PHYS 1443 – Section 003 Lecture #17Angular Displacement, Velocity, and AccelerationRotational KinematicsRolling Motion of a Rigid BodyMore Rolling Motion of a Rigid BodyTorqueExample for TorqueTorque & Angular AccelerationExample for Torque and Angular AccelerationMoment of InertiaExample for Moment of InertiaRotational Kinetic EnergyCalculation of Moments of InertiaExample for Rigid Body Moment of InertiaParallel Axis TheoremExample for Parallel Axis TheoremTorque and Vector ProductProperties of Vector ProductMore Properties of Vector ProductSimilarity Between Linear and Rotational MotionsWednesday, Oct. 29, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu1PHYS 1443 – Section 003Lecture #17Wednesday, Oct. 29, 2002Dr. Jaehoon Yu1. Rolling Motion of a Rigid Body2. Torque3. Moment of Inertia4. Rotational Kinetic Energy5. Torque and Vector ProductsRemember the 2nd term exam (ch 6 – 11), Monday, Nov. 3!Wednesday, Oct. 29, 2003 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu2Using what we have learned in the previous slide, how would you define the angular displacement?Angular Displacement, Velocity, and AccelerationHow about the average angular speed?And the instantaneous angular speed?By the same token, the average angular accelerationAnd the instantaneous angular acceleration?When rotating about a fixed axis, every particle on a rigid object rotates through the same angle and has the same angular speed and angular acceleration.ififififtttttlim0dtdififtttttlim0dtdWednesday, Oct. 29, 2003 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu3Rotational KinematicsThe first type of motion we have learned in linear kinematics was under a constant acceleration. We will learn about the rotational motion under constant angular acceleration, because these are the simplest motions in both cases.fJust like the case in linear motion, one can obtainAngular Speed under constant angular acceleration:Angular displacement under constant angular acceleration:fOne can also obtain 2fti221ttii ifi 22Wednesday, Oct. 29, 2003 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu4Rolling Motion of a Rigid BodyWhat is a rolling motion?To simplify the discussion, let’s make a few assumptionsLet’s consider a cylinder rolling without slipping on a flat surfaceA more generalized case of a motion where the rotational axis moves together with the objectUnder what condition does this “Pure Rolling” happen?The total linear distance the CM of the cylinder moved isThus the linear speed of the CM isA rotational motion about the moving axis1. Limit our discussion on very symmetric objects, such as cylinders, spheres, etc2. The object rolls on a flat surfaceR ss=RRs dtdsvCMCondition for “Pure Rolling”dtdRRWednesday, Oct. 29, 2003 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu5More Rolling Motion of a Rigid BodyAs we learned in the rotational motion, all points in a rigid body moves at the same angular speed but at a different linear speed.At any given time the point that comes to P has 0 linear speed while the point at P’ has twice the speed of CMThe magnitude of the linear acceleration of the CM isA rolling motion can be interpreted as the sum of Translation and RotationCMaWhy??PP’CMvCM2vCMCM is moving at the same speed at all times.PP’CMvCMvCMvCM+PP’CMv=Rv=0v=R=PP’CM2vCMvCMdtdvCMdtdRRWednesday, Oct. 29, 2003 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu6TorqueTorque is the tendency of a force to rotate an object about an axis. Torque, , is a vector quantity.Magnitude of torque is defined as the product of the force exerted on the object to rotate it and the moment arm.FdLine of ActionConsider an object pivoting about the point P by the force F being exerted at a distance r. PrMoment armThe line that extends out of the tail of the force vector is called the line of action. The perpendicular distance from the pivoting point P to the line of action is called Moment arm.When there are more than one force being exerted on certain points of the object, one can sum up the torque generated by each force vectorially. The convention for sign of the torque is positive if rotation is in counter-clockwise and negative if clockwise. d2F2212211dFdF sinrF FdWednesday, Oct. 29, 2003 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu7R1Example for TorqueA one piece cylinder is shaped as in the figure with core section protruding from the larger drum. The cylinder is free to rotate around the central axis shown in the picture. A rope wrapped around the drum whose radius is R1 exerts force F1 to the right on the cylinder, and another force exerts F2 on the core whose radius is R2 downward on the cylinder. A) What is the net torque acting on the cylinder about the rotation axis?The torque due to F1111FRSuppose F1=5.0 N, R1=1.0 m, F2= 15.0 N, and R2=0.50 m. What is the net torque about the rotation axis and which way does the cylinder rotate from the rest?R2F1F2and due to F2222FRUsing the above result21So the total torque acting on the system by the forces is2211FRFR The cylinder rotates in counter-clockwise.2211FRFR mN - 5.250.00.150.10.5Wednesday, Oct. 29, 2003 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu8Torque & Angular AccelerationLet’s consider a point object with mass m rotating on a circle.What does this mean?The tangential force Ft and radial force FrThe tangential force Ft isWhat do you see from the above relationship?mrFtFrWhat forces do you see in this motion?ttmaF The torque due to tangential force Ft isrFtITorque acting on a particle is proportional to the angular acceleration.What law do you see from this relationship?Analogs to Newton’s 2nd law of motion in rotation.How about a rigid object?rdFtdmOThe external tangential force dFt istdFThe torque due to tangential force Ft isThe total torque isdWhat is the contribution due to radial force and why?Contribution from radial force is 0, because its line of action passes through the pivoting point, making the moment arm 0.mrrmat2mrItdmadmrrdFt dmr2dmr2IWednesday, Oct. 29, 2003 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu9Example for Torque and Angular
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