Slide 1Fundamentals on RotationExample 10 – 1Rotational KinematicsAngular Displacement, Velocity, and AccelerationExample for Rotational KinematicsExample for Rotational Kinematics cnt’dRelationship Between Angular and Linear QuantitiesIs the lion faster than the horse?How about the acceleration?ExampleExample for Rotational MotionRolling Motion of a Rigid BodyMore Rolling Motion of a Rigid BodyTorqueExample for TorqueWednesday, Oct. 27, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu11. Fundamentals on Rotational Motion2. Rotational Kinematics3. Relationship between angular and linear quantities4. Rolling Motion of a Rigid Body5. TorquePHYS 1443 – Section 003Lecture #17Wednesday, Oct. 27, 2004Dr. Jaehoon Yu2nd Term Exam Monday, Nov. 1!! Covers CH 6 – 10.5!!No homework today!!Wednesday, Oct. 27, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu2Fundamentals on RotationLinear motions can be described as the motion of the center of mass with all the mass of the object concentrated on it. Is this still true for rotational motions?No, because different parts of the object have different linear velocities and accelerations.Consider a motion of a rigid body – an object that does not change its shape – rotating about the axis protruding out of the slide. One radian is the angle swept by an arc length equal to the radius of the arc.360Since the circumference of a circle is 2r,The relationship between radian and degrees isl =The arc length, or sergita, islRq =Therefore the angle, , is . And the unit of the angle is in radian. It is dimensionless!!rad 1rr /222/360/18057.3@o180 3.14@oRqWednesday, Oct. 27, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu3Example 10 – 1 A particular bird’s eyes can barely distinguish objects that subtend an angle no smaller than about 3x10-4 rad. (a) How many degrees is this? (b) How small an object can the bird just distinguish when flying at a height of 100m? (a) One radian is 360o/2. Thus43 10 rad-�l =( )43 10 rad-= � �( )360 2 radpo0.017=o(b) Since l=r and for small angle arc length is approximately the same as the chord length.rq =4100 3 10m rad-� � =23 10 3m cm-� =Wednesday, Oct. 27, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu4Rotational 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. 27, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu5Using 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.ififififtttttlim0dtdififtttttlim0dtdUnit? rad/sUnit? rad/sUnit? rad/s2Unit? rad/s2Wednesday, Oct. 27, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu6Example for Rotational KinematicsA wheel rotates with a constant angular acceleration of 3.50 rad/s2. If the angular speed of the wheel is 2.00 rad/s at ti=0, a) through what angle does the wheel rotate in 2.00s?ifUsing the angular displacement formula in the previous slide, one getstw= 200.250.32100.200.2 rad0.11.75.1.20.11revrev 212ta+Wednesday, Oct. 27, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu7Example for Rotational Kinematics cnt’dWhat is the angular speed at t=2.00s?tifUsing the angular speed and acceleration relationshipFind the angle through which the wheel rotates between t=2.00 s and t=3.00 s.2tq==srad /00.900.250.300.2 3tq==2rad8.10.72.1.28.10revrev if221ttUsing the angular kinematic formulaAt t=2.00sAt t=3.00sAngular displacement2.00 2.00�13.50 2.002+ �11.0rad=2.00 3.00�( )213.50 3.002+ �21.8rad=Wednesday, Oct. 27, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu8Relationship Between Angular and Linear QuantitiesWhat do we know about a rigid object that rotates about a fixed axis of rotation?When a point rotates, it has both the linear and angular motion components in its motion. What is the linear component of the motion you see?vEvery particle (or masslet) in the object moves in a circle centered at the axis of rotation.Linear velocity along the tangential direction.How do we related this linear component of the motion with angular component?l Rq=The arc-length is So the tangential speed v isWhat does this relationship tell you about the tangential speed of the points in the object and their angular speed?:Although every particle in the object has the same angular speed, its tangential speed differs proportional to its distance from the axis of rotation.The farther away the particle is from the center of rotation, the higher the tangential speed.The direction of follows a right-hand rule.dldt=( )drdtq=drdtq=rWednesday, Oct. 27, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu9Is the lion faster than the horse?A rotating carousel has one child sitting on a horse near the outer edge and another child on a lion halfway out from the center. (a) Which child has the greater linear speed? (b) Which child has the greater angular speed?(a) Linear speed is the distance traveled divided by the time interval. So the child sitting at the outer edge travels more distance within the given time than the child sitting closer to the center. Thus, the horse is faster than the lion.(b) Angular speed is the angle traveled divided by the time interval. The angle both the children travel in the given time interval is the same. Thus, both the horse and the lion have the same angular speed.Wednesday, Oct. 27, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu10How about the acceleration?v rw=TwoHow many different linear
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