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UT Arlington PHYS 1443 - PHYS 1443 – Section 001 Lecture #15

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Thursday, June 30, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 1 PHYS 1443 – Section 001 Lecture #15 Thursday, June 30, 2011 Dr. Jaehoon Yu • Relationship Between Angular and Linear quantities • Torque and Vector Product • Moment of Inertia • Calculation of Moment of Inertia • Torque and Angular Acceleration Today’s homework is homework #8, due 10pm, Monday, July4!!Thursday, June 30, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 2 Reminder: Extra-Credit Special Project • Derive the formula for the final velocity of two objects which underwent an elastic collision as a function of known quantities m1, m2, v01 and v02 in page 8 of lecture note on Tuesday, June 28, in a far greater detail than in the note. – 20 points extra credit • Show mathematically what happens to the final velocities if m1=m2 and explain in detail in words the resulting motion. – 5 point extra credit • NO Credit will be given if the process is too close to the note! • Due: Start of the class Tuesday, July 5Extra Credit: 2-D Collisions • Proton #1 with a speed 5.0x106 m/s collides elastically with proton #2 initially at rest. After the collision, proton #1 moves at an angle of 37o to the horizontal axis and proton #2 deflects at an angle φ to the same axis. Find the final speeds of the two protons and the scattering angle of proton #2, φ. This must be done in much more detail than the book or on page 13 of lecture note on Tuesday, June 28. • 10 points • Due beginning of the class Wednesday, July 6 Thursday, June 30, 2011 3 PHYS 1443-001, Summer 2011 Dr. Jaehoon YuThursday, June 30, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 4 !f=Rotational Kinematics The first type of motion we have learned in linear kinematics was under the constant acceleration. We will learn about the rotational motion under constant angular acceleration, because these are the simplest motions in both cases. Just like the case in linear motion, one can obtain Angular velocity under constant angular acceleration: Angular displacement under constant angular acceleration: !f=One can also obtain !f2= !0+ !0+ !02+ 2" #f$#0( ) !t !0t + 12!t2 v =Linear kinematics Linear kinematics xf= x0+ vot +12at2Linear kinematics vf2= vo2+ 2a xf! xi( ) vo+ atThursday, June 30, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 5 Ex. 10 – 4: Rotational Kinematics A 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? Using the angular displacement formula in the previous slide, one getsThursday, June 30, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 6 Example for Rotational Kinematics cnt’d What is the angular speed at t=2.00s? Using the angular speed and acceleration relationship Find the angle through which the wheel rotates between t=2.00s and t=3.00s. Using the angular kinematic formula At t=2.00s At t=3.00s Angular displacementThursday, June 30, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 7 Relationship Between Angular and Linear Quantities What 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 components in its motion. What is the linear component of the motion in the figure? Every particle (or masslet) in the object moves on a circle centered at the same axis of rotation. Linear velocity along the tangential direction. How do we related this linear component of the motion with angular component? The arc-length is So the tangential speed v is What 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 and is 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 ω follows the right-hand rule.Thursday, June 30, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 8 Is 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.Thursday, June 30, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 9 How about the acceleration? Two How many different linear acceleration components do you see in a circular motion and what are they? Total linear acceleration is Since the tangential speed v is What does this relationship tell you? Although every particle in the object has the same angular acceleration, its tangential acceleration differs proportional to its distance from the axis of rotation. Tangential, at, and the radial acceleration, ar. The magnitude of tangential acceleration at is The radial or centripetal acceleration ar is What does this tell you? The father away the particle is from the rotation axis, the more radial acceleration it receives. In other words, it receives more centripetal force. =v2r=rω( )2r= rω2= at2+ ar2= rα( )2+ rω2( )2= rα2+ω4Thursday, June 30, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 10 Example (a) What is the linear speed of a child seated 1.2m from the center of a steadily rotating merry-go-around that makes one complete revolution in 4.0s? (b) What is her total linear acceleration? First, figure out what the angular speed of the merry-go-around is. Using the formula for linear speed Since the angular speed is constant, there is no angular acceleration. Tangential acceleration is Radial acceleration is Thus the total acceleration is = 1.2m × 1.6rad / s = 1.9m / s = 1.2m × 0rad / s2= 0m / s2 = rω2Thursday, June 30, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 11 Example for Rotational Motion Audio information on compact discs


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