Rotational Dynamics

1
Chapter 9
Rotational Dynamics
2
Torque
3
Direction RHR
4
Note The torque of a force depends upon the
reference point. One can choose the reference
point such that the torque is zero even though
the force is not.
5
Rotational Inertia
Rotational inertia is the measure of resistance a
rigid body offers to the change of its angular
velocity.
Less effort is required to produce the same
change of angular velocity in (a) than in (b)
6
Rotational Inertia of a single particle
Consider a single particle as a rigid body,
constrained to rotate in the x-y plane, about the
z-axis.
Massless rigid rod
7
The net tangential force on the particle is
Equating the tangential force to the tangential
acceleration,
Now, the torque of the force about O is
8
(No Transcript)
9
Two-Particles Rigid Body Rotating about a Fixed
Axis
10
It is only the tangential components of the
forces, that produce torques about O
11
Three Dimensional Rigid Body
12
Equation of Rotational Dynamics
Where, is the rotational inertia about the
axis of rotation.
the ith particle from the axis of rotation
13
The torque of the internal forces is zero
Contribution of the two internal forces to the
net torque is
O
Since all internal forces form pairs as above,
the net torque of all the internal forces is zero
14
Equation for rotational dynamics can, therefore,
be written as
Rotational inertia of continuous mass distribution
Where,
15
and r is the perpendicular distance of dm from
the axis of rotation
Rotational Inertia of a disc
1.
16
2.
Rotational Inertia of a rectangular plate
x
17
(No Transcript)
18
The two theorems of rotational inertia
Prob 9.10 The Perpendicular Axis Theorem
(True only for lamina)
If x, y z are mutually perpendicular axes as
shown, then
Proof
19
Summing (Integrating) over the entire lamina,
2. Parallel Axis Theorem (True in general)
The rotational inertia about any axis is the sum
of rotational inertia about a parallel axis
through the CM and total mass of the body times
square of the distance between the axes
20
cm
21
Integrating over the entire body,
Similarly, the last integral is also zero
22
Torque due to gravity
23
The torque of the gravitational forces on a body
about any point is the same as the torque of the
net weight of the body as if it were acting at
the centre of mass.
Corollary The gravitational torque about the
centre of mass is zero.
24
Centre of Gravity
Suppose acceleration due to gravity is not
constant.
The gravitational torque is then
A point , such that the above torque can be
expressed as
is defined as the Centre of Gravity.
25
Suppose acceleration due to gravity is constant
in direction.
Clearly, CG coincides with CM, when g is uniform
26
Equilibrium Applications of Newtons Laws
For equilibrium, both translational and
rotational, the net force and the net torque must
be zero.
Note Under the above conditions, if torque is
zero about one point, it is zero about every
other point.
27
Proof
28
Ex. 9.32 A 274-N plank, of length L 6.23 m,
rests on a frictionless roller at the top of a
wall of height h 2.87 m. The centre of gravity
of the plank is at its centre. The plank remains
in equilibrium for any value of ? gt 68.00 but
slips if ? lt 68.00. Find the coefficient of
static friction between the plank and the ground.
29
The net torque about O is
Equating this to zero,
Equating the net vertical force to zero,
30
Equating the net horizontal force to zero,
31
Prob. 9.6
Weight of each ball W
Radius of each ball R
Find the force exerted on the spheres by
1) Container bottom
2) Container sides
3) One another
32
System The two spheres
1. F 2W
2. The forces at the container sides are equal
Equating the torque about the centre of the
bottom sphere to zero
33
c) System Bottom sphere
Equating the horizontal component of the net
force to zero
34
Prob. 9.8
L 2.76m
W 315 N
? 320
Tbreak 520 N
a) Find the tension in the wire
b) Horizontal and vertical components of force at
A
c) Find x at which the wire breaks
35
System The horizontal bar
a) Equating the torques of T and W about the
hinge
b) Equating the horizontal and vertical
components of Fext to zero,
36
c) xbreak is obtained from the first eq. by
putting T Tbreak
Non-equilibrium applications of Newtons
equations (Fixed axis rotation)
(z is the fixed axis of rotation)
37
For general rotation,
Torque-free rotation of a spheroid
38
Rolling Motion
Rolling is a combination of pure rotation and
pure translation
a) Pure translation
In rolling
b) Pure rotation
c) Rolling
39
Velocity distribution in a rolling disc
40
Rolling with slipping and sliding
?
Pure Rolling
Rolling with slipping
Rolling with sliding
(Directed backwards)
(Directed forward)
41
Friction in Rolling
1. Rolling with slipping or sliding
Relative motion exists between the two surfaces
(ground and bottom of the wheel) in contact. So
friction is kinetic friction.
2. Pure rolling
No relative motion between the two surfaces. So
friction is static friction.
42
Moreover, the direction of the frictional force
can either be the same as or opposite to the
direction of linear motion.
Friction in the absence of relative motion
Direction of friction on the body on top is
opposite to the direction of attempted relative
motion
43
Rolling without friction
As ? is increased, friction develops forward
As ? is reduced, friction acts backward
44
Prob. 9.23
Condition for slipping µs lt tan?/3
Ans.
Eq. of translational motion along the incline
Eq. of rotation about its axis
45
For rolling, a a/R, and solving for f,
Since the condition for rolling is
46
If µs lt tan?/3, the equations of motion are

The cylinder will slide
47
Angular Impulse
N is the angular impulse
If the torque is due to a force F with moment arm
r,
where J is the linear impulse.
48
Prob. 10.5
F
A billiard ball is given a sharp impulse by a cue
as shown. Its leaves the cue with an initial
velocity of v0 and because of forward English
acquires a final velocity of 9v0/7. Show that h
4R/5
h
R
49
Ans
(Prob. 10.3)
and the ball will roll from the start
50
Let it take time T for the ball to start rolling
after it was hit.
(When rolling starts)
51
However, when the ball starts rolling, 9v0/7 ?R
Solving for h,