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Physics 2211: Lecture 36 Rotational Dynamics and Torque Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Lecture 36, Page 1 Summary (with comparison to 1-D kinematics) Linear Angular a constant constant v v 0 at 0 t s s0 v 0t 21 at 2 0 0t 21 t 2 v 2 v 02 2a s s0 2 02 2 0 And for a point at a distance R from the rotation axis: s = Rv = R Physics 2211 Spring 2005 © 2005 Dr. Bill Holm a = R Lecture 36, Page 2 Rotation & Kinetic Energy The kinetic energy of a rotating system looks similar to that of a point particle: Point Particle K 1 2 mv 2 v is “linear” velocity m is the mass. Rotating System 1 2 K I 2 is angular velocity I is the moment of inertia about the rotation axis. I mi ri 2 i Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Lecture 36, Page 3 Rotational Dynamics Suppose a force acts on a mass constrained to move in a circle. Consider its acceleration in the ˆ direction at some instant. Newton’s 2nd Law in the ˆ direction: a F ma r F mra Physics 2211 Spring 2005 © 2005 Dr. Bill Holm r y Multiply by r : F F m ˆ x Lecture 36, Page 4 rˆ Rotational Dynamics Define torque: r F is the tangential force F times the distance r. Torque has a direction: + z if it tries to make the system turn CCW. - z if it tries to make the system turn CW. F F a r y m ˆ x Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Lecture 36, Page 5 rˆ Rotational Dynamics r F r F sin r sin F F r F rp = “distance of closest approach” or “lever arm” F Fr r rp Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Lecture 36, Page 6 Rotational Dynamics ri Fi mi ri i 2 For a collection of many particles arranged in a rigid configuration: i i i Since the particles are connected rigidly, they all have the same . 2 m r i ii i m4 i I F4 NET I Physics 2211 Spring 2005 © 2005 Dr. Bill Holm r4 m3 F3 r1 F1 m1 r2 m2 r3 F2 Lecture 36, Page 7 Rotational Dynamics NET I 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. Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Lecture 36, Page 8 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 = Izz z Iz z We usually omit the z subscript for simplicity. Physics 2211 Spring 2005 © 2005 Dr. Bill Holm z Lecture 36, Page 9 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 direction of the torque. F y x r Physics 2211 Spring 2005 © 2005 Dr. Bill Holm z Lecture 36, Page 10 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: B AB C The length of C is given by: C = AB sin A C The direction of C is perpendicular to the plane defined by A and B and the “sense” of the direction is defined by the right hand rule. Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Lecture 36, Page 11 The Cross Product Cross product of unit vectors: iˆ i jˆ jˆ kˆ kˆ 0 iˆ ˆj kˆ ˆj iˆ kˆ kˆ i jˆ iˆ kˆ jˆ ˆj kˆ i kˆ ˆj i iˆ + + + k̂ Physics 2211 Spring 2005 © 2005 Dr. Bill Holm ĵ Lecture 36, Page 12 The Cross Product Cartesian components of the cross product: C A B Ax iˆ Ay j Az kˆ Bx iˆ By j Bz kˆ Cx Ay Bz By Az B Cy Az Bx Bz Ax Cz Ax By Bx Ay or iˆ C Ax Bx jˆ Ay By kˆ Az Bz A C Note: B A A B Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Lecture 36, Page 13 Torque & the Cross Product So we can define torque as: r F r F sin X = rY FZ - FY rZ = Y = rZ FX - FZ rX = Z = rX FY - FX rY = F y FZ - FY z z FX - FZ x x FY - FX y r y z Physics 2211 Spring 2005 © 2005 Dr. Bill Holm x Lecture 36, Page 14 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: m2 N RCM m r m1 M r1 i 1 i i r2 RCM m3 y N M mi (total mass) i 1 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm m4 r4 x r3 (In this case, N = 4) Lecture 36, Page 15 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. m1 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm + m2 + m1 m2 Lecture 36, Page 16 System of Particles: Center of Mass For a continuous solid, we have to do an integral. dm y r x Physics 2211 Spring 2005 © 2005 Dr. Bill Holm RCM rdm rdm dm M where dm is an infinitesimal element of mass. Lecture 36, Page 17 System of Particles: Center of Mass We find that the Center of Mass is at the “mass-weighted” center of the object. y RCM Physics 2211 Spring 2005 © 2005 Dr. Bill Holm x 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). Lecture 36, Page 18 System of Particles: Center of Mass The center of mass (CM) of an object is where we can freely pivot that object. pivot + CM 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 object somewhere else, it will orient itself so that the CM is directly below the pivot. This fact can be used to find the CM of odd-shaped objects. Physics 2211 Spring 2005 © 2005 Dr. Bill Holm pivot CM pivot + CM mg Lecture 36, Page 19 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. Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Lecture 36, Page 20 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? M y dF dm g r origin dm x z d r dF r dm g rdm g 1 rdm g rdm g rdm Mg rCM Mg M r dF rCM Mg Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Lecture 36, Page 21 Torque on a body in a uniform gravitation field M y origin dm dF r x rCM y M origin x rCM z z r dF rCM Mg Mg Equivalent Torques Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Lecture 36, Page 22