpendicular axes for which the angular momentum is parallel to the axis. These are
called the Principal Axes of intertia and the moments of inertia about them are the
Principal Moments of Inertia.
For bodies of higher symmetry than potatoes, the Principal Axes are generally Axes
of Symmetry.
Angular Momentum Examples:
Two masses m are going
round the z-axis, at radius R,
in the x-y plane, with speed v
= Rω.
L = r × mv for each mass
r =R
v = Rω
L = 2mR 2 ω for the pair
Direction of L is at right angles to r and v, i.e. in same direction as ω, i.e.
L = 2mR 2 ω = Iω
and I = 2mR
L
Now incline the masses to the
axis.
L = r × mv for each mass
r =R
v = Rω sin ϕ
L = 2mR 2 ω sin ϕ for the pair
Direction of L is at right angles to r and v, i.e. at ϕ to ω, i.e. it precesses about z.
Lz = 2mR2sin2ϕ ω, and I = 2mR2sin2ϕ about the z-axis.
Calculation of Moments of Inertia
I = ∑ mi Ri2
i
Ri is distance to
axis of rotation
If an object is considered to consist of elemental particles of mass dm, then
dm = ρdV
and the sum becomes an integral over the volume. If the density ρ is constant,
it comes out of the integral and
I = ∫ R 2 dm = ρ ∫ R 2 dV
V
Note that
∫R
2
V
dV is a purely geometrical factor
V
Example: Moment of Inertia of a thin rod rotated about one end.
Cross-sectional area S, length L, density ρ.
Element dV is disc of area S, distasnce from axis x, thickness dx.
i.e. dV = Sdx, dm = ρSdx, dI = ρSx 2 dx
So we have
L
1
1
I = ρS ∫ x 2 dx = ρSL3 = ML2
x =0
3
3
(
)
Example: Moment of Inertia of the same thin rod rotated about its centre.
L/2 2
1  L3
L3  1
−−
= ML2
I = ρS ∫ x dx = ρS


x=− L / 2
3  8
8  12
In general, I = Mk2. The length k is a characteristic
length of the object, called the Radius of Gyration –
compare with Centre of Gravity.
Parallel Axis Theorem
Consider an object rotating around its Centre of Mass, with the z-axis as the
axis of rotation. We know the moment of inertia
I CM = ∑ mi Ri2 = ∑ mi xi2 + y i2 .
(
i
)
i
What would be the moment of inertia about the axis P, parallel to the z-axis
but offset at x = a, y = b?
[
I P = ∑ mi ( xi − a )2 + ( yi − b )2
i
[(
]
)
(
= ∑ mi xi2 + yi2 − 2axi − 2byi + a 2 + b 2
)]
i
Recall that the definition of the Centre of Mass gives xCM =
∑ mi xi
i
∑ mi
. Since
i
we have placed the Centre of mass at the origin, our term in xi vanishes.
Similarly for yi. So,
(
) (
I P = ∑ mi xi2 + yi2 + a 2 + b 2
i
)∑ m
i
i
= I CM + Md 2
Compare this result with the thin rod in the earlier example. Iend = ICM + Md2,
and d = ½L.
ML2 ML2 ML2 ML2
=
−
=
4
3
4
12
which is what we found by direct calculation.
I CM = I end −
Torque
To cause a body to rotate we need to apply forces which do not pass through
the axis of rotation. The effect of such a force depends on its magnitude and direction
and also on how far from the axis it is.
In vector terms, this is called the Torque. Its scalar magnistude is often called
the Moment. It is given by
r r r
τ =r×F
r
τ = RF
Torque is the rotational analogue of force in linear mechanics. As force is rate of
change of linear momentum, so torque is rate of change of angular momentum,
r
r
r dL d r
d r
d2 r
τ=
= Iω = I ω = I 2 θ = Iα
dt dt
dt
dt
r
r
r
r
τ = Iα is to be compared with F = ma
Example: The Yo-Yo
Vertical acceleration a
Angular acceleration α
τ = Iα
I = ½ MR2 *
T = ½ MRα
T
*
How would you
calculate this?
Mg
Note that a = αR and that Mg − T = Ma = MRα
So
Mg − ½ MRα = MRα
Mg − ½ Ma = Ma
g = 32 a
a = 23 g = 6.53 m s −1
Note that a < g and is independent of R and M.
Gyroscopes and Precession
z
y
Consider a gyroscope which
is not spinning. It is released
in the position shown. Its
r
centre of mass is at r .
r
Initially L is zero
x
r
F
r
r dL
τ=
Then
dt
r r
dL = τdt
The torque causes the gyroscope to start rotating about the pivot, gaining
angular velocity in the y-direction.
r
Now let the gyroscope be spinning initially. Its angular momentum L is not
zero but lies along the axis of rotation, along the x-direction. Now, after a
time dt,
r
r r
L → L + τdt
which is still in the x-y plane, which has the same magnitude, but has changed
direction by the amount
r
τ dt
dϕ = r
L
r
dϕ τ mgr
Ω=
= r =
dt
Iω
L
This gives us an expression for the well-known rotation in the horizontal
plane of a gyroscope, its PRECESSION.
For a simple disc, the moment of inertia I = ½ mR2 and Ω =
gr
½R 2 ω
Astronomical Example: The Earth is spinning and so acts as a gyroscope. Its
axis of rotation makes an angle of 23º27´ to the normal to the plane of its orbit.
There is a torque due to (tidal) gravitational effects of the Sun and Moon. As
a consequence, the direction of the Earth’s axis precesses (the Precession of
the Equinoxes) with a period of 27725 years.
Consequently, while Isambard Brunel built the Box Tunnel (between Bath and
Bristol on the Great Western Railway) so that the sun would shine through it
at sunrise on his birthday, that condition will not be met for very long.
Kinetic Energy of Rotation.
For a body considered as a large number of point masses,
E K = ½ ∑ mi vi2
i


= ½ ∑ mi Ri2 ω 2 = ½ ∑ mi Ri2 ω 2
i
 i

= ½ Iω 2
Compare E = ½mv2
In the general case of linear translation motion together with rotation about the
centre of mass,
E K = ½ mv 2 + ½ Iω 2
Example: A ring rolls* down an inclined plane, height h0, angle ϕ
* Definition of Rolling: Friction prevents slippage. No work is done
against friction as the bottom of the ring touching the ramp is
instantaneously at rest.
If the ring starts at height h0 at zero linear and angular velocity, and
rolls down to a height h, then at h,
½ E K = ½ mv 2 + ½ Iω 2 = mg (h0 − h)
Now, I for a ring of mass m, radius R is I = mR2, and v = Rω, so
2
v
E = ½ mv 2 + ½ mR 2   = mg (h0 − h)
R
v 2 = g (h0 − h)
Compare frictionless sliding:
E K = ½ mv 2 = mg (h0 − h)
v 2 = 2 g (h0 − h)
The initial Potential Energy is used to supply both the rotational and
translational kinetic energy, so the rolling motion slows down the
tranlational.
Comparison of Dynamical Quantities in Translation and Rotation.
Translational
Position
Velocity
Acceleration
Mass
Momentum
Kinetic Energy
Force
Power
Second Law
r
s
r
r ds
v=
dt
r
r d 2s
a= 2
dt
m
r
r
p = mv
p2
E = ½ mv 2 =
2m
r dpr
F=
dt
v r
P = F .v
r
r
F = ma
Rotational
Angle
Angular
velocity
Angular
acceleration
Moment of
Inertia
Angular
Momentum
.....
Torque
.....
.....
r
θ
r
dθ
ω=
dt
r
r d 2θ
α= 2
dt
I
r
v
L = Iω
L2
2rI
r r r dL
τ = F ×r =
dt
rr
P = τ.ω
r
r
α=Iτ
E = ½ Iω 2 =