Physics 101: Lecture 17 Rotational Dynamics. 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. But first for something completely different ….
Rolling Motion and the Vector Nature of Angular Variables
The Quantum and the Cosmos
By Rocky Kolb
Today, 4:30 pm,
Dark matter ?
How did our universe evolve out of a formless,
shapeless quantum soup of “elementary” particles ?
Dark Energy ?
And for a point at a distance R from the rotation axis:
Rotation Summary (with comparison to 1-D linear kinematics)
See Table 8.1
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 = q 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=q/t r = w r:
v = vT = w r (w in rad/s)
angular velocity and acceleration are vector
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.
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.
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
What is the equivalent of force and mass (inertia) in
rotational motion of a rigid object ?
New Concept: Torque
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
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.
Center of gravity
Center of gravity
Torque about pivot = 0
Torque about pivot 0
Not in equilibrium
A method to find center of gravity of an irregular object
To compute the net torque, choose a rotation axis, e.g. let’s consider the
pivot is located at support A.
Using FTOT = 0: FB = Mg + mg This does not tell us x
t(FB ) = 0 since lever arm is 0
t(Mg) = Mgd1
Total torque = 0 = Mgd1-mgd2
t(mg) = -mgd2
So d2 = Md1 /mand you can use d1 to find x