Lecture-XIII. Angular momentum and Fixed axis rotation

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1 Lecture-XIII Angular momentum and Fixed axis rotation

2 General rigid body motion The general motion of a rigid body of mass m consists of a translation of the center of masswithvelocity andarotationaboutthecenterofmasswithallelementsofthe rigid body rotating with the same angular velocity. Chasles theorem. The center of mass of a thrown rigid rod follows a parabolic traectory while the rod rotates about the center of mass.

3 MotionInvolvingBothTranslationandRotation Consider the body to be an aggregation of N particles with masses m (J = 1,2,..., N) and position vectors r with respect to an inertial coordinate system. The angularmomentumofthebodycanbewrittenas R r and where Mis the total mass and Ris CM position. The first term represents the angular momentum due to the center of mass motion. The second termrepresentsangularmomentumduetomotionaroundthecenterofmass.theonlywayfor the particles of a rigid body to move with respect to the center of mass is for the body as a whole to rotate.

4 Axis of rotation parallel to z-axis Consider motion for which the axis of rotation remains parallel to the z axis. Taking z-component of L, I 0 is the moment of inertia of the body about an axis in the z-direction passing through the center of mass. r = ρ ˆ ρ + z kˆ, r = ρ ωφˆ ˆ ρ ˆ φ kˆ r r = ρ 0 z 0 ρ ω 0 2 ( ) ˆ z = ρ ω ρ + ρ ωkˆ Thus the angular momentum of a rigid obect is the sum of the angular momentum about its center of mass and the angular momentum of the center of mass about the origin.

5 Torque and Angular momentum The first term in is the torque about the center of mass due to the various external forces, and the second term is the torque due to the total external force acting at the center of mass. For fixed axis rotation it can be written where τ 0 is the z component of the torque about the center of mass. But from L z we have

6 Drum Rolling down a Plane Choose a coordinate system whose origin A is on the plane.thetorqueaboutais Alternate 1: Alternate 2: Chosen the origin at the point of contact.

7 Therollingwheel:AwheelofmassM,radiusRandmomentofinertiaIitsaxispassing through the center of mass is spun about this axis with a constant angular velocity ω 0. It is then released in an upright position on a horizontal plane. The coefficient of friction between the wheel and the surface is µ. It slips for a time τ and then rolls without slipping. Find τ and the velocity of the center of mass when slipping stops. R ω(t) F(t)=µMg V(t) x Mx = µ Mg, I ω = µ MgR µ MgR v( t) = µ gt, ω( t) = ω0 t I v µ MgR µ gτ No slipping condition: ω=, ω 0 τ = R I R ω0r ω0r τ =, v( t) = µ gτ = for t > τ 2 2 MR MR µ g I I R

8 Kinetic energy in fixed axis rotation The kinetic energy of a body undergoing pure rotation is given by: Kinetic energy for motions in which translation and rotation occur simultaneously, 1 2 K = m v, 2 v ˆ = V + ρ ωφ v v = V + ρ ω + 2ρ ωφˆ V = V + ρ ω + 2r V K = m V + m ρ ω + m r V Since, m = M, m ρ = I0, m r = K = m V + I0ω 2 2 r = ρ ˆ ρ + z kˆ, r = ρ ωφˆ R r

9 The Work-energy Theorem The work-energy theorem for a particle motion was where One needs to generalize this for a rigid body motion. The work-energy theorem for rigid body motion divides into two parts, one dealing with translational energy and one dealing with rotational energy. To derive the translational part, start with the equation of motion for the center of mass. or To evaluate the work associated with the rotational kinetic energy, start with the equation of motion for fixed axis rotation about the center of mass. or The integral on the left represents the work done by the applied torque. The general work-energy theorem for a rigid body is therefore: where and W ba is the total work done on the body.

10 The Falling Stick A stick of length l and mass M, initially upright on a frictionless table, starts falling. The problem is to find the speedofthecenterofmassasafunctionofposition. Since there are no horizontal forces, the center of mass must fall straight down. As the stick rotated through angle θ, say, the center of mass has fallen distance y. The initial energy is: At a later time: and Constraint equation: or or or

11 Drum Rolling down a Plane: Energy Method The energy equation for the translational motion: The energy equation for the rotational motion: Friction is not dissipative here. Friction decreases the translational energy by an amount fl. However, the torque exerted by friction increases the rotational energy by the same amount. In this motion, friction simply transforms mechanical energy from one mode to another. If slipping occurs, this is no longer the case and some of the mechanical energy is dissipated as heat.

12 Three cylinders: ( ) ( ) ( ) ( ) ( ) I Ω Ω = J 1, I Ω Ω = J + J 2, I Ω Ω = J (3) Since J = J, J = J ( ) ( ) ( ) I ( ) : 3Ω Ω = 0, Ω = Ω IΩ 2 K 2 Ω 1 = = K 1 = 2 Ω 9 3 I Ω 2

PHYSICS 111 HOMEWORK SOLUTION #10. April 8, 2013

PHYSICS 111 HOMEWORK SOLUTION #10. April 8, 2013 PHYSICS HOMEWORK SOLUTION #0 April 8, 203 0. Find the net torque on the wheel in the figure below about the axle through O, taking a = 6.0 cm and b = 30.0 cm. A torque that s produced by a force can be