Chapter 11 Notes: Rolling, Torque, and Angular Momentum

Physics includes the study of rotation, notably in rolling motion of wheels.
Rolling motion can be simplified by treating it as a combination of translation of the center of mass and rotation around that center.
For smooth rolling (without slipping or bouncing), the center of the object moves in a straight line parallel to the surface.
$v{com} = wR$ where $v{com}$ is the linear speed of the wheel's center of mass, $w$ is the angular speed, and $R$ is the radius of the wheel.
The wheel can also be viewed as rotating instantaneously about the point P on the road in contact with the wheel; the angular speed about this point is the same as the angular speed about the center.
Rolling motion is a combination of purely translational and purely rotational motions. In pure rotation, every point rotates about the center with angular speed $w$ , and points on the edge have linear speed $v{com}$ . In pure translation, every point moves to the right with speed $v{com}$ .
In combined rolling motion, the bottom point (P) is stationary, and the top point (T) moves at $2v_{com}$ .

A smoothly rolling wheel has kinetic energy: $K = \frac{1}{2}I{com}w^2 + \frac{1}{2}Mv{com}^2$ where $I_{com}$ is the rotational inertia about its center of mass and $M$ is its mass.
If the wheel accelerates while rolling smoothly, $a_{com} = \alpha R$ , where $\alpha$ is the angular acceleration about the center.
For smooth rolling without sliding, mechanical energy is conserved to relate initial and later energy values.
When a net force acts on a rolling wheel, it causes acceleration $a_{com}$ and angular acceleration $\alpha$ , which tend to make the wheel slide. A frictional force opposes this tendency.
If the wheel does not slide, the frictional force is static friction $f_s$ , and the motion is smooth rolling.
If the wheel slides, kinetic friction $f_k$ acts at point P, and motion is not smooth rolling.
For a body rolling down a ramp of angle $\theta$ , the acceleration along the x-axis (up the ramp) is: $a{com,x} = -\frac{g \sin{\theta}}{1 + I{com}/MR^2}$

A yo-yo can be treated as a wheel rolling along an inclined plane at an angle $\theta = 90^\circ$ .
The linear acceleration of a yo-yo rolling down a string is: $a{com} = \frac{g}{1 + I{com}/MR0^2}$ , where $R0$ is the radius of the axle.

Torque is a vector quantity defined relative to a fixed point (usually an origin).
$\vec{\tau} = \vec{r} \times \vec{F}$ , where $\vec{F}$ is a force applied to a particle and $\vec{r}$ is the position vector locating the particle relative to the fixed point.
The magnitude of $\vec{\tau}$ is given by: $\tau = rF \sin{\phi} = rF\perp = r\perp F$ , where $\phi$ is the angle between $\vec{r}$ and $\vec{F}$ , $F\perp$ is the component of $\vec{F}$ perpendicular to $\vec{r}$ , and $r\perp$ is the moment arm of $\vec{F}$ .
The direction of $\vec{\tau}$ is given by the right-hand rule for cross products.

The angular momentum $\vec{l}$ of a particle with linear momentum $\vec{p}$ , mass $m$ , and linear velocity $\vec{v}$ is a vector quantity defined relative to a fixed point (usually an origin).
$\vec{l} = \vec{r} \times \vec{p} = m(\vec{r} \times \vec{v})$
The magnitude of $\vec{l}$ is given by: $l = rmv \sin{\phi} = rp\perp = rmv\perp = r\perp p = r\perp mv$ , where $\phi$ is the angle between $\vec{r}$ and $\vec{p}$ , $p\perp$ and $v\perp$ are the components of $\vec{p}$ and $\vec{v}$ perpendicular to $\vec{r}$ , and $r_\perp$ is the perpendicular distance between the fixed point and the extension of $\vec{p}$ .
The direction of $\vec{l}$ is given by the right-hand rule: Position your right hand so that the fingers are in the direction of $\vec{r}$ . Then rotate them around the palm to be in the direction of $\vec{p}$ . Your outstretched thumb gives the direction of $\vec{l}$ .

Newton's second law for a particle can be written in angular form as: $\vec{\tau}{net} = \frac{d\vec{l}}{dt}$ , where $\vec{\tau}{net}$ is the net torque acting on the particle and $\vec{l}$ is the angular momentum of the particle.

The angular momentum $\vec{L}$ of a system of particles is the vector sum of the angular momenta of the individual particles: $\vec{L} = \vec{l}1 + \vec{l}2 + … + \vec{l}n = \sum{i=1}^{n} \vec{l}_i$
The time rate of change of this angular momentum is equal to the net external torque on the system: $\vec{\tau}_{net} = \frac{d\vec{L}}{dt}$
For a rigid body rotating about a fixed axis, the component of its angular momentum parallel to the rotation axis is: $L = I\omega$

The angular momentum $\vec{L}$ of a system remains constant if the net external torque acting on the system is zero: $\vec{L} = constant$
This is the law of conservation of angular momentum: $Li = Lf$
If the component of the net external torque on a system along a certain axis is zero, then the component of the angular momentum of the system along that axis cannot change, no matter what changes take place within the system.
For a rigid body with changing rotational inertia: $$Ii\omega