7Rigid bodies
IA Dynamics and Relativity
7.4 Motion of a rigid body
The general motion of a rigid body can be described as a translation of its centre
of mass, following a trajectory
R
(
t
), together with a rotation about an axis
through the center of mass. As before, we write
r
i
= R + r
c
i
.
Then
˙
r
i
=
˙
R +
˙
r
c
i
.
Using this, we can break down the velocity and kinetic energy into translational
and rotational parts.
If the body rotates with angular velocity ω about the center of mass, then
˙
r
c
i
= ω × r
c
i
.
Since r
c
i
= r
i
− R, we have
˙
r
i
=
˙
R + ω × r
c
i
=
˙
R + ω × (r
i
− R).
On the other hand, the kinetic energy, as calculated in previous lectures, is
T =
1
2
M|
˙
R|
2
+
1
2
X
i
m
i
|
˙
r
c
i
|
2
=
1
2
M|
˙
R|
2
| {z }
translational KE
+
1
2
I
c
ω
2
| {z }
rotational KE
.
Sometimes we do not want to use the center of mass as the center. For example,
if an item is held at the edge and spun around, we’d like to study the motion
about the point at which the item is held, and not the center of mass.
So consider any point
Q
, with position vector
Q
(
t
) that is not the center of
mass but moves with the rigid body, i.e.
˙
Q =
˙
R + ω × (Q − R).
Usually this is a point inside the object itself, but we do not assume that in our
calculation.
Then we can write
˙
r
i
=
˙
R + ω × (r
i
− R)
=
˙
Q − ω × (Q − R) + ω × (r
i
− R)
=
˙
Q + ω × (r
i
− Q).
Therefore the motion can be considered as a translation of
Q
(with different
velocity than the center of mass), together with rotation about
Q
(with the same
angular velocity ω).
Equations of motion
As shown previously, the linear and angular momenta evolve according to
˙
P = F (total external force)
˙
L = G (total external torque)
These two equations determine the translational and rotational motion of a rigid
body.
L
and
G
depend on the choice of origin, which could be any point that is
fixed in an inertial frame. More surprisingly, it can also be applied to the center
of mass, even if this is accelerated: If
m
i
¨
r
i
= F
i
,
then
m
i
¨
r
c
i
= F
i
+ m
i
¨
R.
So there is a fictitious force
m
i
¨
R
in the center-of-mass frame. But the total
torque of the fictitious forces about the center of mass is
X
i
r
c
i
×
−m
i
¨
R
= −
X
m
i
r
c
i
×
¨
R = 0 ×
˙
R = 0.
So we can still apply the above two equations.
In summary, the laws of motion apply in any inertial frame, or the center of
mass (possibly non-inertial) frame.
Motion in a uniform gravitational field
In a uniform gravitational field
g
, the total gravitational force and torque are
the same as those that would act on a single particle of mass
M
located at the
center of mass (which is also the center of gravity):
F =
X
i
F
ext
i
=
X
i
m
i
g = Mg,
and
G =
X
i
G
ext
i
=
X
i
r
i
× (m
i
g) =
X
m
i
r
i
× g = MR × g.
In particular, the gravitational torque about the center of mass vanishes:
G
c
=
0
.
We obtain a similar result for gravitational potential energy.
The gravitational potential in a uniform g is
Φ
g
= −r · g.
(since g = −∇Φ
g
by definition)
So
V
ext
=
X
i
V
ext
i
=
X
i
m
i
(−r
i
· g)
= M(−R · g).
Example
(Thrown stick)
.
Suppose we throw a symmetrical stick. So the center
of mass is the actual center. Then the center of the stick moves in a parabola.
Meanwhile, the stick rotates with constant angular velocity about its center due
to the absence of torque.
Example. Swinging bar.
`
Mg
This is an example of a compound pendulum.
Consider the bar to be rotating about the pivot (and not translating). Its
angular momentum is
L
=
I
˙
θ
with
I
=
1
3
M`
2
. The gravitational torque about
the pivot is
G = −M g
`
2
sin θ.
The equation of motion is
˙
L = G.
So
I
¨
θ = −Mg
`
2
sin θ,
or
¨
θ = −
3g
2`
sin θ.
which is exactly equivalent to a simple pendulum of length 2
`/
3, with angular
frequency
q
3g
2`
.
This can also be obtained from an energy argument:
E = T + V =
1
2
I
˙
θ
2
− Mg
`
2
cos θ.
We differentiate to obtain
dE
dt
=
˙
θ(I
¨
θ + Mg
`
2
sin θ) = 0.
So
I
¨
θ = −Mg
`
2
sin θ.
Sliding versus rolling
Consider a cylinder or sphere of radius
a
, moving along a stationary horizontal
surface.
a
v
ω
In general, the motion consists of a translation of the center of mass (with
velocity v) plus a rotation about the center of mass (with angular velocity ω).
The horizontal velocity at the point of contact is
v
slip
= v − aω.
For a pure sliding motion,
v 6
= 0 and
ω
= 0, in which case
v −aω 6
= 0: the point
of contact moves relative to the surface and kinetic friction may occur.
For a pure rolling motion,
v 6
= 0 and
ω 6
= 0 such that
v − aω
= 0: the point
of contact is stationary. This is the no-slip condition.
The rolling body can alternatively be considered to be rotating instanta-
neously about the point of contact (with angular velocity
ω
) and not translating.
Example (Rolling down hill).
a
v
ω
α
Consider a cylinder or sphere of mass
M
and radius
a
rolling down a rough plane
inclined at angle α. The no-slip (rolling) condition is
v − aω = 0.
The kinetic energy is
T =
1
2
Mv
2
+
1
2
Iω
2
=
1
2
M +
I
a
2
v
2
.
The total energy is
E =
1
2
M +
I
a
2
˙x
2
− Mgx sin α,
where
x
is the distance down slope. While there is a frictional force, the
instantaneous velocity is 0, and no work is done. So energy is conserved, and we
have
dE
dt
= ˙x
M +
I
a
2
¨x − Mg sin α
= 0.
So
M +
I
a
2
¨x = M g sin α.
For example, if we have a uniform solid cylinder,
I =
1
2
Ma
2
(as for a disc)
and so
¨x =
2
3
g sin α.
For a thin cylindrical shell,
I = M a
2
.
So
¨x =
1
2
g sin α.
Alternatively, we may do it in terms of forces and torques,
v
α
N
F
Mg
The equations of motion are
M ˙v = Mg sin α −F
and
I ˙ω = aF.
While rolling,
˙v − a ˙ω = 0.
So
M ˙v = Mg sin α −
I
a
2
˙v,
leading to the same result.
Note that even though there is a frictional force, it does no work, since
v
slip
= 0. So energy is still conserved.
Example (Snooker ball).
a
N
Mg
F
ω
It is struck centrally so as to initiate translation, but not rotation. Sliding occurs
initially. Intuitively, we think it will start to roll, and we’ll see that’s the case.
The constant frictional force is
F = µ
k
N = µ
k
Mg,
which applies while v − aω > 0.
The moment of inertia about the center of mass is
I =
2
5
Ma
2
.
The equations of motion are
M ˙v = −F
I ˙ω = aF
Initially, v = v
0
and ω = 0. Then the solution is
v = v
0
− µ
k
gt
ω =
5
2
µ
k
g
a
t
as long as v − aω > 0. The slip velocity is
v
slip
= v − aω = v
0
−
7
2
µ
k
gt = v
0
1 −
t
t
roll
,
where
t
roll
=
2v
0
7µ
k
g
.
This is valid up till
t
=
t
roll
. Then the slip velocity is 0, rolling begins and
friction ceases.
At this point,
v
=
aω
=
5
7
v
0
. The energy is then
5
14
Mv
2
0
<
1
2
Mv
2
0
. So energy
is lost to friction.