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

    KROT =

    =

    It is called MOMENTUM of INERTIA

    Rotation around a fixed axis

    ω M

    ( )

  • Momentum of Inertia of some rigid bodies

    DISK of radius R and Mass M

    DISK of radius R and Mass M

  • Parallel axis theorem

    CM M

    I = ICM + M h2

  • X

    YCM

    yi

    xiRi

    a

    a

  • Work and Kinetic Energy

  • ΔK = W

    z

    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

    m

  • 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?

    Internal forces

    Instantaneous point of contact

    m1m2 m3

    Ftotal = N + Ff +

    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.

    N

    N i

    g

    Rigid body

  • 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

    ij

    g

    mi

    mj

    The

  • =

    =

    + +

    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 .

    )

    Therefore,

    = 0

    d(ri - rj ) ( ri - rj ) d(ri - rj ) Fij

    ) )

    )

  • - mi g ( zfinal, i - zinitial, i )

    regardless of the tra-

    zfinal, i

    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.

    zinitial, i

  • mi gzfinal, i mi gzinitial, i )(

    Adding up all the work done on each particle, one obtains

    -(

    W =

    W = M g zCM, final - M g zCM, initial )

    zCM, final

    zCM, initial

  • Z

    H = 2m

    vi=0

    vf= ?

    zCM

  • 1

    1

  • 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?

    vtip

    X

    Y

    L= 1 m

    Wall

  • 2

    1

  • 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.

    40o

    Solution

    40o