This shows you the differences between two versions of the page.
Both sides previous revision Previous revision Next revision | Previous revision Next revision Both sides next revision | ||
urp:physrot [2021-11-05] nerf_herder |
urp:physrot [2021-11-08] nerf_herder |
||
---|---|---|---|
Line 1: | Line 1: | ||
- | ====Rotation and Oscillation==== | + | =====Rotation===== |
- | ====Spring and Lever==== | ||
- | **Hooke's law** for springs: F=-kx, k=spring constant, x = displacement | ||
- | **Fulcrum**: t = r * f (torque = radius * force) | + | tension on rope being swung = force, centripetal force |
- | just add the torques for multiple objects on one side of a fulcrum | + | F = m * v^2/r |
+ | 4 kg, 2 meter rope, v = 5 m/s | ||
+ | F = 4 * 25/2 = 50 | ||
+ | ---> now need to subtract gravity, for instance at the top of the swing | ||
+ | 50 - 4*9.8 | ||
- | ====SHM - Simple Harmonic Motion==== | + | https://www.wikihow.com/Calculate-Tension-in-Physics |
- | ma = -kx | + | |
- | angular frequency ω = √(k/m) | + | ------- |
- | period of oscillation T = 2π √(m/k) (horizontal or vertical springs) | + | simple pendulum: |
+ | T = 2pi * (L/g)^0.5 (T = time, L = length, g = gravity) | ||
+ | (T/2pi)^2 = L/g | ||
- | In a vertical spring, the weight Mg of the body produces an initial elongation to equilibrium, such that Mg − kyₒ = 0. | + | potential energy: U = 1/2 kx^2 (spring), or P = mgh (at mass at some height) |
+ | kinetic energy: K = 1/2 mv^2 | ||
- | If y is the displacement from this equilibrium position the total restoring force will be Mg − k(yₒ + y) = −ky | + | Change in potential energy is given by |
+ | U=mgh | ||
+ | Joule = kg * m^2/s^2 | ||
+ | ------- | ||
+ | force on a pivot = moment | ||
+ | moment = f * d (distance) | ||
+ | w (greek letter omega) = angular velocity, | ||
+ | measured in rpm, or rads (2pi rads in a circle) | ||
+ | w = 2pi/T = 2pi*f T = time for full rotation, f = frequency | ||
+ | = delta theta / delta t | ||
+ | v = rw (v is distance, not radians) | ||
+ | I = moment of inertia (rotational inertia), resistance to angular acceleration | ||
+ | depends on arrangement of mass about the point of rotation | ||
+ | distance from point of rotation = the radius R (sometime L, or d for distance) | ||
+ | units = kg * m^2 | ||
+ | I for: | ||
+ | point mass I = MR^2 | ||
+ | solid cylinder I = 1/2 * MR^2 | ||
+ | solid sphere I = 2/5 * MR^2 | ||
+ | thin shell sphere I = 2/3 * MR^2 | ||
+ | hoop (around axis) I = MR^2 | ||
+ | hoop (on end?) I = 1/2 * MR^2 | ||
+ | rod (rotating from one end) I = 1/3 * MR^2 | ||
+ | rod (centered on axis) I = 1/12 * MR^2 | ||
+ | t = fr (torque), (technically cross product r x f, or r x (m*alpha x r) | ||
+ | t = I*alpha | ||
+ | |||
+ | If an object is a composite object, simply sum the inertial masses together | ||
+ | |||
+ | t (torque, Greek tau) = Ia (a = acceleration), units are Nm (Newton-meters) | ||
+ | t = Ia is rotational equivalent to f = ma (many parallels to linear forces, etc) | ||
+ | Angular Momentum L = Iw | ||
+ | If L1 is angular momentum of ice skater with arms out: | ||
+ | the velocity (w) is low, but I is big | ||
+ | If L2 is with skater with arms in: | ||
+ | velocity is higher, I is smaller. L1 = L2 for conservation of energy | ||
+ | |||
+ | Oddly, can also have angular momentum of a linearly moving object past another object | ||
+ | L = ->r x ->p (cross product of vectors r and p | ||
+ | = r * p * sin(theta) | ||
+ | r = hypotenuse, p is | ||
+ | |||
+ | cross product of vectors: ->A x ->B = ||->A|| ||->B|| sin(theta) | ||
+ | dot product of vectors: ->A . ->B = ||->A|| ||->B|| cos(theta) | ||
+ | ||->A|| = magnitude (norm) of vector A, sometimes written with single bars | ||
+ | |||
+ | rotational kinetic energy: | ||
+ | E = 1/2*I*w^2 (similar to E = 1/2 mv^2 for linear kinetic energy) | ||
+ | total kinetic energy of a rolling marble is the linear kinetic energy of it moving | ||
+ | plus the rotational energy | ||
+ | |||
+ | A number of similar articles on this on one page: | ||
+ | https://sciencing.com/rotational-kinetic-energy-definition-formula-units-w-examples-13720802.html | ||
+ | |||
+ | tangential acceleration = acceleration * radius | ||
+ | a = delta w/delta t | ||
+ | (a = angular acceleration) in rad/s^2 | ||
+ | a rolling object picks up angular inertia as it accelerates, so an object rolling | ||
+ | down an incline will have a final velocity less than a frictionless object that | ||
+ | does not roll | ||
+ | See: https://www.asc.ohio-state.edu/gan.1/teaching/spring99/C12.pdf | ||
+ | |||
+ | translational motion: movement of the center of mass for a rolling object | ||
+ | Two ways of looking at it: | ||
+ | 1) rolling object has combination of rotational and translational motion | ||
+ | 2) "" object rotates around the contact point with the ground, | ||
+ | but this point continuously changes ... not as easy concept to grasp | ||
+ | distance s = r*theta (theta in radians) | ||
+ | v = delta theta/delta time, (velocity of center of mass) | ||
+ | v = rw | ||
+ | velocity of a point on a disk is velocity relative to center of mass, plus | ||
+ | velocity of center of mass: | ||
+ | Vpt = Vrel + Vcm | ||
+ | If disk is rolling on ground, when point is at the top of the disk, Vrel = Vcm | ||
+ | so Vpt = 2Vcm. Conversely, when in contact with the ground, Vrel = -Vcm, | ||
+ | so vPt = 0. | ||
+ | |||
+ | |||
+ | ---------- | ||
+ | Optics: | ||
+ | refraction on going into a different medium | ||
+ | Snell's law: | ||
+ | sin(theta1) / sin(theta2) = v1/v2 = n2/n1 (note that the n values are reversed) | ||
+ | v = velocity of light in that medium, n = index of refraction | ||
+ | v = c/n (c = speed of light in a vacuum) | ||
+ | it bends towards the normal direction when entering denser material | ||
+ | (and slows down). bend is because photons are waves. | ||
+ | Critical angle : smallest angle that results in total reflection, no refraction | ||
+ | thetaC = arcsin(n2/n1) | ||
+ | |||
+ | |||
+ | ------- | ||
Back to [[Physics]] page or [[00_start|Start]] page. | Back to [[Physics]] page or [[00_start|Start]] page. |