Title: Rigid Bodies
1Rigid Bodies
- Consider a rigid body rotating with angular speed
? about the origin 0.
For a point mass for any point in the body
l rq
vt rw
at ra
and ar rw2 v2/r
The equations of motion apply
w w0 at
q ½(w0 w)t
q w0t ½at2
w2 w02 2aq
2Rigid Bodies
- Example A turntable of a record player, radius
14cm, is rotating at 33 r.p.m. before it is
turned off. It takes 20 s to come to rest.
Calculate - (i) Its angular acceleration.
What do we know?
q ?
t 20 s
w0 33 rpm
r 0.14 m
w 0
a ?
w w0 at
a (w wo)/t
a (0 33)(2p/60)/20
a -0.173 rad/s
3Rigid Bodies
- Example A turntable of a record player, radius
14cm, is rotating at 33 r.p.m. before it is
turned off. It takes 20 s to come to rest.
Calculate - (ii) How many rotations it made before coming to
rest.
What do we know?
q ?
t 20 s
w0 33 rpm
r 0.14 m
w 0
a -0.173 rad/s
w2 w02 2aq
q (w2 - w02)/2a
q - (33?2p/60)2)/2(-0.173)
q 34.5 rad
34.5/2p
5.50 rev
4Rigid Bodies
- Example A turntable of a record player, radius
14cm, is rotating at 33 r.p.m. before it is
turned off. It takes 20 s to come to rest.
Calculate - (iii) The radial and tangential velocity of a
point on its rim at the start of the motion.
What do we know?
q 34.5 rad
t 20 s
w0 33 rpm
r 0.14 m
w 0
a -0.173 rad/s
vr 0
vT rw
vT 0.14?33(2p/60)
vT 0.484 m/s
5Rigid Bodies
- Example A turntable of a record player, radius
14cm, is rotating at 33 r.p.m. before it is
turned off. It takes 20 s to come to rest.
Calculate - (iv) The radial and tangential acceleration of a
point on its rim at the start of the motion.
What do we know?
q 34.5 rad
t 20 s
w0 33 rpm
r 0.14 m
w 0
a -0.173 rad/s
ar rw2
at ra
ar 0.14 ?(33 ?2p/60)2
at 0.14?(-0.173)
at -0.0242 m/s2
ar 1.67 m/s2
6Torque
- Consider a force F applied at an angle ? to a
body r pivoted about axis. The force F will tend
to turn the body clockwise.
The turning effect or TORQUE caused by the force
F is given by (Component of force ?r direction) x
distance from axis.
? Fsin? ? r Nm
7Torque
- Consider a force F applied at an angle ? to a
body r pivoted about axis. The force F will tend
to turn the body clockwise.
? is a maximum when F and r are perpendicular
? is zero when F and r are parallel
8Torque
- Example A constant force of 5.0 N applied to the
outer rim of a pulley (r 1.0 m) is used to
raise a weight of 6.0 N at constant speed
attached to the inner rim (r 0.5 m). What is
the net torque on the pulley?
5N
Clockwise torque 5 x 1.0 5 Nm
Anticlockwise torque 6 x 0.5 3 Nm
Net torque 5 - 3
2 Nm clockwise
6N
9Moment of Inertia
- Consider the case of a particle rotating in a
circle as above at the end of a string whose mass
is negligible.
The applied force creates a torque which gives
rise to its angular acceleration causing the mass
to rotate.
t rF
but
F ma
and
a ra
F mra
so
rF mrra
t mr2a
mr2 is the rotational inertia of the particle and
is called the moment of inertia (I).
I is the rotational analogue of m
10Moment of Inertia
- Consider a rigid body rotating about an axis
through its axis such as a wheel.
The wheel consists of particles at various
distances from the axis of rotation.
Each particles heave a velocity dependent on its
distance to the axis.
Each of these particles has a torque
Total torque of body sum of the individual
torques
11Moment of Inertia
- Moment of Inertia (MOI) depends on the position
of the axis of rotation and the mass
distribution.
General expression for Moment of inertia is
12Moment of Inertia
Consider a cylinder of radius R0.
The density of the entire cylinder is r M/V
M/(pR02L)
So
13Moment of Inertia
- Calculation of Moments of Inertia for a Uniform
Rigid Rod about an axis through its centre of
mass.
For a small element of mass with a length dx
located a distance x from the axis of rotation
through 0.
The mass of the element is given by
The mass per unit length of the rod (assumed
uniform) is
Therefore, the total Moment of Inertia of the
rod about the axis through 0
14Moment of Inertia
15Moment of Inertia
- Example A Wheel of mass 50.0 kg, radius 1.0 m in
the form of a thin disc is mounted on an axis
with negligible friction. A mass of 5.0 kg is
attached to a light cord wrapped round the rim.
Calculate - - 1. the angular acceleration of wheel
For wheel, Torque ? T x R
T x 1
I ?
1/2 x50x(1)2 x ? (I1/2 MR2)
5kg
? 1 / 25 rad/s2
16Moment of Inertia
- Example A Wheel of mass 50.0 kg, radius 1.0 m in
the form of a thin disc is mounted on an axis
with negligible friction. A mass of 5.0 kg is
attached to a light cord wrapped round the rim.
Calculate - - 2. the acceleration of m
a R?
1 x 0.04
0.04 m/s2
5kg
17Moment of Inertia
- Example A Wheel of mass 50.0 kg, radius 1.0 m in
the form of a thin disc is mounted on an axis
with negligible friction. A mass of 5.0 kg is
attached to a light cord wrapped round the rim.
Calculate - - 3. the tension in cord
mg - T ma
T ma mg
m(a g)
5(0.04 9.8)
5kg
49.2 N
18Moment of Inertia
- Example Two masses m2 10kg , m1 5 kg
attached by light string over two identical
pulleys of mass 5kg and radius 0.5m, in the form
of a uniform disc. Calculate acceleration of the
masses.
Note 3 separate tensions
For m1 T1 - m1g m1a
m2 m2g - T3 m2a
For pulley 1 (T2 - T1)R I?
and a r?
For pulley 2 (T3 - T2)R I?
19Parallel Axis Theorem
- If ICM Moment of Inertia of a body, of mass M,
about an axis through its centre of mass, then
the moment of inertia about a parallel axis
distance a away is given by
I ICM Mh2
20Parallel Axis Theorem
- Example Find the MOI of a rod about an axis
passing through one end.
A
G
B
l
ICM 1/12.Ml2
I 1/12Ml2 M(l/2)2
1/12Ml2 1/4Ml2
I 1/3Ml2
21Angular Momentum
- The angular momentum of mass m, moving with the
velocity v about 0 is defined as the 1st moment
of its linear momentum.
Linear momentum m.v
Angular momentum r ? m.v.sin?
Or L r ? m.v which has magnitude
rmvsin? Direction given by R.H. screw rule.
For a particle rotating about a fixed axis with
circular motion ? 90o and L r.m.v but v
r.? L m r2 ?
22Angular Momentum
- For rigid body rotating about 0 each particle
contributes angular momentum - Li ri x mvi
Total angular momentum L ?Li m1r21?2
m2r22?2 m3r32?2 .
???mir2i
I? L I? (compared with P mv)
Also