Rotational Dynamics Lecture Slides presentation.pptx

Chapter 9
Rotational Dynamics
Cause of rotational motion

9.1 The Action of Forces and Torques on Rigid Objects
In pure translational motion, all
points on an object travel on parallel
paths.
The most general motion is a
combination of translation and
rotation.

What causes an object to have an angular acceleration?
TORQUE,
𝑟
According to Newton’s second law, a net force
causes an object to have an acceleration.
𝐚
=
∑
𝐅
�
�

The amount of torque depends on where and in what direction the
force is applied, as well as the location of the axis of rotation.

DEFINITION OF TORQUE
Magnitude of Torque = (Magnitude of the force) x (Lever
arm)
 
F𝑙
Direction: The torque is positive when the force tends to produce
a
counterclockwise rotation about the axis.
Lever arm, 𝑃 is minimum distance
between line of action and axis of
rotation.
SI Unit of Torque? Newton meter (N·m)

Example: The Achilles Tendon
The tendon exerts a force of magnitude
790 N. Determine the torque
(magnitude and direction) of this force
about the ankle joint.
  F𝑙
𝑙  (3.6 102
m) cos 55o
2.06 ×
10−2𝑚
  (790N )(2.06 102
m) 16.27 N.m

9.2 Rigid Objects in Equilibrium
If a rigid body is in equilibrium, neither its linear
motion nor its rotational motion changes.
Equilibrium
 Fx 
0
 Fy 
0

 0
Linear motion
ax  ay 
0
Rotational motion
  0

1. Write down what is known and what is unknown.
2. Select the object to which the equations for equilibrium are to be
applied.
3. Draw a free-body diagram that shows all of the external forces acting
on the object.
4. Choose a convenient set of x, y axes and resolve all forces into
components that lie along these axes. (chose + and – directions)
5. Apply the equations that specify the balance of forces at
equilibrium. (Set the net force in the x and y directions equal to
zero.)
6. Select a convenient axis of rotation. Set the sum of the torques
about
this axis equal to zero. (Chose counter clockwise + and clockwise -)
Reasoning Strategy

Example: A Diving Board
A woman whose weight is 530 N is
poised at the right end of a diving board
with length 3.90 m. The board has
negligible weight and is supported by
a fulcrum 1.40 m away from the
left end.
Find the forces that the bolt and the
fulcrum exert on the board.
  0
𝐹2𝑃2 − 𝑤𝑃𝑤
= 0
1476.4 N
What if weight of board is 150
N?

Example : Bodybuilding
The arm is horizontal and weighs 31.0 N. The deltoid muscle can supply
1840 N of force. What is the weight of the heaviest dumbbell he can
hold?
86.1 N

9.3 Center of Gravity
DEFINITION OF CENTER OF GRAVITY
The center of gravity of a rigid body is the point at
which its weight can be considered to act when
the torque due to the weight is being calculated.
When an object has a symmetrical shape and
its weight is distributed uniformly, the
center of gravity lies at its geometrical
center.

W1  W2 

W1x1  W2 x2 
xcg
When weight is not
uniformly distributed

• the upper arm (17 N),
• the lower arm (11 N), and
• the hand (4.2 N).
Find the center of gravity of
the arm relative to the shoulder
joint.
Example: The Center of Gravity of an
Arm
The horizontal arm is
composed of three parts:
0.278 m
W1  W2  W3

W1 x1  W2 x2  W3 x3
xcg

Conceptual Example: Overloading a Cargo Plane
This accident occurred because the plane was overloaded toward
the rear. How did a shift in the center of gravity of the plane cause
the accident?

Finding the center of gravity of an irregular
shape.

In rotational acceleration
According to Newton’s second law, a net force
causes an object to have an acceleration.
∑ 𝐅
= 𝐦𝐚

 I  

 

 

acceleration
 Moment of 
 Angular
Net externaltorque  
inertia
I  mr 2

I (moment of inertia) depends on how the mass
is distributed around the axis of rotation.
Units of I ? 𝑘𝑔 ∙ 𝑚2
If an object is not in equilibrium
For translation acceleration:

9.4 Newton’s Second Law for Rotational Motion About a Fixed Axis
Example: The Moment of Inertia Depends on Where
the Axis Is.
Two particles each have mass m and are fixed at the ends of a thin
rigid rod. The length of the rod is L.
Find the moment of inertia when this object rotates relative to
an axis that is perpendicular to the rod at
(a) one end and
(b) the center.
𝐼 = 𝑚2𝑟2
2
𝐼 = 𝑚1𝑟1
2 +
𝑚2𝑟2
2
(a)
(b)
Table 9.1: Moment of inertia for various rigid bodies.

9.5 Rotational Work and Energy
WR 

s  r
Work
W  Fs
W  Fr
  Fr
The rotational work is done by a constant torque in turning
an object through an angle is
SI Unit of Rotational Work? N.m = joule (J)

2
1
2
2
2
 mr 
1
2
2 2
2
1
 mr    
I
KE  

2 2
T
KE  1
mv2
 1
mr
2
 2
vT 
r
Kinetic Energy
1
2
𝐾𝐸 =
𝑚𝑣
2
�
�
2
1
𝐾𝐸 =
𝐼𝜔2
SI Unit of Rotational Kinetic Energy? joule (J)

Mechanical energy
Total mechanical Energy is equal to sum of translational kinetic
energy, rotational kinetic energy, and gravitational energy
E = KET + KER +PE
E  1
mv2
 1
I 2
 mgh
2 2

9.6 Angular Momentum
DEFINITION OF ANGULAR MOMENTUM
The angular momentum L of a body rotating about a
fixed axis is the product of the body’s moment of inertia
and its angular velocity with respect to that axis:
L 
I
SI Unit of Angular
Momentum? kg·m2/s
𝒑 =
𝑚𝒗

9.6 Angular Momentum
PRINCIPLE OF CONSERVATION OF
ANGULAR MOMENTUM
The angular momentum of a system remains
constant (is conserved) if the net external torque
acting on the system is zero.

Rotational Dynamics Lecture Slides presentation.pptx

Rotational Dynamics Lecture Slides presentation.pptx

More Related Content

Similar to Rotational Dynamics Lecture Slides presentation.pptx (20)

Recently uploaded (20)

Rotational Dynamics Lecture Slides presentation.pptx