Monday, Nov. 29, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 1 PHYS 1441 – Section 002 Lecture #21 Monday, Nov. 29, 2010 Dr. Jaehoon Yu • Rolling Motion of a Rigid Body • Rotational Dynamics – Torque – Moment of Inertia – Torque and Angular Acceleration • Rotational Kinetic Energy • Angular Momentum Today’s homework is homework #12, due 10pm, Thursday, Dec. 9!!Announcements • The Final Exam – Date and time: 11am, Monday Dec. 13 – Place: SH103 – Comprehensive exam • Covers from CH1.1 – what we finish Wednesday, Dec. 8 • Plus appendices A.1 – A.8 • Combination of multiple choice and free response problems • Bring your Planetarium extra credit sheet to the class next Wednesday, Dec. 8, with your name clearly marked on the sheet! • No more quizzes and no more home works! • Colloquium this Wednesday Monday, Nov. 29, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 2Monday, Nov. 29, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 4 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=Rθ s = vCM=ΔsΔtThe condition for a “Pure Rolling motion”Monday, Nov. 29, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 5 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 Why?? P P’ CM vCM 2vCM CM is moving at the same speed at all times. P P’ CM vCM vCM vCM + P P’ CM v=Rω v=0 v=Rω = P P’ CM 2vCM vCMMonday, Nov. 29, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 6 Starting from rest, the car accelerates for 20.0 s with a constant linear acceleration of 0.800 m/s2. The radius of the tires is 0.330 m. What is the angle through which each wheel has rotated? Ex. An Accelerating Car θ α ω ωo t -2.42 rad/s2 0 rad/s 20.0 s ?Monday, Nov. 29, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 7 Torque Torque is the tendency of a force to rotate an object about an axis. Torque, τ, 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. F φ"l1 The line of Action Consider an object pivoting about the point P by the force F being exerted at a distance r from P. P r Moment arm The 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 the 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. l2 F2 Unit?Monday, Nov. 29, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 8 The tendon exerts a force of magnitude 790 N on the point P. Determine the torque (magnitude and direction) of this force about the ankle joint which is located 3.6x10-2m away from point P. Ex. The Achilles Tendon 790 N First, let’s find the lever arm length So the torque is Since the rotation is in clock-wise 3.6x10-2mMonday, Nov. 29, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 9 Moment of Inertia Rotational Inertia: What are the dimension and unit of Moment of Inertia? kg ⋅ m2ML2⎡⎣⎤⎦Measure of resistance of an object to changes in its rotational motion. Equivalent to mass in linear motion. Determining Moment of Inertia is extremely important for computing equilibrium of a rigid body, such as a building. For a group of objects For a rigid body miri2i∑ r2∫dmDependent on the axis of rotation!!!Monday, Nov. 29, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 10 Two particles each have mass m1 and m2 and are fixed at the ends of a thin rigid rod. The length of the rod is L. Find the moment of inertia when this object rotates relative to an axis that is perpendicular to the rod at (a) one end and (b) the center. Ex. The Moment of Inertia Depends on Where the Axis Is. (a) (b) mL2 m 0( )2+ m L( )2= mr2( )∑= m1r12+ m2r22 mr2( )∑= m1r12+ m2r22 m L 2( )2+ m L 2( )2= 12mL2Which case is easier to spin? Case (b) Why? Because the moment of inertia is smallerMonday, Nov. 29, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 11 Example for Moment of Inertia In a system of four small spheres as shown in the figure, assuming the radii are negligible and the rods connecting the particles are massless, compute the moment of inertia and the rotational kinetic energy when the system rotates about the y-axis at angular speed w. Since the rotation is about y axis, the moment of inertia about y axis, Iy, is Thus, the rotational kinetic energy is Find the moment of inertia and rotational kinetic energy when the system rotates on the x-y plane about the z-axis that goes through the origin O. x y This is because the rotation is done about y axis, and the radii of the spheres are negligible. Why are some 0s? M M l l m m b b O = mii∑ri2 = Ml2= 2Ml2=12Iω2=122Ml2( )ω2= Ml2ω2= mii∑ri2 = Ml2= 2 Ml2+ mb2( )=12Iω2=122Ml2+ 2mb2( )ω2= Ml2+ mb2( )ω2 + Ml2 +m ⋅ 02 +m ⋅ 02 + Ml2 +mb2 +mb2Monday, Nov. 29, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 12 Check out Figure 8 – 21 for moment of inertia for various shaped objectsMonday, Nov. 29, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 13 Torque & Angular Acceleration Let’s consider a point object with mass m rotating on a circle. What does this mean? The tangential force Ft and the radial force Fr The tangential force Ft is What do you see from the above relationship? m r Ft Fr What forces do you see in this motion? The torque due to tangential force Ft is Torque
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