# Physics 2

John Niclasen

## Contents

1. Symbols in Newtonian Mechanics
2. Week 1
2.1 Angular Momentum
2.2 Conservation of Angular Momentum
2.3 Torque and Angular Momentum Around an Axis
2.4 System of Particles
2.5 Center of Gravity
2.6 Angular Momentum Around the Center of Mass
2.7 Equations of Motion for a System of Particles
2.8 Kinetic Energy of a Rotating Disk
3. Week 2
3.1 Angular Momentum of an Arbitrary Rigid Body in Rotation Around a Fixed Axis
3.2 Calculation of the Moment of Inertia for Simple Bodies
3.3 Equation of Motion for a Rigid Body Rotating Around a Fixed Axis
3.4 The Angular Momentum Theorem Referred to Various Points

## 1. Symbols in Newtonian Mechanics

Symbol

Meaning

\bar A

Area

I

Moment of Inertia

\bar L

Angular Momentum

\bar N

Torque

T

Kinetic Energy

\bar \omega

Rotation vector

## 2. Week 1

### 2.1 Angular Momentum

We define the angular momentum \bar L relative to an origin O as:

\displaylines{\bar L\equiv \bar r\times \bar p=\bar r\times m\bar v}

Differentiate \bar L with respect to time:

\displaylines{{d\bar L\over dt}=\bar r\times \bar F}

The product \bar N=\bar r\times \bar F is called torque.

\displaylines{{d\bar L\over dt} =\bar N\cr \bar N=\bar r\times \bar F\cr }

 Angular momentum theorem The rate of change of the angular momentum of a particle around some point O equals the torque on the particle, with respect to O.

### 2.2 Conservation of Angular Momentum

 Theorem of conservation of angular momentum If no torque acts on a particle, the angular momentum of that particle with respect to O is constant in time.

 Central force field A central force field is a field where the force on the particle is always directed towards (or away from) a point fixed in an inertial frame. This point is called the force center.

Consider a particle being accelerated by a force of type \bar F=f(r)\bar e_r, where \bar e_r is a unit vector along the position vector \bar r relative to the coordinate center O, and f(r) is some function depending only on the magnitude of \bar F. Such a force han no torque around O:

\displaylines{\bar N\equiv \bar r\times \bar F=\bar r\times f(r)\bar e_r=0}

From this Kepler's second law emerge.

\displaylines{{d\bar A\over dt}={\bar r\times m\bar v\over 2m}={\bar L\over 2m}}

, where \bar A represents the area swept out by \bar r and m is the mass of the planet.

### 2.3 Torque and Angular Momentum Around an Axis

Torque and angular momentum with respect to a line is the projection of the vector on that line.

The torque \bar N\equiv \bar r\times \bar F, in cartesian coordinates, is

The projection of the torque \bar N on, say the z-axis, is the component N_z.

### 2.4 System of Particles

\displaylines{{d\over dt}\bar L^{tot}=\sum _i\bar N_i^{ext}}

 Angular Momentum Theorem for Particle Systems The time derivative of the total angular momentum for a particle system relative to a point O fixed in an inertial frame, equals the sum of the torques of the external forces, around O.

 No external torques \bar L^{tot} = constant vector: When the sum of external torques around a point O (fixed in an inertial frame) vanishes, the total angular momentum of the particle system relative to the point O is a constant of the motion.

 Closed System For a closed system the total angular momentum is conserved.

### 2.5 Center of Gravity

\displaylines{\bar N_O=\bar R_{CM}\times M\bar g}

### 2.6 Angular Momentum Around the Center of Mass

The total angular momentum around some point O is the angular momentum around CM plus "the angular momentum of CM" around the point O:

\displaylines{\bar L_O=\bar L_{CM}+\bar R_{CM}\times \bar P}

The angular momentum theorem with respect to the origin of the inertial frame is:

\displaylines{{d\bar L_O\over dt}=\bar N_O^{ext}}

Relative to the center of mass:

\displaylines{{d\bar L_{CM}\over dt}=\bar N_{CM}^{ext}}

 When the external torque about CM vanishes the spin-angular momentum of the system of particles (the body) is a constant of the motion.

### 2.7 Equations of Motion for a System of Particles

#### 2.7.2 Angular Momentum Theorem

\displaylines{{d\bar L_{CM}\over dt} =\sum \bar N_{CM}^{ext}}

### 2.8 Kinetic Energy of a Rotating Disk

A masss element m_i in the disk at the position \bar r_i has the velocity

\displaylines{\bar v_i=\bar \omega \times \bar r_i}

The kinetic energy is

\displaylines{T={1\over 2}I_z\omega ^2}

where

\displaylines{I_z=\sum _im_ir_i^2}

 Moment of Inertia The moment of inertia about an arbitrary axis z equals the moment of inertia about an axis parallel to z through the center of mass, plus the total mass M of the body times the square of the distance d between the two axes:\displaylines{I_z=I_{CM}+Md^2}

\displaylines{T={1\over 2}I_z\omega ^2={1\over 2}I_{CM}\omega ^2+{1\over 2}MR_{CM}^2\omega ^2}

Since v_{CM}=R_{CM}\omega :

\displaylines{T={1\over 2}I_{CM}\omega ^2+{1\over 2}Mv_{CM}^2=T_{rot}+T_{trans}}

## 3. Week 2

### 3.1 Angular Momentum of an Arbitrary Rigid Body in Rotation Around a Fixed Axis

\displaylines{\bar L_O=\sum _i\bar r_i\times m_i\bar v_i}

The kinetic energy is

\displaylines{T={1\over 2}\sum _im_i\dot r_i^2}

We find that

\displaylines{\,\vert\, \dot {\bar r_i}\,\vert\, =\,\vert\, \bar \omega \times \bar r_i\,\vert\, =\omega a_i}

and consequently

\displaylines{T={1\over 2}\left ( \sum _im_ia_i^2\right ) \omega ^2={1\over 2}I_z\omega ^2}

### 3.2 Calculation of the Moment of Inertia for Simple Bodies

A homogeneous thin rod has length L and total mass M. A small segment of length dx has the mass (M/L)dx. The segment, at distance x from the rotational axis A, will contribute to the moment of inertia with an amount

\displaylines{dI=x^2(M/L)dx}

The total moment of inertia:

\displaylines{I_A={M\over L}\int _0^Lx^2dx={1\over 3}ML^2}

If rotation is through center of mass:

\displaylines{I_{CM}={1\over 3}ML^2-M{L\overwithdelims () 2}^2={1\over 12}ML^2}

### 3.3 Equation of Motion for a Rigid Body Rotating Around a Fixed Axis

\displaylines{I_A={d\omega \over dt}=N_A}

 When the external torque around a given fixed axis A is zero, eht angular momentum about the axis A is a constant of the motion.

### 3.4 The Angular Momentum Theorem Referred to Various Points

\displaylines{\dot {\bar L}_Q=\bar N_Q-\bar v_Q\times M\bar v_{CM}}

The angular momentum theorem is valid in the simple form

\displaylines{\dot {\bar L}_Q=\bar N_Q}

whenever \bar v_Q\times \bar v_{CM}=\bar 0, that is, only if one of the following conditions are satisfied:

1. \bar v_Q=\bar 0 (Q is a point at rest in the inertial frame);
2. Q=CM (Q coincides with the center of mass);
3. \bar v_Q is parallel to \bar v_{CM}.

NicomDoc - 17-Nov-2006 - niclasen@fys.ku.dk