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General Physics(ppt) Ch 11

# General Physics(ppt) Ch 11 - Chapter 11 Rolling Torque...

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Unformatted text preview: Chapter 11 Rolling, Torque, Angular Momentum Rolling Motion of A Rigid Object The linear speed of the center of mass for pure rolling motion is given by Consider a uniform cylinder of radius R rolling without slipping on a horizontal surface ω θ R dt d R dt ds v CM = = = The magnitude of the linear acceleration of the center of mass for pure rolling motion is α ω R dt d R dt dv a CM CM = = = Chapter 11 Rolling, Torque, Angular Momentum Rolling Motion of A Rigid Object Chapter 11 Rolling, Torque, Angular Momentum The total kinetic energy of the rolling cylinder 2 2 1 ω P I K = Where I p is the moment of inertia about a rotation axis through P. Applying the parallel-axis theorem: 2 MR I I CM P + = 2 2 2 2 1 2 1 ω ω MR I K CM + = 2 2 2 1 2 1 CM CM Mv I + = ω Rotational kinetic energy Translational kinetic energy Chapter 11 Rolling, Torque, Angular Momentum A sphere rolling down an incline. Mechanical energy is conserved if no slipping occurs. ω R v CM = For pure rolling motion 2 2 2 1 ) ( 2 1 CM CM CM Mv R v I K + = 2 2 R 2 2 ) ( 2 1 CM CM v M R I + = Mgh v M R I CM CM = + 2 2 ) ( 2 1 2 1 2 ) 1 2 ( MR I gh v CM CM + = Chapter 11 Rolling, Torque, Angular Momentum...
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General Physics(ppt) Ch 11 - Chapter 11 Rolling Torque...

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