Chapter 8- Rotational Motion

Chapter 8- Rotational Motion

Assignment 8 Textbook (Giancoli, 6 th edition), Chapter 7-8: Due on Thursday, November 13, 2008 - Problem 28 - page 189 of the textbook - Problem 40 - page 190 of the textbook - Problems 7 and 17 - page 219 of the textbook

Old assignments and midterm exams (solutions have been posted on the web) can be picked up in my office (LB-212)

All marks, including assignments, have been posted on the web. http://ilc2.phys.uregina.ca/~barbi/academic/phys109/marks.pdf Please, verify that all your marks have been entered in the list.

Chapter 8 Angular Quantities Constant Angular Acceleration Rolling Motion (Without Slipping) Torque Rotational Dynamics; Torque and Rotational Inertia Collisions in Two or Three Dimensions Rotational Kinetic Energy Angular Momentum and Its Conservation Vector Nature of Angular Quantities

Recalling Last Lecture

Momentum and Its Relation to Force Momentum is a vector symbolized by the symbol p, and is defined as (7-20) The momentum of an object tells how hard (or easy) is to change its state of motion.

Momentum and Its Relation to Force (7-22) Eq. 6-22 is another way of expressing Newton s second law. However, it is a more general definition because it introduces the situation where the mass may change.

Collision and Impulse their surrounding environment, and act for a very short period of time t. We can use eq. 6-22 and define the impulse on an object as: (7-23)

Conservation of Momentum (7-25) Equation 6-25 tells that the total momentum of the system (the sum of the momentum of the two balls) before the collision is equal to the total momentum of the system after the collision IF the net force acting on the system is zero isolated system. This is known as Conservation of Total Momentum. The above equation can be extended to include any number of objects such that the only forces are the interaction between the objects in the system.

Conservation of Energy and Momentum in Collisions In general, we can identify two different types of collisions: 1. Elastic collision 2. Inelastic collision In elastic collisions the total kinetic energy of a system is conserved. There is no energy dissipate in form of heat or other form of energy. An example is the collision between the two billiard balls discussed in the previous slides: (7-26) In inelastic collision, there is NO conservation of kinetic energy. Some of the total initial kinetic energy is transformed into some other form of energy.

Note: Conservation of Energy and Momentum in Collisions Also The total energy (the sum of all energies) in a closed (isolated system) is ALWAYS conserved.

Conservation of Energy and Momentum in Collisions Collision in two or more dimensions We then have: (i) (ii) It follows then, using the above expressions in (i) and (ii), that: This gives a system of two equations and three variables. if you can measure any of these variables, the other two can be calculated from system of equations.

Today

Linear Momentum Problem 7-34 (textbook) An internal explosion breaks an object, initially at rest, into two pieces, one of which has 1.5 times the mass of the other. If 7500 J were released in the explosion, how much kinetic energy did each piece acquire?.

Linear Momentum Problem 7-34 Use conservation of momentum in one dimension, since the particles will separate and travel in opposite directions. Call the direction of the heavier particle s motion the positive direction. Let A represent the heavier particle, and B represent the lighter particle. We have m A = 1.5m B v = v = 0 A B m v m B B 2 p = p 0 = m v + m v v = = v initial final A A B B A 3 B The negative sign indicates direction. Since there was no mechanical energy before the explosion, the kinetic energy of the particles after the explosion must equal the energy added. ( ) 2 ( 1.5 )( ) ( ) E = KE + KE = m v + m v = m v + m v = m v = KE 1 2 1 2 1 2 1 2 5 1 2 5 added A B 2 A A 2 B B 2 B 3 B 2 B B 3 2 B B 3 B KE = E = 7500 J = 4500 J KE = E KE = 7500 J 4500 J = 3000 J 3 3 B 5 added 5 A added B A Thus: KE = = A 3 3 3.0 10 J KE 4.5 10 J B

Angular Quantities We have extensively discussed translational (linear) motion of an object in terms of its kinematics (displacement), dynamics (forces) and energy. But we also know that objects can also move following some circular path. This is called rotational motion. The basis for the discussions on rotational motion is what we have seen so far concerning translational motion. So, I will use the definitions introduced in the previous chapters and apply then to introduce you to rotational motion. We will consider only rigid objects in other words, objects that do not change shape (or the distances between points in it do not change)

Angular Quantities In purely rotational motion, all points on the object move in circles around the axis of rotation ( O ) which is perpendicular to this slide. The radius of the circle is r. All points on a straight line drawn through the axis move through the same angle in the same time. The angle θ in radians is defined: (8-1) Where r = radius of the circle l = arc length covered by the angle θ The angular displacement is what characterizes the rotational motion. For mathematical reasons, it is more convenient

Angular Quantities For mathematical reasons, it is more convenient to define angle not in terms of degrees but in terms of radians. One radian is defined such that it corresponds to an arc of circle equal to the radius of the circle. Or, if we use eq. 8.1: Note that radians are dimensionless. Radian can be related to degrees by observing the the full length of a circle corresponds to the maximum arc length, or 2πr. It comprises an angle of 360 0. Using eq. 8.1, we have: (8-2)

Angular Quantities We can also define an object revolution as the length in radians that the object has travelled. A complete revolution will correspond to the total length of the circle, or: (8-3) As mentioned before, the object rotational displacement is defined in terms of the angle θ. Defining a coordinate system as in the figure, the displacement of a certain point P in the object can be given by: (8-4)

Angular Quantities In a similar way we did for translational motion, we can define the average angular velocity and instantaneous angular velocity of this point as: (8-5) (8-6) Similarly, the average acceleration and instantaneous acceleration can be defined as: (8-7) (8-8) Note that given the fact that each point in the object will be displaced by the same angle in the same interval of time, both the velocity and acceleration are the same for any point in the object.

Angular Quantities Now, the rotational motion of an object or a point in the object can be related to its translational motion. For realizing that, you should observe that a point rotating around a circle will also be subjected to a translational motion as depicted in the figure. At each angular position, this point will have a linear velocity whose directions are tangent to its circular path. Note: the direction of the linear velocity changes as the point undergoes a rotational motion. This is due to the so called centripetal acceleration. We will come back to this acceleration in the next slides.

Angular Quantities Back to the linear velocity, the figure shows that a change in the rotation angle θ corresponds to a linear distance traveled l. With the help of eq. 8-1: (8-9)

Angular Quantities (8-9) Eq. 8-9 says that although ω is the same for every point in the rotating object, the linear velocity changes with the distance r of the point to the axis of rotation. Therefore, objects farther from the axis of rotation will move faster. If there is an angular acceleration α, the angular velocity ω changes. Therefore, there is also a change in the linear velocity and thus an acceleration involved in the process. This acceleration is in the direction of the velocity and is called tangential linear acceleration: (8-10)

Angular Quantities As already mentioned a couple of slides ago, the velocity changes direction. But the tangential acceleration is parallel to the velocity, so it is not responsible for the change in the velocity s direction. It turns out that this acceleration is called radial acceleration, or centripetal acceleration. It is always perpendicular to the direction of the velocity. The magnitude of the centripetal acceleration is given by magnitude (see textbook, page 107 for details): (8-11) (8-12) Therefore, objects farther from the axis of rotation will have greater centripetal acceleration.

Angular Quantities The total acceleration of the point at a distance r from the axis of the rotation of the object will be the vector sum the radial (centripetal) and linear (tangential) accelerations: (8-13) We can summarize this discussion with the following table:

Angular Quantities Another important quality in rotational motion the frequency of rotation of an object. The frequency is the number of complete revolutions per second: (8-14) Frequencies are measured in hertz. The time required for a complete revolution is called period, or in other words: period is the time one revolution takes: (8-15)

Constant Angular Acceleration The equations of motion for constant angular acceleration are the same as those for linear motion, with the substitution of the angular quantities for the linear ones.

Linear Momentum Problem 8-8 (textbook) A rotating merry-go-round makes one complete revolution in 4.0 s (Fig. 8 38). (a) What is the linear speed of a child seated 1.2 m from the center? (b) What is her acceleration (give components)?

Linear Momentum Problem 8-8 The angular speed of the merry-go-round is 2 π rad 4.0 s= 1.57 rad s (a) v ( )( ) = ωr = 1.5 7 ra d s e c 1.2 m = 1.9 m s (b) Ignoring air our other resistance, there is no tangential acceleration (no tangential forces are applied).therefore, the acceleration is purely radial. a and a tan = 0 ω 2 ( ) ( ) 2 2 = r = 1.57 rad sec 1.2 m = 3.0 m s towards the center R

Linear Momentum Problem 8-13 (textbook) A turntable of radius R 1 is turned by a circular rubber roller of radius R 2 in contact with it at their outer edges. What is the ratio of their angular velocities, ω 1 / ω 2.

Linear Momentum Problem 8-13 The tangential speed of the turntable must be equal to the tangential speed of the roller, if there is no slippage. v = v ω R = ω R ω ω = R R 1 2 1 1 2 2 1 2 2 1

Linear Momentum Problem 8-19 (textbook) A cooling fan is turned off when it is running at 850 rev/min. It turns 1500 revolutions before it comes to a stop. (a) What was the fan s angular acceleration, assumed constant? (b) How long did it take the fan to come to a complete stop?

Linear Momentum Problem 8-19 (a) The angular acceleration can be found from ω = ω + 2αθ 2 2 o ( ) 2 π 2 ( ) 2 2 ω ω 0 850 rev min rev 2 rad 1 min rad o α = = = 241 = 0.42 2 2 2θ 2 1500 rev min 1 rev 60 s s (b) The time to come to a stop can be found from ( ) θ = ω + ω t 1 t 2 o ( ) 2θ 2 1 5 0 0 rev 6 0 s = = = ω + ω 8 5 0 rev m in 1 m in o 2 1 0 s