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