Physics 101: Lecture 17
Rotational Dynamics
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Review of the last lecture on Rotational Kinematics and
Rolling Motion and the Vector Nature of Angular Variables
Today’s lecture will cover Textbook Sections 9.1 - 9.3
Physics 101: Lecture 17, Pg 1
But first for something completely different …
The Quantum and the Cosmos
Today, 4:30 pm,
By Rocky Kolb
114 Hochstetter
Dark matter ?
How did our universe evolve out of a formless,
Dark Energy ?
shapeless quantum soup of “elementary” particles ?
Physics 101: Lecture 17, Pg 2
See text: chapter 8
Rotation Summary
(with comparison to 1-D linear kinematics)
Angular
Linear
  constant
a  constant
ω  ω0  αt
v  v 0  at
1
  0   0 t  t 2
2
x  x 0  v 0t 
 2   02  2Δθ
v2  v20  2ax
1 2
at
2
And for a point at a distance R from the rotation axis:
x = Rv = R
a = R
See Table 8.1
Physics 101: Lecture 17, Pg 3
Rolling Motion
Rolling motion involves both linear and rotational
motion.
If there is no slipping at the point of contact with
the ground, the motion of an automobile tire is an
example for rolling motion.
While the car is moving with a linear speed v
covering a distance d, a point on the outer edge of
the tire moves the same distance along the circular
path s =  r = d.
Thus, the linear speed v=d/t of the car is the same
as the tangential speed of a point on the outer edge
on the tire vT=s/t=/t r =  r:
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v = vT =  r ( in rad/s)
Physics 101: Lecture 17, Pg 4
Vector Nature of Angular Variables
Like linear velocity and acceleration, also
angular velocity and acceleration are vector
quantities.
So far we only talked about the magnitude
of these vectors. But as vectors they also
have a direction. Both angular velocity and
acceleration point along the rotation axis.
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As in linear motion, if the velocity is
decreasing, the acceleration points opposite
to the angular velocity and if the velocity is
increasing, velocity and acceleration point
in the same direction.
Physics 101: Lecture 17, Pg 5
Rotational Dynamics
Remember Newtons first and second laws:
An object does not change its state of motion unless
a net force is applied. If a net force is applied on an
object with mass m it experiences an acceleration:
a = Fnet/m or Fnet=m a
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What is the equivalent of force and mass (inertia) in
rotational motion of a rigid object ?
Physics 101: Lecture 17, Pg 6
See text: chapter 9
New Concept: Torque
Torque is the rotational analog of force:
Torque = (magnitude of force) x (lever arm)
t=Fl
SI unit: [N m]
The lever arm is the distance between the axis of
rotation and the “line of action”, i.e. the line drawn
along the direction of the force.
The torque is positive, if resulting into a counter
clockwise rotation and negative for a
clockwise rotation.
Physics 101: Lecture 17, Pg 7
Concept Question
The picture below shows three different ways of using a wrench
to loosen a stuck nut. Assume the applied force F is the same in
each case.
In which of the cases is the torque on the nut the biggest?
1. Case 1
2. Case 2
3. Case 3
CORRECT
Physics 101: Lecture 17, Pg 8
Rigid Objects in Equilibrium
A system is in equilibrium if and only if:
 a = 0  S Fext = 0
  = 0  Stext = 0 (about any axis)
pivot
Torque about pivot due to
gravity:
tg = m g d
d
Center of gravity
W=-mg
(gravity acts at center of gravity)
This object is NOT in equilibrium.
Physics 101: Lecture 17, Pg 9
Center of Gravity
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The center of gravity of a rigid body is the point at
which the gravitational force is considered to act when
the torque due to gravity (weight) is calculated.
If an object is of symmetric shape and the weight is
distributed uniformly, the center of gravity is located at
the geometrical center of the object.
If the object is made of a group of objects with known
weights, then its center of gravity is determined
by the sum of the torques around an arbitrary axis:
xcg =(W1 x1 + W2 x2 +W3 x3 + …)/(W1+W2+W3+..)
xcg is the point at which the weight W1+W2+W3+… acts.
Physics 101: Lecture 17, Pg 10
pivot
Not in equilibrium
Equilibrium
pivot
d
Center of gravity
W=-mg
Center of gravity
Torque about pivot  0
Torque about pivot = 0
A method to find center of gravity of an irregular object
Physics 101: Lecture 17, Pg 11
Torque
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Painter on plank (uniform) which sits on two supports.
Free body diagram for plank:
FA
FB
-Mg
-mg
To compute the net torque, choose a rotation axis, e.g. let’s consider the
pivot is located at support A.
Physics 101: Lecture 17, Pg 12
Torque
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If its just balancing on “B”, then FA = 0
the only forces on the beam are:
x
FB
Mg
Using FTOT = 0:
FB = Mg + mg
mg
This does not tell us x
Physics 101: Lecture 17, Pg 13
Torque
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Find net torque around pivot B: (or any other place)
FB
d1
Mg
d2
mg
t (FB ) = 0 since lever arm is 0
t (Mg ) = Mgd1
t (mg ) = -mgd2
Total torque = 0 = Mgd1 -mgd2
So d2 = Md1 /m and you can use d1 to find x
Physics 101: Lecture 17, Pg 14