Rotation of a Rigid Body
PHY1012F
NEWTON’S LAWS
ROTATION OF A RIGID BODY
PHY1012F
NEWTON’S LAWS
ROTATION OF A RIGID BODY
ROTATIONAL KINEMATICS
ROTATION OF A RIGID BODY
Rigid body model:
Learning outcomes:
A rigid body is an extended object whose shape and size
do not change as it moves. Neither does it flex or bend.
At the end of this chapter you should be able to…
Extend the ideas, skills and problem-solving techniques
developed in kinematics and dynamics (particularly
w.r.t. circular motion) to the rotation of rigid bodies.
Types of motion:
Translational motion:
Rotational motion:
Combination motion:
Calculate torques and moments of inertia.
Use… - appropriate mathematical representations
- the laws of conservation of mechanical energy
and angular momentum
- vector mathematics
…to solve problems involving rotation.
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NEWTON’S LAWS
ROTATION OF A RIGID BODY
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NEWTON’S LAWS
ANGULAR VELOCITY and
ANGULAR ACCELERATION
ANGULAR ACCELERATION
relationship
Rotational motion
s
s = r

ds  v
t
dt
dv t
 at
dt
vt =  r
  d
Angular acceleration, 
Angular velocity, 
Linear motion
dt
  d
dt
at =  r
ROTATION OF A RIGID BODY
A body’s angular acceleration, , is the rate at which its
angular velocity changes.
Units: [rad/s2]
+
–
+
–
>0
<0
faster
slower
>0
>0
>0
<0
slower
faster
<0
<0
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NEWTON’S LAWS
ROTATION OF A RIGID BODY
VELOCITY GRAPHS  ACCELERATION GRAPHS
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NEWTON’S LAWS
ROTATION OF A RIGID BODY
VELOCITY GRAPHS  ACCELERATION GRAPHS
Angular acceleration is equivalent to the slope of a -vs-t graph.
Angular acceleration is equivalent to the slope of a -vs-t graph.
 (rad/s)
3
 (rad/s)
3
0
Eg: A wheel rotates about
its axle…
3
6
9
12 t (s)
–3
 (rad/s2)
For the first 6 s the
slope
acceleration


is
 3  0  0.5 rad/s 2
t
60s
0
–½
3
6
9
12 t (s)
3
6
9
12 t (s)
–3
 (rad/s2)
½
0
Eg: A wheel rotates about
its axle…
For the last 6 s the
slope
acceleration
is   3  3   rad/s 2
t
12  6 s
½
3
6
9
0
12 t (s)
–½
–
–
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Rotation of a Rigid Body
PHY1012F
NEWTON’S LAWS
ROTATION OF A RIGID BODY
PHY1012F
NEWTON’S LAWS
KINEMATIC EQUATIONS
CENTRE OF MASS
The following equations apply for constant acceleration:
Linear motion
ROTATION OF A RIGID BODY
While the various points on a flipped spanner describe
different (complicated) trajectories, one special point, the
centre of mass, follows the usual, simple parabolic path.
Rotational motion
vf = vi + at
f = i + t
xf = xi + vit + ½a (t)2
f = i + it + ½ (t)2
vf2 = vi2 + 2ax
f2 = i2 + 2
The centre of mass (CM) of a system of particles…
is the weighted mean position of the system’s mass;
is the point which behaves as though all of the system’s
mass were concentrated there, and all external forces
were applied there (so, in terms of Newton I…);
is the point around which an unconstrained system
(i.e. one without an axle or pivot) will naturally rotate.
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NEWTON’S LAWS
ROTATION OF A RIGID BODY
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NEWTON’S LAWS
CENTRE OF MASS
ROTATION OF A RIGID BODY
TORQUE
To find the centre of mass of
two uniform masses…
m1
1. place the masses on an x-axis,
x1 = 0
with one of the masses at the
origin;
m x  m2 x2
2. apply the formula: xCM  1 1
m1  m 2
Torque may be regarded as…
x2
the “amount of turning” required to rotate a body around a
certain point called an axis or a pivot;
the effectiveness of a force at causing turning.
sCM 
In general, for many particles on any axis:
And for a continuous distribution of mass
in which mi = M…
xCM
The rotational analogue of force is called torque, .
x
m2
sCM
E.g. To push open a heavy door
around its hinge (as seen from
the top) requires a force
applied at some point on the door.
 m i si
 mi

F3
Consider the effectiveness of each
of the (equal) forces shown…
 1  s dm
M

F4
hinge

F2

F1
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NEWTON’S LAWS
ROTATION OF A RIGID BODY
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NEWTON’S LAWS
ROTATION OF A RIGID BODY
TORQUE
TORQUE
Three factors determine the amount of torque achieved:
Torque can also be defined as the product of the force, F,
and the perpendicular distance between the pivot and
the line of action of the force, d, known as the torque arm,
moment arm, or lever arm:
magnitude of the applied force;
distance between the point of application and the pivot;
angle at which the force is applied.
Only the tangential
component of the applied
force produces any turning…
| | = dF
y

F
 = rFt
Hence:
  rFsin
Units: [N m]
( joule!)
Ft = Fsin
Fr

Torque…
radial
line
r
point of
application
pivot
x
is positive if it tries to rotate
the object anticlockwise
about the pivot; negative if
rotation is clockwise;
torque
arm
d = r sin

F
r

line of
action
pivot
x
must be measured relative to a specific pivot point.
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Rotation of a Rigid Body
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NEWTON’S LAWS
ROTATION OF A RIGID BODY
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NEWTON’S LAWS
NET TORQUE
The axle prevents any
translational movement,
so…





Fnet  F1  F2  F3    Faxle  0
GRAVITATIONAL TORQUE

F1

F4
Several forces act on an
extended object which is free
to rotate around an axle.
ROTATION OF A RIGID BODY

Faxle
For an object to be balanced,
its centre of mass must lie
either directly above or directly
below the point of support.

F2
axle
centre of
mass
If this is not so, the body’s own
weight, acting through its centre of mass (as if all its
mass were concentrated there), causes a net torque
due to gravity:
grav = –MgxCM

F3
And the net torque is given by:
CM
pivot
net = 1 + 2 + 3 + … = i
where xCM is the distance between
the centre of mass and the pivot.

(Faxle causes no torque since it is applied at the axle.)
Mg
x
xCM
0
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NEWTON’S LAWS
ROTATION OF A RIGID BODY
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NEWTON’S LAWS
COUPLES
ROTATIONAL DYNAMICS
Rotation without translation is
achieved by the application of a pair
of equal but opposite forces at two
different points on the object.
d1
Notes:
d2
A rocket is constrained to move in
a circle by a lightweight rod…

F2
Force causes linear acceleration.

F1
(sign by inspection)
y
Linear acceleration is
Fnet
“limited” by inertial mass. a  m
l
 Ft = mr
Ft = mat
The pivot is immaterial – a couple will exert the
same net torque lF about any point on the object.
 rFt = mr2

Fthrust
Ft
Newton II (linear):
pivot
|net | = d1F + d2F = (d1 + d2)F
 |net | = lF
ROTATION OF A RIGID BODY
rod


T
Fr
pivot
x
  = mr2
Newton II (rotational):
Torque causes angular acceleration.
Unless the rotation is constrained to act around a
specific pivot, it will occur around the body’s
centre of mass.

Angular acceleration is “limited” by the
  net2
particle’s rotational inertia, mr2, about the pivot.
mr
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NEWTON’S LAWS
ROTATION OF A RIGID BODY
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ROTATIONAL INERTIA
A rotating extended object can be
modelled as a collection of particles,
each a certain distance from the pivot.
ROTATIONAL DYNAMICS
m1
m2
r2
r1
pivot
The body’s rotational inertia…
ROTATION OF A RIGID BODY
Summary of corresponding quantities and relationships:
m3
Linear dynamics
r3
force
is also known as its moment of inertia, I;
is the aggregate of the individual (mr2)’s:
I = m1r12 + m2r22 + m3r32 + … =  miri2
gives an indication of how the mass of the body is
distributed about its axis of rotation (pivot);
is the rotational equivalent of mass.
Thus:

Fnet
net
torque
inertial mass
m
moment of inertia
I
acceleration
a
angular acceleration

Newton II
 net
Rotational dynamics
Fnet = ma
Newton II
net = I
I
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Rotation of a Rigid Body
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NEWTON’S LAWS
ROTATION OF A RIGID BODY
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NEWTON’S LAWS
MOMENTS OF INERTIA
MOMENTS OF INERTIA
For an extended system with a continuous distribution of
mass, the system is divided into equal-mass elements, m.
Then, by allowing these to shrink in size, the moment of
inertia summation is converted to an integration:
2
I   ri 2m 
m  0 I   r dm
m
r
 I    x 2  y 2  dm
E.g. in a wheel, or hoop, where all the
mass lies at a distance R from the axis…
R
I  R 2  dm ,
becomes
where the integral  dm is simply the sum of all the mass
elements, i.e. the total mass, M, of the wheel.
y
pivot x
For simple, uniform distributions of mass,
however, the integration can be trivial.
I   r 2 dm
y
For complex distributions of mass, r is
usually replaced by x and y components…
ROTATION OF A RIGID BODY
x
So for open wheels (hoops) I = MR2.
…and, before integration, dm is replaced by an expression
involving a coordinate differential such as dx or dy.
In practice, the rotational inertias of certain common
shapes (of uniform density) are looked up in tables…
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ROTATION OF A RIGID BODY
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MOMENTS OF INERTIA
NEWTON’S LAWS
ROTATION OF A RIGID BODY
ROTATION ABOUT A FIXED AXIS
Knight 2Ed
p347:
Problem-solving strategy:
1 MR 2
2
1 ML2
12
Knight 3Ed
p362:
1 ML2
3
1.
Model the system as a simple shape or group of shapes.
2.
Identify the axis around which the system rotates.
3.
Draw a picture of the situation, including coordinate axes,
symbols and known information.
4.
Identify all the significant forces acting on the system and
determine the distance of each force from the axis.
MR 2
2 MR 2
5
1 Ma 2
12
1 Ma 2
3
2 MR 2
3
5.
Determine all torques, including their signs.
6.
Apply Newton II: net = I.
7.
Use rotational kinematics to find angular positions and
velocities.
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NEWTON’S LAWS
ROTATION OF A RIGID BODY
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ROTATION OF A RIGID BODY
ROTATION ABOUT A FIXED AXIS
ROTATION ABOUT A FIXED AXIS
A 25.0 kg child sits on the edge of a 640 kg roundabout which
has a diameter of 4.00 m while her father applies a steady
tangential force of 100 N to a corner pole.
(a) Determine the child’s angular acceleration.
(b) After how many revolutions will the child slide off?
A 25.0 kg child sits on the edge of a 640 kg roundabout which
has a diameter of 4.00 m while her father applies a steady
tangential force of 100 N to a corner pole.
(a) Determine the child’s angular acceleration.
(b) After how many revolutions will the child slide off?
F = 100 N
pivot
R=2m
pivot
R=2m
roundabout
child
roundabout
child
M = 640 kg
m = 25 kg
M = 640 kg
m = 25 kg
1-3. Model the system as a point mass on the edge of a
disc rotating about the disc’s centre (with data).
4. Identify all significant forces acting on the system and
determine the distance of each force from the axis.
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Rotation of a Rigid Body
PHY1012F
NEWTON’S LAWS
ROTATION OF A RIGID BODY
PHY1012F
NEWTON’S LAWS
ROTATION OF A RIGID BODY
ROTATION ABOUT A FIXED AXIS
ROTATION ABOUT A FIXED AXIS
A 25.0 kg child sits on the edge of a 640 kg roundabout which
has a diameter of 4.00 m while her father applies a steady
tangential force of 100 N to a corner pole.
(a) Determine the child’s angular acceleration.
(b) After how many revolutions will the child slide off?
A 25.0 kg child sits on the edge of a 640 kg roundabout which
has a diameter of 4.00 m while her father applies a steady
tangential force of 100 N to a corner pole.
(a) Determine the child’s angular acceleration.
(b) After how many revolutions will the child slide off?
F = 100 N
+ve
F = 100 N
pivot
R=2m
+ve
pivot
R=2m
roundabout
child
roundabout
child
M = 640 kg
m = 25 kg
M = 640 kg
m = 25 kg
6. Apply Newton II: net = I.
5. Determine all torques, including their signs.
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PHY1012F
NEWTON’S LAWS
ROTATION OF A RIGID BODY
A 25.0 kg child sits on the edge of a 640 kg roundabout which
has a diameter of 4.00 m while her father applies a steady
tangential force of 100 N to a corner pole.
(a) Determine the child’s angular acceleration.
(b) After how many revolutions will the child slide off?
R=2m
NEWTON’S LAWS
Provided it does not slip,
a rope passing over a
pulley moves in the same
way as the pulley’s rim,
and thus also objects
attached to the rope.
r
Fr
PHY1012F
ROTATION OF A RIGID BODY
CONSTRAINTS DUE TO ROPES and PULLEYS
t
F
26
Ft = 100 N
rim acceleration = | |R
rim speed = | |R
non-slipping
rope
R
vobj = | |R

aobj = | |R
As before, the constraints
are given as magnitudes.
Actual signs are chosen
by inspection.
m = 25 kg
 = 0.14 rad/s
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PHY1012F
NEWTON’S LAWS
ROTATION OF A RIGID BODY
T2
Two blocks are connected by a light
string which passes over two
identical pulleys, each with a moment
of inertia I, as shown.
Find the acceleration of each mass
and the tensions T1 , T2 and T3.
T1
m1
m1g
–ve
T1
w
T3
y
m2
x
T2
T2
n1
+ve
T3
(2)
(3)
net = T2R – T3R = –I
(4)
(1) – (2):
(1)
F2y = T3 – m2g = –m2a
(2)
net = T1R – T2R = –I
(3)
net = T2R – T3R = –I
(4)
T2
T1
T3
y
m1
x
m2
T1R – T3R = –2I
 T1  T3   2I    2I2 a
R
R
T1 – T3 + m2g – m1g = m1a + m2a
  2I2 a   m 2  m1  g   m1  m 2  a
R
 m 2  m1  g
…etc
a 
m1  m 2  2 I2
R
–ve
F1y = T1 – m1g = m1a
ROTATION OF A RIGID BODY
(1)
net = T1R – T2R = –I
m2
m2g
+ve
NEWTON’S LAWS
F2y = T3 – m2g = –m2a
(3) + (4):
T3
n2
w
PHY1012F
F1y = T1 – m1g = m1a
T1
m1
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Rotation of a Rigid Body
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ROTATION OF A RIGID BODY
PHY1012F
ROTATIONAL ENERGY
Each particle in a rigid rotating body
has kinetic energy.
m2
The sum of all the individual kinetic
energies of each of the particles is
the rotational kinetic energy of the
body:
Krot = ½ m1v12 + ½ m2v22 + …
 Krot =
 Krot =
½ I2
ROTATION OF A RIGID BODY
CONSERVATION OF ENERGY
r2
r1
m1
r3
axle
As usual, energy is conserved
(in frictionless systems).
m3
If, however, a horizontal axis
of rotation does not coincide
with the centre of mass, the
object’s potential energy will
vary.

 Krot = ½ m1r122 + ½ m2r222 + …
½ (m1r12)2
NEWTON’S LAWS
axle
So we write:
Emech = Krot + Ug = ½ I2 + MgyCM
K
U
Emech = K + U
Emech = 0
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
NEWTON’S LAWS
ROTATION OF A RIGID BODY
A 70 g metre stick pivoted freely at one end is released from a
horizontal position. At what speed does the far end swing
through its lowest position?
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NEWTON’S LAWS
A 70 g metre stick pivoted freely at one end is released from a
horizontal position. At what speed does the far end swing
through its lowest position?
x
0
ROTATION OF A RIGID BODY
xCM = 0.5 m
0
x
pivot
pivot
M = 0.07 kg
Mg
1-3. Model the stick as a uniform rod rotating around one
end, and draw a picture with pivot, x-axis and data.
M = 0.07 kg
4. Identify all significant forces acting on the object and
determine the distance of each force from the axis.
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PHY1012F
NEWTON’S LAWS
ROTATION OF A RIGID BODY
A 70 g metre stick pivoted freely at one end is released from a
horizontal position. At what speed does the far end swing
through its lowest position?
0
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PHY1012F
NEWTON’S LAWS
A 70 g metre stick pivoted freely at one end is released from a
horizontal position. At what speed does the far end swing
through its lowest position?
x
xCM = 0.5 m
pivot
pivot
Mg
x
M = 0.07 kg
grav = –0.34 Nm
M = 0.07 kg
Mg
–ve
ROTATION OF A RIGID BODY
net = I

 0.34 
grav = –MgxCM

 0.34  1 ML2 
3
1
3
 0.07  12
  = –15 rad/s2
 grav = –0.07(9.8)(0.5) = –0.34 Nm
6. Apply Newton II: net = I.
5. Determine all torques, including their signs.
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Rotation of a Rigid Body
PHY1012F
NEWTON’S LAWS
ROTATION OF A RIGID BODY
A 70 g metre stick pivoted freely at one end is released from a
horizontal position. At what speed does the far end swing
through its lowest position?
PHY1012F
NEWTON’S LAWS
A 70 g metre stick pivoted freely at one end is released from a
horizontal position. At what speed does the far end swing
through its lowest position?
y
pivot
 i = 0 rad
i = 0 rad/s
 = –15 rad/s2
(Not constant)
 f = –0.5 rad
f = ?
L=1m
pivot
f2 = i2 + 2ional
t
rota oblem!
OTf2u=se
0 + 2(–15)(–0.5
)
pr
s
i
N
h
t
can
l ve
o
s
Y ou
o



15



6.8
rad/s
f
i cs t h
not?
m at
W y
kine
vt =  r
ROTATION OF A RIGID BODY
Before: M = 0.07 kg
yCM = 0 m
0 = 0 rad/s
After: yCM = –0.5 m
1 = ?
vtip = ?
 vt = –6.8  1 = 6.8 m/s
7. Use rotational kinematics to find angular positions
and velocities. (Not this time!)
x
½ I02 + MgyCM 0 = ½ I12 + MgyCM 1
0 
1
2
 1 3 ML2 12  Mg  1 2 L
 1 
3g
L
v tip  1L  3 gL  5.42 m/s
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NEWTON’S LAWS
ROTATION OF A RIGID BODY
VECTOR DESCRIPTION
OF ROTATIONAL MOTION
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ROTATION OF A RIGID BODY
THE CROSS PRODUCT

The magnitude of the torque exerted by force F applied

at displacement r from the turning point is:   rFsin
Using only “clockwise” and “counterclockwise” is the
rotational analogue of using “backwards” and
“forwards” in rectilinear kinematics. A more general
handling of rotational motion requires vector quantities.
Once again, the quantity rF sin is the product of two

vectors, r and F , at an angle  to each other. This time,
however, we use the orthogonal components to

determine the cross product of the vectors: r  F .
iˆ  iˆ = ˆj  ˆj = kˆ  kˆ = 0
The vector associated with a rotational quantity…
has magnitude equal to the magnitude of that
quantity;


has direction as given by the right-hand rule.
In RH system:

E.g. The angular velocity vector, , of this
anticlockwise-turning disc points…
in the positive z-direction.
 
ˆj  kˆ  iˆ   kˆ  ˆj
kˆ  iˆ  ˆj   iˆ  kˆ
y
iˆ  ˆj  kˆ   ˆj  iˆ
1
ĵ
k̂
z
x
î
1
1
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Notes:
NEWTON’S LAWS
ROTATION OF A RIGID BODY
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ROTATION OF A RIGID BODY
THE CROSS PRODUCT
ANGULAR MOMENTUM
 
r  F  rF sin 

The momentum, p, of a particle in circular motion is NOT
conserved (since its velocity changes continuously).
The more orthogonal the
vectors, the greater the
cross product; the more
parallel, the smaller…
Nevertheless, an orbiting or rotating body
does have a tendency to “keep going”
– due to its angular momentum, L:

F

r
Since it is a vector quantity, the cross product is
also known as the vector product.
L = mrvt
Units: [kg m2/s]
L = I
Law of conservation of angular momentum:
The angular momentum of a system remains constant
– provided the net external torque on it is zero.
A   B  C   A  B  A  C.
Derivative of a cross product:


d  r  p   dr  p  r  dp
dt
dt
dt
i.e. Lf = Li
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(if t = 0)
42
Rotation of a Rigid Body
PHY1012F
NEWTON’S LAWS
ROTATION OF A RIGID BODY
A student sits on a stool which can
rotate freely about a vertical axis.
Initially at rest, the student holds
vertically the axle of a bicycle
wheel which has a rotational
inertia IW = 1.2 kg m2 and which is
rotating anticlockwise (viewed
from above) atiW = 3.9 rev/s.
The student and the stool together have a combined rotational
inertia IB = 6.8 kg m2 (and iB = 0 rev/s).
The student now inverts the wheel so that (viewed from
above) it is now rotating clockwise…
PHY1012F
NEWTON’S LAWS
ROTATION OF A RIGID BODY
Determine the student’s final angular velocity.
iW = 3.9 rev/s
iB = 0 rev/s
fW = 3.9 rev/s fB = ?
IW = 1.2 kg m2
IB = 6.8 kg m2
(Lf)B + (Lf)W = (Li)B + (Li)W
(Lf)B + (–Li)W = 0 + (Li)W
 (Lf)B = 2(Li)W
 (If)B = 2(Ii)W
 ( f )B  2( I  i ) W  2(1.2)(3.9) = 1.4 rev/s , anticlockwise
6.8
IB
Determine the student’s final angular velocity.
43
PHY1012F
NEWTON’S LAWS
ROTATION OF A RIGID BODY
44
PHY1012F
NEWTON’S LAWS
ROTATION OF A RIGID BODY
ANGULAR MOMENTUM
ROTATIONAL MOMENTUM & ENERGY
We have shown that in circular motion (where vt and r are
perpendicular) a particle has angular momentum L = mrvt.
Summary of corresponding quantities and relationships:
mvt = p
 L = rp
Linear
z

r


p  mv

p
KCM = ½ MvCM2


P  MvCM


Fnet  dP
dt
More generally (and allowing



for r and p to be at any angle )…
  
L  r  p = (mrv sin, direction from RH rule)



dL  d r  p   dr  p  r  dp  v  p  r  F
net
dt dt
dt
dt



dp

I.e. dL   net
(Cf. in linear motion: Fnet 
)
dt
dt

Linear momentum, P, is conserved if there is no net force
45
Rotational
Krot = ½ I2


L  I

(around an axis of
symmetry)

 net  dL
dt

Angular momentum, L, is conserved if there is no net torque
46