Tension on rope being swung = force, centripetal force F = m * v^2/r Example:
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
simple pendulum: T = 2π * √(L/g) (T = time, L = length, g = gravity) (T/2π)² = L/g
potential energy: U = 1/2 kx² (spring), or P = mgh (at mass at some height) kinetic energy: K = 1/2 mv²
Change in potential energy is given by U=mgh Joule = kg * m²/s²
ω (greek letter omega) = angular velocity, measured in rpm, or rads (2π rads in a circle) ω = 2π/T = 2π*f T = time for full rotation, f = frequency = delta theta / delta t v = rω (v is distance, not radians)
I = moment of inertia (rotational inertia), resistance to angular acceleration, units = kg * m²
Depends on arrangement of mass about the point of rotation, distance from point of rotation = the radius R (sometime L, or d for distance)
I for:
point mass I = MR² solid cylinder I = 1/2 * MR² solid sphere I = 2/5 * MR² thin shell sphere I = 2/3 * MR² hoop (around axis) I = MR² hoop (on end?) I = 1/2 * MR² rod (rotating from one end) I = 1/3 * MR² rod (centered on axis) I = 1/12 * MR²
τ = f*r (torque), (technically cross product r x f, or r x (m*α x r)
τ = I*α
If an object is a composite object, simply sum the inertial masses together
τ (torque, Greek tau) = Ia (a = acceleration), units are Nm (Newton-meters)
τ = Iα is rotational equivalent to f = ma (many parallels to linear forces, etc)
Angular Momentum L = Iω
If L₁ is angular momentum of ice skater with arms out: the velocity (ω) 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(θ) r = hypotenuse, p is
cross product of vectors: →A x →B = ||→A|| ||→B|| sin(θ) dot product of vectors: →A . →B = ||→A|| ||→B|| cos(θ)
||->A|| = magnitude (norm) of vector A, sometimes written with single bars
Rotational kinetic energy: Total kinetic energy of a rolling marble is the linear kinetic energy of it moving plus the rotational energy
E = 1/2*I*ω² (similar to E = 1/2 mv² for linear kinetic 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 = Δω/Δt (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*θ (θ in radians) v = Δ θ/Δ 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.