Momentum of Inertia of some symmetric bodies 2018-08-07¢ Momentum of Inertia of some...
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Momentum of Inertia of some symmetric bodies
Rotational kinetic energy
It is called MOMENTUM of INERTIA
Rotation around a fixed axis
Momentum of Inertia of some rigid bodies
DISK of radius R and Mass M
DISK of radius R and Mass M
Parallel axis theorem
I = ICM + M h2
Work and Kinetic Energy
ΔK = W
E = K + mg z Mechanical energy is conserved
E = m vCM2 + I ω2 + mg z E is conserved throughout the motion
In particular , if the system of particles moves under the action of a conservative force, like the gravitational force for example, we obtain
Kf + mg zf = Ki + mg zi
APPENDIX Conservation of the Mechanical Energy Rigid Body Motion
While the ball is rolling down, what is the work done by all these 4 types of forces?
Instantaneous point of contact
Ftotal = N + Ff +
N : It acts on a different mi each time. It acts on the so called "point of contact." The only possible displacement of the point of contact would be perpendiculat to N. Therefore, the work done by N is zero.
Ff : It acts on the rolling ball through the instantaneous point of contact (ipc). If we assume no slipping, the instantaneous velocity of the ipc is zero. Therefore, there will not be kinetic friction; i.e. Ff does not do work. There is static friction, but such a force does not produce work
Fij Let's evaluate the work done by the internal forces. Consider two arbitrary small masses mi and mj. There will be a couple of action and reaction forces Fij and Fji acting on these two particles. Let's calculate the work done by this couple of internal forces
Since we are dealing with a rigid body, the magnitude of the vector ( ri - rj ) remains constant throughout the motion.
Then, the only possible orientation of d(ri - rj ) is to be perpendicular to the line that passes through mi and mj .
d(ri - rj ) ( ri - rj ) d(ri - rj ) Fij
- mi g ( zfinal, i - zinitial, i )
regardless of the tra-
If we extend this argument to all pairs of particles of the rigid body we conclude: The work done by the internal forces in a rigid body is zero
Thus, the only force that produces work on the rolling block is the gravitational force
We already know that regardless of the trajectory followed by the "small" particle mi the work done by the gravitational force is,
As the block rolls down, a particular "small" mass mi undergoes a complicated trajectory.
mi gzfinal, i mi gzinitial, i )(
Adding up all the work done on each particle, one obtains
W = M g zCM, final - M g zCM, initial )
H = 2m
Practice problem A 1 m long rod of mass m = 0.2 Kg is hinged at one end and
connected to a wall. It is held out horizontally , then released. With respect to the reference XY, whose origin is at the hinge (as shown in the figure), calculate:
The speed of the tip of the rod at the moment it hits the wall? The kinetic energy of the rod at the moment it hits the wall?
L= 1 m
A thin uniform rod of length L = 2.0 m and can pivot about a horizontal, frictionless pin through one end. It is released from rest at angle q = 40o above the horizontal. Use the principle of conservation of mechanical energy to determine the angular speed of the rod as it passes through the horizontal position.