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Following points are the summary of these Lecture Slides : Rigid Objects and Torque, Vector Quantity, Torque, Body Resulting, Rotation, Rotational State, Torque Depends, Component, Force Perpendicular, Lever arm
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Torque is a vector quantity that represents the application of
force to a body resulting in rotation
(or change in rotational state)
Definition: •^
The SI units for torque are N
. m
(not to be confused with joules)
Torque depends on:– The Lever arm (leverage),– The component of force perpendicular to lever arm,
^
ˆ
F =
F sin
i
^
F r
r
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Newton’s 2
nd
Law
(for Rotational Motion)
When there is a net torque exerted on a rigid body its stateof rotation (
) will change depending on:
Amount of net torque,
Rotational inertia of object,
I
Note:
No net torque means
= 0 and
= constant}
Alternatively, the net torque exerted on a body is equal tothe product of the rotational inertia and the angularacceleration:
This is Newton’s 2
nd
Law (for rotation)!!
n
1
2
i
i=
=
... =
= I
^
^
^
n
1
2
i i=
...
1
=
=
=
I^
I^
I
Center of Gravity (COG)
The COG is the location where all of an
object’s weight can be considered to actwhen calculating torque
-^
The COG may or may not actually be onthe object (i.e. consider a hollow sphereor a ring)
-^
When an object’s weight produces atorque on itself, it acts at its center ofmass To calculate center of gravity:
n
i^
i
1
1
2
2
i=
cg
1
2
Wx tot
W x
W x
...
x^
=
=
W + W + ...
W^ n
i^
i
1
1
2
2
i=
cg
1
2
Wy tot
W y
W y
...
y^
=
=
W + W + ...
W
Moment of Inertia
The resistance of a rigid body to changes in its state ofrotational motion (
) is called the moment of inertia
(^ or
rotational inertia)The Moment of Inertia (I) depends on:
Mass of the object, m
The axis of rotation
Distribution (position) of mass about the axis of rotation
Definition
(For a discrete distribution of mass):
Where:
m
is the mass of a small segment of the object i r^ i^
is the distance of the mass mi from the axis of rotation
The SI units for
are kg
.m
2
Note:
For a continuous distribution of mass there are more
sophisticated techniques for calculating the moment of inertia
n
2
2
2
2
1 1
2 2
n n
i i
i=
I = m r
m r
... m r
=
mr
ω
Moment of Inertia (common examples, cont.)
A hollow sphere:
A solid sphere:
A thin rod:
.^
m
2
rod
mL
I^
=
12
m R
m R
2
hollow sphere
2mR
I^
=
3
2
solid sphere
2mR
I^
=
5
L
The Parallel Axis Theorem
-^
The Parallel Axis Theorem is used to determine themoment of inertia for a body rotated about an axis adistance, l, from the center of mass: Example:
A 0.2 kg cylinder (r=0.1 m) rotated about an
axis located 0.5 m from its center:
2
cm
I = I
L
^
2
cm
2
2
2
2
I =
2
kg m
I = 0.052 kg m
Rolling objects & Inclined Planes
Consider a solid disc & a hoop rolling down a hill (inclined plane):1.
Apply Newton’s 2
nd
Law
(force)
to each object
Apply Newton’s 2
nd
Law
(torque)
to each object
What is the acceleration of each object as they roll down thehill?
Which one reaches the bottom first?
Soliddisc
Hoop
Angular Momentum
Angular momentum is the rotational analog of linearmomentum
-^
It represents the “quantity of rotational motion” for an object(or its inertia in rotation)
-^
Angular Momentum (a vector we will treat as a scalar) isdefined as:
Note:Angular Momentum is related to Linear Momentum:
L = r
.p sin
r is distance from the axis of rotation p^
is the linear momentum ^
is the angle between r and
p
SI units of angular momentum are kg
.m
2 /s
r
p
Conservation of Angular Momentum
(Examples)
-^
A figure skater
-^
A high diver
-^
Water flowing down a drain
Archimedes
(287-212 BC)
Possibly the greatest mathematician inhistory
-^
Invented an early form of calculus
-^
Discovered the Principle of Buoyancy
(now
called Archimedes’ Principle)
-^
Introduced the Principle of Leverage(Torque)
and built several machines based
on it.
“Give me a point ofsupport and I willmove the Earth”
