Physics 1408-00 Principles of Physics Lecture 17 Chapter 10 March 1, 009 Sung-Won Lee Sungwon.Lee@ttu.edu Chapter 10 Announcement I Lecture note is on the web Handout (6 slides/page) http://highenergy.phys.ttu.edu/~slee/1408/ *** Class attendance is strongly encouraged and will be taken randomly. Also it will be used for extra credits. HW Assignment #7 will be placed on MateringPHYSICS today, and is due by 11:59pm on Wendseday, 3/5 (after spring break) 10-1 Angular Quantities Every point on a rotating body has an angular velocity and a linear velocity v. Rotational Motion They are related: Angular Quantities Constant Angular Acceleration Torque Rotational Dynamics; Torque and Rotational Inertia Determining Moments of Inertia Rotational Kinetic Energy Rotational Plus Translational Motion; Rolling Why Does a Rolling Sphere Slow Down? 10-1 Angular Quantities If the angular velocity of a rotating object changes, it has a tangential acceleration also: Even if the angular velocity is constant, each point on the object has a centripetal (radial) acceleration: A point P on a rotating wheel has a linear velocity v at any moment. 10-1 Angular Quantities Here is the correspondence between linear and rotational quantities:
10-3 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. Rotational Kinetic Energy An object rotating about some axis with an angular speed,, has rotational kinetic energy Each particle has a kinetic energy of K i = m i v i Since the tangential velocity depends on the distance, r, we can substitute v i = I r The total rotational kinetic energy of the rigid object is the sum of the energies of all its particles where I is called the moment of inertia" Moment of Inertia The definition of moment of inertia is Moments of Inertia of Various Rigid Objects The dimensions of moment of inertia: [kg. m ] We can rewrite the expression for I in terms of m With the small volume segment assumption, Some examples see next slides The Parallel-Axis Theorem The I depends on the rotation axis. Suppose you need to know the I for rotation about the off-center axis in Fig. If axis of interest is distance d from a parallel axis through the center of mass, the I is I = I + Md cm The parallel-axis theorem 10.4 Torque The ability of a force to cause a rotation or twisting motion depends on 3 factors: 1.The magnitude F of the force..the distance r from the point of application to the pivot. 3.The angle at which the force is applied. Try to make these ideas specific. Fig shows a force F applied at one point on a rigid body. We define a new quantity called torque, denoted by the symbol : = rf sin" where " is the angle from r to F.
Torque 10.5 Torque and Angular Acceleration A longer lever arm is very helpful in rotating objects. We ve found that toque is the rotational equivalent of force. we need to learn what toque does. at = r F = ma = mr t t Now Fig shows a model rocket engine attached to one end of lightweight, rigid rod. Tangential & angular a are related. Tangential component of thrust. Multiplying both sides by r: rft = mr ( mr ) = " = rf t Therefore, torque causes angular acceleration. The relation is analogous to F=ma. Newton s nd Law for Rotation Fig: A rigid body undergoes pure rotational motion about a fixed, frictionless, and unmoving axis. Net torque on the object is the sum of the torques on all the individual particles. net = ) i = )( miri ") = # mir $ % ) i & " i i ' i ( Definition: Moment of Inertia I m r + m r + m r + = " m r 1 1 3 3 i i i I net " = or net = I" Newton s nd law for rotational motion Work in Rotational Motion Find the work done by F on the object as it rotates through an infinitesimal distance ds = rd# dw = F. ds = (F sin $) r d# dw = %d# The radial component of F does no work because it is perpendicular to the displacement Power in Rotational Motion The rate at which work is being done in a time interval "t is This is analogous to P = Fv in a linear system Work-Kinetic Energy Theorem in Rotational Motion The work-kinetic energy theorem for rotational motion states that the net work done by external forces in rotating a symmetrical rigid object about a fixed axis equals the change in the object s rotational kinetic energy
Summary of Useful Equations Rolling Object The red curve shows the path moved by a point on the rim of the object. This path is called a cycloid The green line shows the path of the center of mass of the object In pure rolling motion, an object rolls without slipping In such a case, there is a simple relationship between its rotational and translational motions Rolling Object, Center of Mass The velocity of the center of mass: Rotational Kinetic Energy The kinetic energy of a rotating object is given by By substituting the rotational quantities, we find that the rotational kinetic energy can be written: The acceleration of the center of mass: A object that has both translational and rotational motion also has both translational and rotational kinetic energy: Total Kinetic Energy of a Rolling Object The total kinetic energy of a rolling object is the sum of the translational energy of its center of mass and the rotational kinetic energy about its center of mass K = 1/ I CM w + 1/ Mv CM Accelerated rolling motion is possible only if friction is present between the sphere and the incline K f + U f = K i + U i K f = 1/ (I CM /R ) v CM +1/ Mv CM U i = Mgh U f = K i = 0 10-8 Rotational Kinetic Energy The kinetic energy of a rotating object is given by By substituting the rotational quantities, we find that the rotational kinetic energy can be written: A object that both translational and rotational motion also has both translational and rotational kinetic energy:
10-8 Rotational Kinetic Energy The Great Downhill Race When using conservation of energy, both rotational and translational kinetic energy must be taken into account. All these objects have the same potential energy at the top, but the time it takes them to get down the incline depends on how much rotational inertia they have A sphere, a cylinder, and a hoop, all of mass M and radius R, are released from rest and roll down a ramp of height h and slope #. They are joined by a particle of mass M that slides down the ramp without friction. Who wins the race? Who is the big loser? v I + Mv = Mgh 1 1 cm cm = v R Icm = cmr cm / Mgh = ( cmr )( v / R) + Mv cm Energy Conservation: Kf = Ui Depends on the object s geometry 1 1 cm cm = M (1 + c) v 1 cm The Winners gh = 1+ c Finishing speed of an object w/ I Therefore, v > v > v > v particle sphere cylinder hoop and a > a > a > a particle sphere cylinder hoop Chapter 11 Angular Momentum General Rotation Angular Quantities Angular Momentum Objects Vector Cross Product; Torque as a Vector Angular Momentum of a Particle Angular Momentum and Torque Conservation of Angular Momentum The Spinning Top and Gyroscope Rotating Frames of Reference; Inertial Forces; The Coriolis Effect 11-1 Angular Momentum Objects The rotational analog of linear momentum is angular momentum, L: Then the rotational analog of Newton s second law is: This form of Newton s second law is valid even if I is not constant. 11-1 Angular Momentum Objects In the absence of an external torque, angular momentum is conserved: dl = 0 and L = I = constant. dt More formally, the total angular momentum of a rotating object remains constant if the net external torque acting on it is zero.
11-1 Angular Momentum Objects This means: Therefore, if an object s moment of inertia changes, its angular speed changes as well. A skater doing a spin on ice, illustrating conservation of angular momentum: (a) I is large and is small; (b) I is smaller so is larger. A diver rotates faster when arms and legs are tucked in than when they are outstretched. Angular momentum is conserved. 11-1 Angular Momentum Objects Angular momentum is a vector; for a symmetrical object rotating about a symmetry axis it is in the same direction as the angular velocity vector. A person on a circular platform, both initially at rest, begins walking along the edge at speed v. The platform begins rotating in the opposite direction, so that the total angular momentum remains zero, as shown in (b). 11- Vector Cross Product; Torque as a Vector Torque can be defined as the vector product of the force and the vector from the point of action of the force to the axis of rotation: The torque due to the force F (in the plane of the wheel) starts the wheel rotating counterclockwise so and " point out of the page. 11- Vector Cross Product; Torque as a Vector For a particle, the torque can be defined around a point O: Here, r is the position vector from the particle relative to O. The Vector Product & Torque There are instances where the product of vectors is another vector. Earlier we saw where the product of vectors was a scalar (this was called the dot product) The vector product of vectors is also called the cross product. The torque vector lies in a direction perpendicular to the plane formed by the position vector and the force vector " = r x F The torque is the vector (or cross) product of the position vector and the force vector The Vector Product Defined Given two vectors, A and B Vector (cross) product of A and B is defined as a third vector, C C (=A x B) is read as A cross B The magnitude of C = AB sin# is the angle b/w A and B The direction of C is perpendicular to the plane formed by A and B The best way to determine this direction is to use the right-hand rule (see Fig)
Properties of the Vector Product The vector product is not commutative. A x B = - B x A If A is parallel to B (# = 0 o or 180 o ), then A x B = 0 Therefore A x A = 0 If A is perpendicular to B, then A x B = AB The vector product obeys the distributive law A x (B + C) = A x B + A x C 11- Vector Cross Product; Torque as a Vector Some properties of the cross product: The derivative of the cross product with respect to some variable such as t : where it is important to preserve the multiplicative order of A, B Vector Products of Unit Vectors Using Determinants The cross product can be expressed as" Expanding the determinants gives" Signs are interchangeable in cross products" A x (-B) = - A x B " ˆ i (" ˆj ) = " ˆi ˆj Cross Product with Unit Vectors A = iˆ + 3 ˆj B = iˆ + 3 ˆj + kˆ Calculate A B. iˆ ˆj kˆ A B = 3 0 3 = iˆ (6 " 0) + ˆj (0 " 4) + kˆ (6 " 6) = 6iˆ " 4 ˆj " =?" Torque Vector Example Given the force" F = (.00ˆi + 3.00ˆ) j N r = (4.00ˆi + 5.00ˆ) j m j"