$(\frac{d\vec{p}}{dt})\parallel = \vec{F}_{net}\parallel$ is the rate of change of the amgnitude of the momentum.
$(\frac{d\vec{p}}{dt})\perp = \vec{F}_{net}\perp$ is the rate of change of the direction of the momentum, which is numerically equal to the sideways (perpendicular) component of net force. Other associations:$$|(\frac{d\vec{p}}{dt})\perp| = |\vec{p}|\frac{|\vec{p}|}{R}$$$$|(\frac{d\vec{p}}{dt})\perp| = |\vec{p}||\frac{d\hat{p}}{dt}| = |\vec{p}|\frac{|\vec{p}|}{R} = \frac{\gamma mv^2}{R}$$$$|\vec{a}\perp| = |(\frac{d\vec{v}}{dt})\perp| = |\vec{v}||\frac{d\hat{v}}{dt}| = |\vec{v}|\frac{|\vec{v}|}{R} = \frac{v^2}{R}$$$$\frac{d\vec{p}}{dt}\leftarrow\vec{F}_{net}$$
Sunday, May 5, 2013
Sunday, April 7, 2013
Speed of Sound in a Solid
The Speed of Sound in a Solid can be determined through the equation:$$v = \omega d = \sqrt{\frac{k_{s,i}}{m_a}d}$$
$v$ is the speed of sound in m/s.
$k_{s,i}$ is the stiffness of the interatomic bond.
$m_a$ is the mass of one atom.
$d$ is the length of the interatomic bond.
$v$ is the speed of sound in m/s.
$k_{s,i}$ is the stiffness of the interatomic bond.
$m_a$ is the mass of one atom.
$d$ is the length of the interatomic bond.
Sunday, March 17, 2013
Friction
"Friction is the dissipation of kinetic energy into internal energy in the form of agitation of atoms throughout the objects. We say that friction is a "dissipative" process."
Friction can either speed up an object or slow it down, depending on its motion. For instance, on an airport conveyor belt your bags initially have no momentum, but with the addition of the movement of the conveyor belt and friction forces opposing the bag staying in its zero position, the bag begins to move in the direction of the conveyor belt. We can model this sliding friction with the equation:$$F_{KF} = \mu_k F_N$$$F_N$ is the "normal force," or what is causing the objects to contact each other.
$\mu_k$ is the coefficient of kinetic friction, a value that will range depending on the material.
The speed of an object does not effect the sliding friction force.
If a force is applied to an object that does not exceed the friction force, $F_{friction}$, we say that you are being counteracted by the static friction. In order for movement to occur, the force applied to an object must be greater than or equal to the frictional force in the direction of the application.$$F_{SF} <= \mu_s F_N$$$\mu_s$ is the coefficient of static friction. $\mu_s$ will more than likely be larger than $\mu_k$ because once an object is already moving, it requires only as much force as it already has to continue its movement. You can feel this effect when you manage to slide a heavy object across the floor, and although it is difficult initially it becomes much easier once it is moving.
Any force of at least $F_{SF}$ in value can be applied to an object, but once it is exceeded, the equations governing its frictional status changes to sliding, $F_{KF}$.
Friction can either speed up an object or slow it down, depending on its motion. For instance, on an airport conveyor belt your bags initially have no momentum, but with the addition of the movement of the conveyor belt and friction forces opposing the bag staying in its zero position, the bag begins to move in the direction of the conveyor belt. We can model this sliding friction with the equation:$$F_{KF} = \mu_k F_N$$$F_N$ is the "normal force," or what is causing the objects to contact each other.
$\mu_k$ is the coefficient of kinetic friction, a value that will range depending on the material.
The speed of an object does not effect the sliding friction force.
If a force is applied to an object that does not exceed the friction force, $F_{friction}$, we say that you are being counteracted by the static friction. In order for movement to occur, the force applied to an object must be greater than or equal to the frictional force in the direction of the application.$$F_{SF} <= \mu_s F_N$$$\mu_s$ is the coefficient of static friction. $\mu_s$ will more than likely be larger than $\mu_k$ because once an object is already moving, it requires only as much force as it already has to continue its movement. You can feel this effect when you manage to slide a heavy object across the floor, and although it is difficult initially it becomes much easier once it is moving.
Any force of at least $F_{SF}$ in value can be applied to an object, but once it is exceeded, the equations governing its frictional status changes to sliding, $F_{KF}$.
Young's Modulus
"Young's modulus ($Y$) is the ratio of stress to strain for a particular material. young's modulus is a property of the material, and does not depend on the size or shape of an object. The stiffer the material, the larger is Young's modulus."
The strain on a material is the fractional stretch, or the change in length divided by the original length. This gives a ratio defined as $$strain = \frac{\Delta L}{L}$$The stress on a material is the tension force per unit of area. This eliminates the dependency of thickness of the material by using both the tension force ($F_T$) and cross-sectional area of a wire ($A$).$$stress = \frac{F_T}{A}$$These ratios can be related down to the atomic level, the stress ($\frac{F_T}{A}$) being the force that each chain of atomic bonds must exert, and the strain ($\frac{\Delta L}{L}$) being the stretch of the interatomic bond. The stiffer the material, the larger the modulus. We write young's modulus as$$\frac{F_T}{A} = Y\frac{\Delta L}{L}$$$$Y = \frac{stress}{strain} = \frac{\frac{F_T}{A}}{\frac{\Delta L}{L}}$$This is dependent upon the fact that too large a stress will result in the material coming apart and breaking, a process known as "yielding." The Young's modulus in terms of atomic quantities can be defined as:$$Y = \frac{\frac{k_{s,i}s}{d^2}}{\frac{s}{d}} = \frac{k_{s,i}}{d}$$$d$ represents the relaxed length of an interatomic bond, and the diameter of one atom. The cross-sectional area of an atom is $d^2$, considering we are viewing the atom as occupying a cube of space in the crystal lattice versus its approximately spherical shape.
The stretch of the interatomic bond is $s$, or the $\Delta L$.
The interatomic force is $k_{s,i}$.
The strain on a material is the fractional stretch, or the change in length divided by the original length. This gives a ratio defined as $$strain = \frac{\Delta L}{L}$$The stress on a material is the tension force per unit of area. This eliminates the dependency of thickness of the material by using both the tension force ($F_T$) and cross-sectional area of a wire ($A$).$$stress = \frac{F_T}{A}$$These ratios can be related down to the atomic level, the stress ($\frac{F_T}{A}$) being the force that each chain of atomic bonds must exert, and the strain ($\frac{\Delta L}{L}$) being the stretch of the interatomic bond. The stiffer the material, the larger the modulus. We write young's modulus as$$\frac{F_T}{A} = Y\frac{\Delta L}{L}$$$$Y = \frac{stress}{strain} = \frac{\frac{F_T}{A}}{\frac{\Delta L}{L}}$$This is dependent upon the fact that too large a stress will result in the material coming apart and breaking, a process known as "yielding." The Young's modulus in terms of atomic quantities can be defined as:$$Y = \frac{\frac{k_{s,i}s}{d^2}}{\frac{s}{d}} = \frac{k_{s,i}}{d}$$$d$ represents the relaxed length of an interatomic bond, and the diameter of one atom. The cross-sectional area of an atom is $d^2$, considering we are viewing the atom as occupying a cube of space in the crystal lattice versus its approximately spherical shape.
The stretch of the interatomic bond is $s$, or the $\Delta L$.
The interatomic force is $k_{s,i}$.
Tension Forces
Tension Forces are just forces that compensate for other forces. For instance, when a ball is hanging motionless on a wire, the net force on the ball must be zero. The gravity of the Earth is pulling it down, therefore there must be another force in the opposite direction compensating for this.
The force exerted by an object such as a wire or string is called a tension.$$\Delta \vec{p} = \vec{F}_{net}\Delta t$$$$0 = (\vec{F}_T - \vec{F}_{ext})\Delta t$$
The force exerted by an object such as a wire or string is called a tension.$$\Delta \vec{p} = \vec{F}_{net}\Delta t$$$$0 = (\vec{F}_T - \vec{F}_{ext})\Delta t$$
Properties of Atoms
- All matter consists of atoms, the typical radius of an atom is about $r = 1 \times 10^-10 meters$.
- Atoms attract each other when they are close, but not too close.
- Atoms repel each other when they get to close together.
- Atoms move even at very low temperatures. This applies to solids, liquids, and gases.
Center of Mass
The center of mass is the center of a system of masses, located at a point that is central in relation to the masses of all of the objects within the system. To describe the position of the center of mass, we use the equation$$\vec{r}_{CM} = \frac{\sum\limits_{i=0}^n m_i \vec{r}_i}{\sum\limits_{i=0}^n m_i}$$$$\vec{r}_{CM} = \frac{\sum\limits_{i=0}^n m_i \vec{r}_i}{M_{total}}$$
The velocity of the center of mass is the same thing, except finding velocity instead of position.$$\vec{v}_{CM} = \frac{\sum\limits_{i=0}^n m_i \vec{v}_i}{M_{total}}$$
If the velocity is very small compared to the speed of light, $\gamma = 1$ we can say that the total velocities and masses of the system are equal based off of the equation for momentum. This helps us find the momentum of the center of mass:$$\vec{p} = \gamma\times m\vec{v}$$$$\vec{p}_{sys} = M_{total}\vec{v}_{CM}$$$$\vec{p}_{sys} = \sum\limits_{i=0}^n m_i \vec{v}_i$$
The velocity of the center of mass is the same thing, except finding velocity instead of position.$$\vec{v}_{CM} = \frac{\sum\limits_{i=0}^n m_i \vec{v}_i}{M_{total}}$$
If the velocity is very small compared to the speed of light, $\gamma = 1$ we can say that the total velocities and masses of the system are equal based off of the equation for momentum. This helps us find the momentum of the center of mass:$$\vec{p} = \gamma\times m\vec{v}$$$$\vec{p}_{sys} = M_{total}\vec{v}_{CM}$$$$\vec{p}_{sys} = \sum\limits_{i=0}^n m_i \vec{v}_i$$
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