PHYS 218 Chapter 9

« previous | Monday, March 28, 2011 | next »

Rotation of Rigid bodies

Examples:

• CDs
• Fans
• Circular Saws
• Wheels
• Gyroscopes

Rotating objects also have energy. We can express this as a function of the angle (in radians) it makes with a coordinate axis centered at the axis of rotation.

Angular Velocity

Change in angle over time. Represented as Greek letter omega (ω)

$\omega ={\frac {\theta _{f}-\theta _{i}}{t_{f}-t_{i}}}={\frac {d\theta }{dt}}$

Units in radians/second or just /second, since radians have no units.

ω is actually a vector:

• magnitude: $\left|{\frac {d\theta }{dt}}\right|$
• direction: right-hand rule (perpendicular to plane of rotation)

Angular Acceleration

Change in angular velocity over time. Represented as Greek letter alpha (α)

$\alpha ={\frac {d\mathbf {\omega } }{dt}}={\frac {d^{2}\theta }{dt^{2}}}$

Angular acceleration also follows the right-hand rule (perpendicular to plane of rotation).

Dynamics with Angular Variables

${\begin{aligned}\omega (t)&=\omega _{0}+\alpha t\\\theta (t)&=\theta _{0}+\omega _{0}t+{\frac {1}{2}}\alpha t^{2}\\\omega ^{2}(t)={\omega _{0}}^{2}2\alpha (\theta -\theta _{0})\end{aligned}}$

Point Movement Inside a Body

Velocity

$\left|\mathbf {v} _{p}\right|=\left|\mathbf {r} _{p}\right|\cdot \mathbf {\omega }$

Using the Cross Product:

$\mathbf {v} _{p}=\mathbf {r} _{p}\times \mathbf {\omega }$

Acceleration

${\begin{aligned}a_{\mathrm {tangent} }&=r_{p}\alpha \\a_{\mathrm {radial} }&={\frac {\mathbf {v} ^{2}}{r}}=r_{p}\omega \end{aligned}}$

Energy in Rotational Motion

Imagine a rigid system of particles that rotates about the origin. What is the total kinetic energy?

where $r_{i}$ is the distance from the axis to the particle.

For one particle:

$K_{i}={\frac {1}{2}}m_{i}{v_{i}}^{2}={\frac {1}{2}}m_{i}(r_{i}\omega )^{2}={\frac {1}{2}}I\omega ^{2}$

Sum over all particles for total energy:

$E=\sum K_{i}=\sum {\frac {1}{2}}m_{i}(r_{i}\omega )^{2}={\frac {\omega ^{2}}{2}}\underbrace {\sum m_{i}{r_{i}}^{2}} _{\mathrm {moment\ of\ inertia} }={\frac {\omega ^{2}}{2}}I_{a}$

Moment of Inertia ($I$) is defined below. In last form, $a$ represents that $I$ is taken with respect to the axis of rotation (Moment of inertia can be different for different axes)

Moment of Inertia

Represented as $I$.

For a discrete system of particles (separate and countable):

$I=\sum m_{i}\left|\mathbf {r} _{i}\right|^{2}$

For a continuum density (ρ) of particles:

$I=\int _{V}\rho (r)r^{2}\ dV$

Moment of inertia depends on the mass distribution and the position of the origin (centered at rotational axis).

When two objects stick together, their total moment of inertia is the sum of their individual moments of inertia

Moments of Inertia for Common Objects

(See Table 9.2 on P. 299)

Example

A disk and a ring have the same mass. In the ring, all of the mass is concentrated farther away from the axis of rotation. Therefore, the ring has a higher moment of inertia than the disk and will be harder to rotate.

Summary

${\begin{aligned}v_{\perp }&=\omega r\\a_{\perp }&=\alpha r\\K&={\frac {1}{2}}\omega ^{2}I\\I&=\sum m_{i}{r_{i}}^{2}\end{aligned}}$

Parallel Axis Theorem

Given the moment of inertia for an axis that passes through the center of mass ($I_{cm}$) of an object of mass $m$, the moment of inertia for a parallel axis ($I_{A}$) $d$ meters away is:

$I_{A}=I_{cm}+md^{2}\,\!$

Corollary

If the distance $d$ is significantly large, the object behaves like a particle rather than a solid body.

Example

Suppose we have a disk with a moment of inertia $I_{A}$ and a nut with moment of inertia $I_{N}$:

If we glue the nut onto the disk at a distance $d$,

$I=I_{A}+I_{N/A}={\frac {1}{2}}m_{d}r_{d}^{2}+{\frac {1}{2}}m_{n}(r_{1}+r_{2})^{2}+m_{n}d^{2}\,\!$