Chapter 9: Rotational MotionAngular Quantities“R” from the Axis (O)Linear and Angular QuantitiesKinematical EquationsChapter 10: Rotational Motion (II)Angular Quantities: VectorRotational Dynamics: tNote: t = F R sinqNote: sign of tRotational Dynamics: ISlide 12Parallel-axis TheoremSlide 14Example 1Example 1 (cont’d)Rotational Motion Chapter 9: Rotational MotionRigid body instead of a particleRotational motion about a fixed axisRolling motion (without slipping)Rotational Motion Angular QuantitiesKinematical variables to describe the rotational motion:Angular position, velocity and acceleration)()()(20 t0 trad/s dtdt lim rad/s dtdt lim rad Rl aveaveRotational Motion “R” from the Axis (O)Solid Disk Solid CylinderRotational Motion Linear and Angular QuantitiesRRRRaRaRaRRll Rl 222radtantan( (3) (2))( (1))ddddvv vvtttt like t )(atanaradRotational Motion Kinematical EquationsconstantNote ax Conversion :)(:02020200-2 (3) (2)21 (1) , , t t tvRotational Motion Chapter 10: Rotational Motion (II)Rigid body instead of a particleRotational motion about a fixed axisRotational dynamicsRolling motion (without slipping)Rotational Motion Angular Quantities: VectorKinematical variables to describe the rotational motion:Angular position, velocity and accelerationVector natures)()()(2rad/s k dtdkrad/s k dtdkrad Rl ˆˆ ˆˆ xyzR.-H. RuleRotational Motion Rotational Dynamics: xxamcos Fcos F a )(magnitude (torque) lF I l Fl F l F;mI(a)(b)lblaaxRotational Motion Note: = F R sin...) distanceular (Perpendic : R or l arm Lever R F sin R F )( RFR sin F )(Rotational Motion Note: sign of (c.w.) mN 6.7 mN 6.7- m)N (21.7 -m)N (15.0 )1()1( )c.w.()c.c.w(2121 netmN 21.7 m)(0.866) N)(0.500 (50.06022 sin R F )(2mN 15.0 m) N)(0.300 (50.09011 sin R F )(1Rotational Motion Rotational Dynamics: Iparticles) of group afor particle) singlefor inertia of(moment inertia of(moment )( (torque) )( 2ii22Rm I R mI R m RF R ma m F(b)(a) m1m2m3Rotational Motion Rotational Dynamics: Ic.m.) about the intertia of(moment cmII theorem)axis-(parallel 2cmMdII dRotational Motion Parallel-axis Theoremd202020 2cmMR23MRMR21MdII Rotational Motion 2222l M314lMl M 2lM I I 121fgParallel-axis TheoremRotational Motion Example 1Calculate the torque on the 2.00-m longbeam due to a 50.0 N force (top) about (a) point C (= c.m.)(b) point PCalculate the torque on the 2.00-m longbeam due to a 60.0 N force about (a) point C (= c.m.)(b) point PCalculate the torque on the 2.00-m longbeam due to a 50.0 N force (bottom) about (a) point C (= c.m.)(b) point PRotational Motion Example 1 (cont’d)Calculate the net torque on the 2.00-mlong beam about(a) point C (= c.m.)(b) point
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