Rotation
Rotation
Rotation
1. Angular kinematics
1.1 Angular velocity and angular acceleration
1.2 Constant-angular-acceleration formulae
1.3 Displacement, velocity and acceleration in circular motion
2. Angular dynamics
2.1 Torque
2.2 Angular momentum
2.3 The angular-momentum principle for motion in a circle
2.4 The angular-momentum principle for arbitrary motion
3. Rigid-body rotation
3.1 Moment of inertia
3.2 Second moments and the radius of gyration
3.3 The equation of rotational motion
3.4 Comparing translation and rotation
3.5 Examples
If s is length of arc and r is radius then the angle θ in radians is defined such that
s rθ (1)
It is common to use a dot to indicate differentiation w.r.t. time; e.g. θ means dθ/dt.
Linear Angular
Displacement s θ
ds dθ
Velocity v ω
dt dt
dv d 2 s dv dω d θ 2
dω
Acceleration a 2 v α 2 ω
dt dt ds dt dt dθ
s 12 (u v)t θ 2 (ω0 ω)t
1
Constant-acceleration v u at ω ω 0 αt
formulae s ut 12 at 2 θ ω 0 t 12 αt 2
v 2 u 2 2as ω 2 ω 02 2αθ
Example 2.
A turbine starts from rest and has a constant angular acceleration of 0.1 rad s–2. How many
revolutions does it make in reaching a rotation rate of 50 rpm?
Consider a particle moving at a fixed radius r. The following have already been derived in
Topic A (Kinematics), as a special case of motion in general polar coordinates.
Velocity r
dv
dt
Acceleration
v2
Because it is not moving in a straight line, the particle has two components
r
of acceleration: r
The latter is called the centripetal acceleration. A centripetal force is necessary to maintain
this and keep the particle moving in a circular path. This force can be provided in many ways:
for example, the tension in a cable, a normal reaction from an outer boundary or friction.
Example 4.
Find the minimum coefficient of friction necessary to prevent slipping for a particle which is
placed:
(a) 100 mm from the rotation axis on a turntable rotating at 78 rpm;
(b) on the inside of a cylindrical drum, radius 0.3 m, rotating about a vertical axis at
200 rpm.
O 15 o
3 kg
20 m
(a) Find an expression for the velocity v of the object when its position vector makes an
angle θ with the vertical through the centre of the dome (see the figure).
(b) While it is in contact with the roof the object is undergoing motion in a circular arc.
Write down an expression for its centripetal acceleration as a function of angle θ.
(c) Find an expression for the normal contact force as a function of angle θ and the mass
m of the object.
(d) Hence determine the angle θ at which the object leaves the roof, as well as its height
and speed at this point.
(e) Find the distance from the outside wall of the dome at which the object hits the
ground.
Torque measures the turning effect of a force. When the force is not perpendicular to the
radius vector then only the component perpendicular to the radius vector contributes torque.
(For non-circular motion, v is the transverse component of velocity – see Section 2.4.)
Force-momentum principle:
d
F (mv)
dt
If F is the tangential component of force and r is constant (i.e. circular motion) then
d
Fr (mvr)
dt
torque = rate of change of angular momentum (8)
In fact, (8) holds for non-circular motion, but the proof requires more effort; see Section 2.4.
Equation (8) is the rotational analogue of the momentum principle for translational motion:
force = rate of change of momentum
For single particles the angular-momentum equation offers no advantage over the momentum
equation. However, it is invaluable in the treatment of systems of particles and, in particular,
rigid-body rotation, to which it may be applied by summing over all masses in the system.
The torque is then the sum of the moments of the external forces only, since the internal
forces between particles are equal and opposite and cancel in pairs.
1
Whilst one can have a moment of any physical quantity, torque is used almost exclusively for moment of force.
In the absence of an external torque, a direct corollary of the angular-momentum principle is:
Using a vector cross product (denoted by ), both angular momentum and torque may be
represented by vectors oriented along the rotation axis (in the sense of a right-hand screw):
angular momentum: h r mv sin α or h r mv (9a)
torque: T r F sin α or T r F (9b)
Differentiating the vector expression for angular momentum, using the product rule:
dh dr d
mv r (mv)
dt dt dt
v mv r F
0 r F
Hence,
dh
T (10)
dt
which is, in vector form, the angular-momentum principle:
rate of change of angular momentum = torque.
By summing over all particles this can be applied to the whole system, with T the torque due
to external forces only.
Example.
A bicycle wheel and a flat disk have the same mass, the
same radius and are spinning at the same rate. Which has
the greater angular momentum and kinetic energy?
For rotating rigid bodies, different particles lie at different radii and hence have different
speeds. Particles at greater radius move faster and contribute more to the body’s angular
momentum and kinetic energy. Thus, the angular momentum and kinetic energy depend on
the distribution of mass relative to the axis of rotation.
The total angular momentum and kinetic energy may be obtained by summing over
individual particles of mass m at radius r. Most importantly, although particles at different
radii have different speeds v, they all have the same angular velocity ω.
Angular Momentum
H mvr
m(rω)r r
r
( mr 2 ) ω m
Kinetic Energy
K 12 mv 2
12 m(rω) 2
12 ( mr 2 ) ω 2
The quantity
I mr 2 (11)
is the moment of inertia (or second moment of mass) of the body about the specified axis.
“Moment” describes the distribution of mass relative to the selected axis. (It gives higher
weighting to masses at greater radii.) “Inertia” refers to a resistance to a change in motion
(acceleration). In this sense, the moment of inertia fulfils the same role for rotation as the
mass of a body in translation.
For an extended body the “distance” varies, so we must sum over constituent parts; e.g.
I mr 2 second moment of mass (14)
(In your Hydraulics and Structures courses you will come across an exactly analogous
quantity – the second moment of area – in connection with hydrostatic forces on dams and
resistance to bending of beams, respectively.)
The centre of mass is that point at which the same concentrated mass would have the same
first moment:
M x mx
The radius of gyration, k, is that radius at which the same mass would have the same second
moment:
M k 2 I mr 2 (15)
R k
Circular disc of mass M and radius R (axis through centre, perpendicular to plane)
R
Moment of inertia I 12 MR 2 radius of gyration k
2
The radius of gyration is less than the geometric radius because mass is distributed over a
range of distances from the axis.
For rigid-body rotation the equation of motion is the angular momentum equation:
d
T ( Iω)
dt (16)
torque rate of change of angular momentum
For solid bodies, since moment of inertia and mass are usually constant we usually write
these in terms of acceleration:
dω dv
T I (rotation), FM (translation) (17)
dt dt
Alternatively we can integrate (16) with respect to angle to obtain an energy equation. First
rewrite it as
d d( Iω) dθ d( Iω) d 1 2
T ( Iω) ω ( Iω )
dt dθ dt dθ dθ 2
A (19)
work done (torque angle ) change in kinetic energy
Translation Rotation
Displacement x θ
Velocity v ω
Acceleration a α
Inertia m I
Effective location of mass centre of mass radius of gyration
Cause of motion force torque
Translation Rotation
Momentum mv Iω
Kinetic energy 1
2 mv 2 1
2 Iω 2
Power Fv Tω
A A
Example 8.
A bucket of mass M is fastened to one end of a light inextensible rope. The rope is coiled
round a windlass in the form of a circular cylinder (radius r, moment of inertia I) which is left
g
free to rotate about its axis. Prove that the bucket descends with acceleration .
I
1
Mr 2
Mg
Example 9.
A flywheel whose axial moment of inertia is 1000 kg m2 rotates with an angular velocity of
300 rpm. Find the angular impulse which would be required to bring the flywheel to rest.
Hence, find the frictional torque at the bearings if the flywheel comes to rest in 10 min under
friction alone.
Example 10.
A flywheel of radius 500 mm is attached to a shaft of radius 100 mm, the combined assembly
having a moment of inertia of 500 kg m2. Long cables are wrapped around flywheel and shaft
in opposite directions and are attached to masses of 10 kg and 20 kg respectively, which are
initially at rest as shown. Calculate:
(a) how far the 10 kg mass must drop in order to raise the 20 kg mass by 1 m;
(b) the angular velocity of the shaft at this point.
500 mm
100 mm
10 kg
20 kg
15
kg
Example 12.
A square plate of mass 6 kg and side 0.2 m is suspended vertically from a frictionless hinge
along one side. It is struck dead centre by a lump of clay of mass 1 kg which is moving at
10 m s–1 horizontally and remains stuck (totally inelastic collision). To what height will the
bottom of the plate rise after impact?
(The moment of inertia of a square lamina, side a and mass M, about one side, is 1
3
Ma 2 )
rod rectangle
Method 4. Change of Axis
Calculations may be performed first about some convenient (typically symmetry) axis; the
moment of inertia about the actual axis is then determined by one of two techniques for
changing axes: the parallel-axis theorem and the perpendicular-axes theorem.
4.2.1 Hoop
R
For a hoop (an infinitesimally-thin circular arc) of mass M and
radius R, rotating about its symmetry axis, all the mass is
concentrated at the single distance R from the axis. Hence,
For a hoop of mass M and radius R, about the symmetry axis perpendicular to its plane:
I MR 2 (20)
4.2.2 Disc
Consider the moment of inertia of a uniform circular disc (an infinitesimally-thin, circular
plane lamina) of mass M and radius R, about the axis of symmetry perpendicular to its plane.
The disc can be broken down into sub-elements which are hoops of
radius r and thickness r. Let ρ be the mass per unit area.
m ρ (2πr δr )
r
Sum over all elements:
r
R
π
I m r 2 ρ 2πr 3 dr ρ R 4
0 2 R
M ρπR 2
I
12 R 2
M
For a disc of mass M and radius R, about the symmetry axis perpendicular to its plane:
I 12 MR 2 (21)
4.2.3 Rod
Consider the moment of inertia of a rod (an infinitesimally-thin line segment) of mass M and
length L, about its axis of symmetry.
x
The rod can be broken down into sub-elements
of length δx, distance x from the axis. Let ρ be x
the mass per unit length.
L
m ρ δx
Summing:
L/ 2
I m x2 ρ x 2 dx 121 ρL3
L/ 2
M ρL
I
121 L2
M
For a rod of mass M and length L, about its axis of symmetry:
I 121 ML2 (22)
The distribution of mass about the axis and hence the moment of inertia is not changed by
stretching parallel to the axis of rotation without change of mass. Hence, for the axes shown:
Example 13.
outer steel rim
(part (b)) 40 cm
6 cm
flywheel
30 cm
shaft
10 cm
(a) A flywheel consists of an aluminium disc of diameter 40 cm and thickness 6 cm,
mounted on an aluminium shaft of diameter 10 cm and length 30 cm as shown.
Calculate the moment of inertia of flywheel + shaft.
(b) To increase the moment of inertia a steel rim is fixed to the outside of the flywheel.
Calculate the outer radius of the steel rim required to double the moment of inertia of
the assembly.
(c) If the flywheel is initially rotating at 100 rpm, calculate the constant frictional braking
force which needs to be applied to the outside of the steel rim in part (b) if the
flywheel is to be brought to rest in 30 seconds.
Let ρ be the mass per unit volume. Then the elemental mass
and moment of inertia are, respectively: x
mass: δm ρ (πy 2 δx)
π 4
moment of inertia: δI 12 mass disk radius 2 12 δm y 2 ρ y δx
2
Example.
Find the moment of inertia of a solid sphere, mass M and radius R about an axis of symmetry.
Hence, R
I axis
52 R 2
M
whence
I 25 MR 2
If the moment of inertia of a body of mass M about an axis through its centre of mass is IG,
then the moment of inertia about a parallel axis A is given by
I A I G Md 2
where M is the mass of the body and d is the distance between axes.
Proof. P(x,y)
Take (x,y,z) coordinates relative to the centre of mass,
with the z direction parallel to the axes of rotation. By d
Pythagoras, G A(x A, y A)
(0,0)
I G mr 2 m( x 2 y 2 )
I A m AP m[( x x A ) 2 ( y y A ) 2 )]
2
Corollary 2. For a set of parallel axes, the smallest moment of inertia is about an axis
through the centre of mass.
Example.
The M.I. for a rod of mass M and length L about an axis through its centre and normal to the
rod is I G 121 ML2 . Hence the M.I. about a parallel axis through the end of the rod is
I A I G M ( 12 L) 2
121 ML2 14 ML2 A G
13 ML2
1 1
2L 2L
Important note. This is applicable to plane laminae only. However, it can often be combined
with stretching parallel to the axis to give 3-d shapes.
If a plane body has moments of inertia Ix and Iy about perpendicular axes Ox and Oy in the
plane of the body, then its moment of inertia about an axis Oz, perpendicular to the plane, is:
Iz Ix Iy
Proof. z
By Pythagoras,
y
r 2 x2 y2
x
Hence r
y
mr 2 mx 2 my 2 x
Iz Iy Ix
Example.
Find the moment of inertia of a rectangular lamina, mass M and sides a and b, about an axis
through the centre, perpendicular to the lamina.
Solution.
From the earlier examples, the moments of inertia about axes in the plane of the lamina are
I x 121 Mb 2 , I y 121 Ma 2
z
b y
Hence, x
I z I x I y 121 M (a 2 b 2 )
a
Example.
Find the moment of inertia of a circular disc, radius R, about a diameter.
Solution.
In this case we use the perpendicular-axis theorem in reverse, Iz 21 MR 2
because we already know the moment of inertia about an axis
through the centre, perpendicular to the plane of the disc: I
I z 12 MR 2 . By rotational symmetry the unknown moment of
inertia I about a diameter is the same for both x and y axes. I
Hence,
2 MR I I
1 2
whence
I 14 MR 2
axis axis
0.1 m
0.5 m
0.1 m
0.5 m
Example 15.
A rigid framework consists of four rods, each of length L and mass M, connected in the form
of a square ABCD as shown. Find expressions, in terms of L and M, for the moment of inertia
of the framework about:
(a) the axis of symmetry SS;
(b) the side AB;
(c) an axis perpendicular to the framework and passing through centre O;
(d) an axis perpendicular to the framework and passing through vertex A;
(e) the diagonal AC.
Data: the moment of inertia of a rod, length L and mass M, about an axis through its centre
and perpendicular to the rod is 121 ML2 .
S
B C
A D
S
The motion of a rigid body which is allowed to rotate as well as translate (e.g. a rolling body)
can be decomposed into:
motion of the centre of mass rotation o f the body
under the resultant external f orce relative t o the cent re of mass
It may be shown (optional exercise) that, for a system of particles (e.g. a rigid body):
(1) The centre of mass moves like a single particle of mass M under the resultant of the
external forces:
dV dx
FM where V
dt dt
For a rigid body, motion relative to the centre of mass must be rotation and hence:
K 1
2 MV
2
2 IGω
1 2
The instantaneous point of contact with the plane has zero velocity; hence the friction force
does no work … but it is responsible for rotating the body!
A common example is of a spherical or cylindrical body rolling down an inclined plane. The
forces on the body are its weight Mg, the normal reaction force R and the friction force F.
F
v
mg
Consider the linear motion of the centre of mass and the rotational motion about it.
“torque = rate of change of angular momentum” for rotation about the centre of mass:
dω
Fr I
dt
Solution.
Linear motion:
Mg sin θ F Ma (i)
Rotation about centre of mass:
Fr Iα (ii)
For the friction force use either linear or rotational equation of motion; e.g. from (i):
F Mg sin θ Ma
M ( g sin θ a)
360 (9.81 sin 30 2.453) 882.7 N
mg
Example. (Synge and Griffiths)
A hollow spherical ball of radius 5 cm is set spinning about a horizontal axis with an angular
velocity of 10 rad s–1. It is then gently placed on a horizontal plane and released. If the
coefficient of friction between the ball and the plane is 0.34, find the distance traversed by the
ball before slipping ceases.
[The moment of inertia of a spherical shell of mass m and radius r is 23 mr 2 ].
Solution.
Initially slipping must occur, because the ball is not moving forward but it is rotating. Whilst
slipping it is friction which (a) accelerates the translational motion from rest; (b) decelerates
the rotation. Slipping ceases when v = rω, but until this point friction is maximal and given
by F μR μmg .
Linear motion
dv
F m
dt
Whilst slipping, F μR μmg . Hence,
dv
μmg m
dt
dv
μg , with v = 0 at t = 0.
dt
v μgt (i)
Rotational motion
dω
Fr I
dt
Whilst slipping, F μR μmg . Also, I 23 mr 2 . Hence,
Slipping stops when v = rω. From (i) and (ii), this occurs when
μgt rω0 32 μgt
2 μgt rω 0
5
2 rω 0
t
5 μg
This distance travelled may be determined from the linear constant-acceleration formula
s ut 12 at 2 , with
2 rω 0
u = 0, a = μg, t
5 μg
Hence,
2 rω 0 2 2 r 2 ω 02
s 0 12 μg ( )
5 μg 25 μg
Using consistent length units of metres:
2 0.05 2 10 2
s 6.00 10 3 m
25 0.34 9.81
Many formulae are given in the textbooks of Meriam and Kraige or Gere and Timoshenko.
Only some of the more common ones are summarised here.
Geometric figures are assumed to have a uniform density and have a total mass M.
Geometry Axis I
Rod, length L (1) Through centre 1
12 ML2
(2) End 1
3 ML2
Rectangular lamina, sides L (1) In-plane; symmetry 1
12 ML2
(perpendicular to axis) and W (2) Side 1
3 ML2
(3) Perpendicular to plane; symmetry 1
12 M ( L2 W 2 )
Triangular lamina, base B, altitude H Base 1
6 MH 2
Circular ring, radius R (1) Perpendicular to plane; symmetry MR 2
(2) Diameter 1
2 MR
2
Moments of inertia for many different shapes or axes can be constructed from these by:
use of the parallel-axis or perpendicular-axes rules;
stretching parallel to an axis (without change of mass distribution);
combination of fundamental elements.
Second moment of area rather than second moment of mass appears in structural engineering
(resistance to beam bending) and hydrostatics (pressure force). The formulae for second
moments of area of plane figures are the same as those in the table above except that mass M
is replaced by area A. The same symbol (I) is used.
Dependence on a length dimension parallel to the axis is often hidden inside M or A; e.g.
second moment of area of a rectangular lamina about an in-plane symmetry axis:
I 121 AL2 121 WL3 (since A WL )
second moment of area of a triangular lamina about a side of length B:
I 16 AH 2 121 BH 3 (since A 12 BH )
You will meet second moments of area a great deal in your Structures courses.
Example 1.
1.7510–3 rad s–1
Example 2.
21.8
Example 3.
0.393
Example 4.
(a) 0.680; (b) 0.0746
Example 5.
(a) 114 N; (b) 1.26 m; (c) 9.88 m s–1
Example 6.
v2
(a) v 2 gr (1 cos θ) 392.4(1 cos θ) ; (b) 2 g (1 cos θ) 19.62(1 cos θ)
r
(c) R mg(3 cos θ 2) ; (d) 48.2°; 13.3 m; 11.4 m s–1; (e) 2.49 m
Example 7.
4.79105 N m; 12600 N
Example 9.
3.14104 N m s; 52.4 N m
Example 10.
(a) 5 m; (b) 1.08 rad s–1
Example 11.
(a) rω 2v ; (b) 2.68 m s–2; (c) 3.27 m s–1; (d) 4.41 revs
Example 12.
0.162 m
Example 13.
(a) 0.407 kg m2; (b) 215 mm; (c) 1.32 N
Example 14.
0.301 m
Example 15.
(a) 23 ML2 ; (b) 53 ML2 ; (c) 4
3 ML2 ; (d) 10
3 ML2 ; (b) 2
3 ML2