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Physics 2211: Lecture 36

Physics 2211: Lecture 36. Rotational Dynamics and Torque. Linear. Angular. Summary (with comparison to 1-D kinematics). And for a point at a distance R from the rotation axis:. s = R  v = R   a = R . Rotation & Kinetic Energy.

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Physics 2211: Lecture 36

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  1. Physics 2211: Lecture 36 • Rotational Dynamics and Torque

  2. Linear Angular Summary (with comparison to 1-D kinematics) And for a point at a distance R from the rotation axis: • s = Rv = R a = R

  3. Rotation & Kinetic Energy • The kinetic energy of a rotating system looks similar to that of a point particle:Point Particle Rotating System v is “linear” velocity m is the mass.  is angular velocity I is the moment of inertia about the rotation axis.

  4. Newton’s 2nd Law in the direction: m  Rotational Dynamics • Suppose a force acts on a mass constrained to move in a circle. Consider its acceleration in the direction at some instant. • Multiply by r :

  5. Define torque: •  is the tangential force Ftimes the distancer. m  Rotational Dynamics • Torque has a direction: • +zif it tries to make the systemturn CCW. • - z if it tries to make the systemturn CW.

  6. Fr F  rp Rotational Dynamics rp = “distance of closest approach” or “lever arm”

  7. m4 m1 a m3 m2 Rotational Dynamics • For a collection of many particles arranged in a rigid configuration: • Since the particles are connected rigidly,they all have the same .

  8. Rotational Dynamics • This is the rotational analog of FNET = ma • Torque is the rotational analog of force: • The amount of “twist” provided by a force. • Moment of inertia I is the rotational analog of mass, i.e., “rotational inertial.” • If I is big, more torque is required to achieve a given angular acceleration. • Torque has units of kg m2/s2 = (kg m/s2) m = N-m.

  9. Comment on=I • When we write =I we are really talking about the z component of a more general vector equation. (More on this later.) We normally choose the z-axis to be the rotation axis.) z=Izz • We usually omit the zsubscript for simplicity. z Iz z z

  10. y x  z Torque and the Right Hand Rule • The right hand rule can tell you the direction of torque: • Point your hand along the direction from the axis to the point where the force is applied. • Curl your fingers in the direction of the force. • Your thumb will point in the directionof the torque.

  11. • The length of is given by: • C = AB sin • The direction of is perpendicular to the plane defined by and and the “sense” of the direction is defined by the right hand rule. The Cross (or Vector) Product • We can describe the vectorial nature of torque in a compact form by introducing the “cross product”. • The cross product of two vectors is a third vector:

  12. + + + The Cross Product • Cross product of unit vectors:

  13. or Note: The Cross Product • Cartesian components of the cross product:

  14. y x z Torque & the Cross Product • So we can define torque as: X = rY FZ - FY rZ = yFZ - FY z Y = rZ FX - FZ rX = zFX - FZ x Z = rX FY - FX rY = xFY - FX y

  15. m2 m1 m3 y x (total mass) m4 (In this case, N = 4) Center of Mass Revisited • Define the Center of Mass (“average” position): • For a collection of N individual pointlike particles whose masses and positions we know:

  16. + + m2 m1 m2 m1 System of Particles: Center of Mass • The center of mass is where the system is balanced! • Building a mobile is an exercise in finding centers of mass. • Therefore, the “center of mass” is the “center of gravity” of an object.

  17. System of Particles: Center of Mass • For a continuous solid, we have to do an integral. dm y where dm is an infinitesimal element of mass. x

  18. y x System of Particles: Center of Mass • We find that the Center of Mass is at the “mass-weighted” center of the object. The location of the center of mass is an intrinsic property of the object!! (it does not depend on where you choose the origin or coordinates when calculating it).

  19. pivot + CM pivot pivot + + CM CM mg System of Particles: Center of Mass • The center of mass (CM) of an object is where we can freely pivot that object. • Force of gravity acts on the object as though all the mass were located at the CM of the object. (Proof coming up!) • If we pivot the objectsomewhere else, it willorient itself so that theCM is directly below the pivot. • This fact can be used to findthe CM of odd-shaped objects.

  20. System of Particles: Center of Mass • Hang the object from several pivots and see where the vertical lines through each pivot intersect! pivot pivot pivot + CM • The intersection point must be at the CM.

  21. Torque on a body in a uniform gravitation field • What is the torque exerted by the force of gravity on a body of total mass M about the origin? origin

  22. Equivalent Torques Torque on a body in a uniform gravitation field origin origin

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