1. Lecture 36, Page 1 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Physics 2211: Lecture 36 l Rotational Dynamics and Torque

2. Lecture 36, Page 2 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm 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  Angular Linear

3. Lecture 36, Page 3 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Rotation & Kinetic Energy Point Particle Rotating System l 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. Lecture 36, Page 4 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm l Suppose a force acts on a mass constrained to move in a circle. Consider its acceleration in the direction at some instant. l Multiply by r : l Newton’s 2nd Law in the direction:  m Rotational Dynamics

5. Lecture 36, Page 5 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm l Torque has a direction: ç+ z if it tries to make the system turn CCW. ç- z if it tries to make the system turn CW. l Define torque:   is the tangential force F  times the distance r.  m Rotational Dynamics

6. Lecture 36, Page 6 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm r p = “distance of closest approach” or “lever arm” FF FrFr   rprp Rotational Dynamics

7. Lecture 36, Page 7 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm l For a collection of many particles arranged in a rigid configuration: Since the particles are connected rigidly, they all have the same . m4m4 m1m1 m2m2 m3m3  Rotational Dynamics

8. Lecture 36, Page 8 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm l This is the rotational analog of F NET = ma l Torque is the rotational analog of force: ç The amount of “twist” provided by a force. Moment of inertia I is the rotational analog of mass, i.e., Moment of inertia I is the rotational analog of mass, i.e., “rotational inertial.” “rotational inertial.”  If I is big, more torque is required to achieve a given angular acceleration. l Torque has units of kg m 2 /s 2 = (kg m/s 2 ) m = N-m. Rotational Dynamics

9. Lecture 36, Page 9 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm 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 = I z  z l We usually omit the z subscript for simplicity. Comment on  = I  z zz zz IzIz

10. Lecture 36, Page 10 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Torque and the Right Hand Rule l The right hand rule can tell you the direction of torque: çPoint your hand along the direction from the axis to the point where the force is applied. çCurl your fingers in the direction of the force. çYour thumb will point in the direction of the torque. x y z 

11. Lecture 36, Page 11 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm l We can describe the vectorial nature of torque in a compact form by introducing the “cross product”. çThe cross product of two vectors is a third vector: The Cross (or Vector) Product l The direction of is perpendicular to the plane defined by and and the “sense” of the direction is defined by the right hand rule.  l The length of is given by: C = AB sin 

12. Lecture 36, Page 12 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm l Cross product of unit vectors: The Cross Product + + +

13. Lecture 36, Page 13 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm The Cross Product l Cartesian components of the cross product:  Note: or

14. Lecture 36, Page 14 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Torque & the Cross Product l So we can define torque as: x y z  X = r Y F Z - F Y r Z = y F Z - F Y z  Y = r Z F X - F Z r X = z F X - F Z x  Z = r X F Y - F X r Y = x F Y - F X y

15. Lecture 36, Page 15 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Center of Mass Revisited l Define the Center of Mass (“average” position): çFor a collection of N individual pointlike particles whose masses and positions we know: (In this case, N = 4) y x m1m1 m4m4 m2m2 m3m3 (total mass)

16. Lecture 36, Page 16 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm System of Particles: Center of Mass l The center of mass is where the system is balanced! çBuilding a mobile is an exercise in finding centers of mass. çTherefore, the “center of mass” is the “center of gravity” of an object. m1m1 m2m2 + m1m1 m2m2 +

17. Lecture 36, Page 17 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm System of Particles: Center of Mass l For a continuous solid, we have to do an integral. y x dm where dm is an infinitesimal element of mass.

18. Lecture 36, Page 18 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm System of Particles: Center of Mass 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). y x l We find that the Center of Mass is at the “mass-weighted” center of the object.

19. Lecture 36, Page 19 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm System of Particles: Center of Mass l The center of mass (CM) of an object is where we can freely pivot that object. l Force of gravity acts on the object as though all the mass were located at the CM of the object. (Proof coming up!) l If we pivot the object somewhere else, it will orient itself so that the CM is directly below the pivot. l This fact can be used to find the CM of odd-shaped objects. + CM pivot + CM pivot + CM mg

20. Lecture 36, Page 20 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm System of Particles: Center of Mass l Hang the object from several pivots and see where the vertical lines through each pivot intersect! pivot + CM l The intersection point must be at the CM.

21. Lecture 36, Page 21 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Torque on a body in a uniform gravitation field l What is the torque exerted by the force of gravity on a body of total mass M about the origin? origin

22. Lecture 36, Page 22 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Torque on a body in a uniform gravitation field origin Equivalent Torques