Chapter 9 Rotational Dynamics - University of Georgia：9章旋转动力学-乔治亚大学.pptVIP

Chapter 11: Rotational Dynamics Newton’s 2nd Law for Rotational Motion Example Problem Rotational Work * * As we did for linear (or translational) motion, we studied kinematics (motion without regard to the cause) and then dynamics (motion with regard to the cause), we now proceed in a similar fashion We know that forces are responsible for linear motion We will now see that rotational motion is caused by torques Consider a wrench of length r F Torque Units of N m Not work or energy, do not use Joules is called the lever arm. Must always be perpendicular to the force If the force is not perpendicular to the level arm, we need to find the component that is perpendicular (either the force or the lever arm) ? F F sin? If ?=0, then the torque is zero Therefore, torques (force times length) are responsible for rotational motion F=Ft r m The torque for a point-particle of mass m a distance r from the rotation axis is Axis of rotation Define I = mr2 = Moment of Inertia for a point particle; a scalar, units of kg m2 A rigid body is composed of many, many particles of mass mi which are ri from the axis of rotation Each of these masses creates a torque about the axis of rotation m1 m2 r1 r2 Rotation axis Sum up all torques due to all particles is the same for all particles I =moment of inertia for the rigid body. It is different for different shaped objects and for different axes of rotation. Table 10-1. For a thin rod of mass M and length L L/2 L The last equation is Newton’s 2nd Law for rotation. Compare to the translational form Example Problem A rotating door is made of 4 rectangular panes each with a mass of 85 kg. A person pushes on the outer edge of one pane with a force of 68 N, directed perpendicular to the pane. Determine the door’s ?. L F Given: L = 1.2 m, mpane=85 kg, F=68 N From Table 10-1, moment of inertia for a thin rectangular rod is (same as a thin sheet) Since there are 4 panes. The parallel axis theorem prov