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 ROTATIONAL EQUILIBRIUM AND DYNAMICS

 * __Section 8-1: Torque__**

Like a net external force acting on an object produces linear acceleration, a net **torque** produces angular acceleration. The torque caused by a force acting on an object is represented by the equation:

math \displaystyle \tau=rF\sin \theta math

where //F// is the force causing the torque, //r// is the distance from the center to the point where the force acts on an object, and //theta// is the angle between the force and a line from the object's center through the point where the force is acting. Torque is measured in Newton meters, written as: math \displaystyle N\cdot m math The convention used in our class is that torque in a counterclockwise direction is positive and torque in a clockwise direction is negative (corresponding to right-hand rule). If more that one force is acting on an object, the net torque can be found by adding up all of the torques:

math \displaystyle \tau_{net}=\Sigma\tau math

In a canyon between two mountains, a spherical boulder with a radius of 1.4 m is just set in motion by a force of 1600 N. The force is applied at an angle of 53.5 degrees measured with respect to the vertical radius of the boulder. What is the magnitude of the torque on the boulder? (Page 265, Holt Physics 2006)
 * Torque practice:**

__**Section 8-2: Rotation and Inertia**__

The **center of mass** of an object is the point at which all the mass of an object is concentrated. A freely rotating object will rotate about the center of mass. The **center of gravity** of an object is the point through which a gravitational force acts on the object. For most objects the center of mass and center of gravity will be the same point. The **moment of inertia** (//I//) of an object is the object's resistance to changes in rotational motion about an axis. Moment of inertia in rotational motion is similar to mass in translational motion. The moments of inertia for some common shapes are:

\displaystyle I math || \displaystyle mr^{2} math || \displaystyle\frac{1}{2}mr^{2} math || \displaystyle\frac{2}{5}mr^{2} math || \displaystyle\frac{2}{3}mr^{2} math || \displaystyle mr^{2} math || \displaystyle\frac{1}{2}mr^{2} math || \displaystyle\frac{1}{12}ml^{2} math || \displaystyle\frac{1}{3}ml^{2} math ||
 * Shape || math
 * Point mass at a distance //r// from the axis || math
 * Solid disk or cylinder of radius //r// about the axis || math
 * Solid sphere of radius //r// about its diameter || math
 * Thin spherical shell of radius //r// about its diameter || math
 * Thin hoop of radius //r// about the axis || math
 * Thin hoop of radius //r// about the diameter || math
 * Thin rod of length //l// about its center || math
 * Thin rod of length //l// about its end || math

An object is in **rotational equilibrium** when there is **no** net torque acting on the object. If there is also no net force acting on the object (which is referred to as **translational equilibrium**) then the object is in **equilibrium**.

Equilibrium || math \displaystyle \Sigma\vec{F}=0 math || The net force on the object is zero || Equilibrium || math \displaystyle \Sigma\tau=0 math || The net torque on the object is zero ||
 * Type || Equation || Meaning ||
 * Translational
 * Rotational

A force of 25 N is applied to the end of a uniform rod that is 0.50 m long and has a mass of 0.75 kg. Find the moment of inertia of the rod, and the torque applied to the rod. (Pg 54, Review Packet)
 * Moment of Inertia Practice:**


 * __Section 8-3: Rotational Dynamics__**

Newton's second law can be applied to angular motion using the equation:

math \displaystyle \tau_{net}=I\alpha math

This is related to the equation for translational motion as follows:

\displaystyle \vec{F}=m\vec{a} math || \displaystyle \tau = I\alpha math ||
 * Type of Motion || Equation ||
 * Translational || math
 * Rotational || math

Just as moment of inertia was similar to mass, the **angular momentum** (//L//) in rotational motion is similar to the momentum of an object in translational motion. This is related to the equation for translational motion as follows:

\displaystyle \vec{p}=m\vec{v} math || \displaystyle L=I\omega math ||
 * Type of Motion || Equation ||
 * Translational || math
 * Rotational || math

The angular momentum of an object is conserved when there is no acting external force or torque. Rotating objects have **rotational kinetic energy** according to the following equation:

math \displaystyle KE_{r}=\frac{1}{2}I\omega^{2}=\frac{L^{2}}{2I} math

Just as other types of mechanical energy can be conserved, rotational kinetic energy is also conserved when there is no friction.

A satellite in orbit around Earth is initially at a constant angular speed of 7.27 x 10^-5 rad/s. The mass of the satellite is 45 kg, and it has an orbital radius of 4.23 x 10^7 m. Find the rotational kinetic energy of the satellite around Earth. (Pg 54, Review Packet)
 * Rotational Kinetic Energy Practice:**


 * __Section 8-4: Simple Machines__**

In physics, there are six basic types of machines, referred to as **simple machines**, which are used as the tools to build other complex machines. The six simple machines are **levers**, **inclined planes**, **wheels**, **wedges**, **pulleys**, and **screws**. The main purpose of a machine is to maximize the ratio of the output of the machine and the input force. That ratio is called the **mechanical advantage** (//MA//) of a machine. //MA// is a unitless number according to the following equation:

math \displaystyle MA=\frac{output force}{input force}=\frac{F_{out}}{F_{in}} math

Sadly, when frictional forces are brought into the mix, some output force is lost, causing less work to be done by the machine than the original force. The ratio of work done by the machine to work put in to the machine is called the **efficiency** (//eff//) of the machine:

math \displaystyle eff=\frac{W_{out}}{W_{in}} math

If a machine is perfectly efficient (//eff// = 1) then the **ideal mechanical advantage** (//IMA//) can be found by comparing the input and output distances:

math \displaystyle IMA=\frac{d_{in}}{d_{out}} math

Combining the equations for //MA// and //IMA// provide us with another equation to find the efficiency of a machine:

math \displaystyle eff=\frac{MA}{IMA} math

A pulley system is used to lift a piano 3.0 m. If a force of 2200 N is applied to the rope as the rope is pulled in 14 m, what is the efficiency of the machine? Assume the mass of the piano is 750 kg. (Pg 266, Holt Physics 2006)
 * Effiency Practice:**


 * Practice Problems (Methods/Answers):**

Torque:

//r// = 1.4 m //F// = 1600 N //Theta// = 53.5 degrees math \displaystyle\tau=rF\sin \theta math math \displaystyle\tau=(1.4m)(1600N)(\sin 53.5^{o})=(2240N\cdot m)(\sin 53.5^{o})=1800.639N\cdot m math In accordance with the laws of sig figs, your answer should be math \displaystyle\tau=1800 N\cdot m math

Moment of Inertia:

//F// = 25 N //l// = 0.50 m //m// = 0.75 kg math \displaystyle I=\frac{1}{12}ml^{2} math math \displaystyle I=\frac{1}{12}(0.75kg)(0.50m)^{2}=\frac{1}{12}(0.375kg\cdot m^{2}=0.03125kg\cdot m^{2} math math \displaystyle I=0.031kg\cdot m^{2} math math \displaystyle\tau=rF math math \displaystyle\tau=(0.25m)(25N)=6.25N\cdot m math math \displaystyle\tau=6.2N\cdot m math


 * Rotational Dynamics Practice:**

Angular speed = 7.25 x 10^-5 rad/s //m// = 45 kg //r// = 4.23 x 10^7 m math \displaystyle I=mr^{2} math math \displaystyle I=(45kg)(4.23x10^{7}m)^{2}=8.051805x10^{16}N\cdot m^{2} math math \displaystyle KE_{r}=\frac{1}{2}I\omega^{2}=\frac{L^{2}}{2I} math math \displaystyle KE_{r}=\frac{1}{2}(8.051805x10^{16}N\cdot m^{2})(7.25x10^{-5}rad/s)=2.9187x10^{12}J math math \displaystyle KE_{r}=2.9x10^{12}J math


 * Efficiency Practice:**

Piano distance = 3.0 m Applied force = 2200 N Rope distance = 14 m Piano mass = 750 kg math \displaystyle\vec{F}=m\vec{a} math math \displaystyle\vec{F}=(750kg)(9.81m/s^{2})=7357.5N math math \displaystyle W=Fd math math \displaystyle W_{in}=(2200N)(14m)=30800J math math \displaystyle W_{out}=(7357.5N)(3.0m)=22072.5J math math \displaystyle eff=\frac{W_{out}}{W_{in}} math math \displaystyle eff=\frac{22072.5J}{30800J}=0.7166 math math \displaystyle eff=0.72 math

All information was gathered from the Honors Physics Review Notes 2008-2009, distributed by Tom Strong.