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Phys1441-spring08-04 - PHYS 1441 Section 002 Lecture#21 Monday Apr 14 2008 Dr Jaehoon Yu Rolling Motion Rotational Dynamics Torque Equilibrium

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Monday, Apr. 14, 2008 PHYS 1441-002, Spring 2008 Dr. Jaehoon Yu 1 PHYS 1441 – Section 002 Lecture #21 Monday, Apr. 14, 2008 Dr. Jae hoon Yu Rolling Motion Rotational Dynamics Torque Equilibrium Moment of Inertia Torque and Angular Acceleration Rotational Kinetic Energy Today’s homework is HW #11, due 9pm, Wednesday, Apr. 23!!

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Monday, Apr. 14, 2008 PHYS 1441-002, Spring 2008 Dr. Jaehoon Yu 2 Announcements •3 rd term exam – Next Monday, Apr. 21, in class – Covers: Ch. 6.7 – what we complete this Wednesday, Apr. 16 – This is the final term exam in the semester – The worst of the three term exams will be dropped from the final grading • Quiz results – Class average: 2/6 • Equivalent to 33/100 • How did you do before?: 50, 44 and 47 – Top score: 4.2/6
Monday, Apr. 14, 2008 PHYS 1441-002, Spring 2008 Dr. Jaehoon Yu 3 Rolling Motion of a Rigid Body What is a rolling motion? To simplify the discussion, let’s make a few assumptions Let’s consider a cylinder rolling on a flat surface, without slipping. A more generalized case of a motion where the rotational axis moves together with an object Under what condition does this “Pure Rolling” happen? The total linear distance the CM of the cylinder moved is Thus the linear speed of the CM is A rotational motion about a moving axis 1. Limit our discussion on very symmetric objects, such as cylinders, spheres, etc 2. The object rolls on a flat surface R θ s s=R θ s = CM s v t Δ = Δ The condition for a “Pure Rolling motion” R t θ Δ = Δ ω R = R

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Monday, Apr. 14, 2008 PHYS 1441-002, Spring 2008 Dr. Jaehoon Yu 4 More Rolling Motion of a Rigid Body As we learned in rotational motion, all points in a rigid body moves at the same angular speed but at different linear speeds. At any given time, the point that comes to P has 0 linear speed while the point at P’ has twice the speed of CM The magnitude of the linear acceleration of the CM is A rolling motion can be interpreted as the sum of Translation and Rotation CM a Why?? P P’ CM v CM 2v CM CM is moving at the same speed at all times. P P’ CM v CM v CM v CM + P P’ CM v=R ω v=0 v=R ω = P P’ CM 2v CM v CM CM v t Δ = Δ R t ω Δ = Δ α R =
Monday, Apr. 14, 2008 PHYS 1441-002, Spring 2008 Dr. Jaehoon Yu 5 Starting from rest, the car accelerates for 20.0 s with a constant linear acceleration of 0.800 m/s 2 .

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This note was uploaded on 10/21/2009 for the course PHYS phys1441 taught by Professor Dr.jhaehoonyou during the Spring '08 term at UT Arlington.

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Phys1441-spring08-04 - PHYS 1441 Section 002 Lecture#21 Monday Apr 14 2008 Dr Jaehoon Yu Rolling Motion Rotational Dynamics Torque Equilibrium

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