In physics, angular momentum (rarely, moment of momentum or rotational momentum) is the rotational equivalent of linear momentum. It is an important quantity in physics because it is a conserved quantity—the total angular momentum of a closed system remains constant.
In three dimensions, the angular momentum for a point particle is a pseudovector r × p, the cross product of the particle's position vector r (relative to some origin) and its momentum vector; the latter is p = mv in Newtonian mechanics. Unlike momentum, angular momentum depends on where the origin is chosen, since the particle's position is measured from it.
Just as for angular velocity, there are two special types of angular momentum of an object: the spin angular momentum is the angular momentum about the object's centre of mass, while the orbital angular momentum is the angular momentum about a chosen center of rotation. The total angular momentum is the sum of the spin and orbital angular momenta. The orbital angular momentum vector of a point particle is always parallel and directly proportional to its orbital angular velocity vector ω, where the constant of proportionality depends on both the mass of the particle and its distance from origin. The spin angular momentum vector of a rigid body is proportional but not always parallel to the spin angular velocity vector Ω, making the constant of proportionality a second-rank tensor rather than a scalar.
Angular momentum is an extensive quantity; i.e. the total angular momentum of any composite system is the sum of the angular momenta of its constituent parts. For a continuous rigid body or a fluid the total angular momentum is the volume integral of angular momentum density (i.e. angular momentum per unit volume in the limit as volume shrinks to zero) over the entire body.
Torque can be defined as the rate of change of angular momentum, analogous to force. The net external torque on any system is always equal to the total torque on the system; in other words, the sum of all internal torques of any system is always 0 (this is the rotational analogue of Newton's Third Law). Therefore, for a closed system (where there is no net external torque), the total torque on the system must be 0, which means that the total angular momentum of the system is constant. The conservation of angular momentum helps explain many observed phenomena, for example the increase in rotational speed of a spinning figure skater as the skater's arms are contracted, the high rotational rates of neutron stars, the Coriolis effect, and the precession of gyroscopes. In general, conservation limits the possible motion of a system but does not uniquely determine it.
In quantum mechanics, angular momentum (like other quantities) is expressed as an operator, and its one-dimensional projections have quantized eigenvalues. Angular momentum is subject to the Heisenberg uncertainty principle, implying that at any time, only one projection (also called "component") can be measured with definite precision; the other two then remain uncertain. Because of this, the axis of rotation of a quantum particle is undefined. Quantum particles do possess a type of non-orbital angular momentum called "spin", but this angular momentum does not correspond to a spinning motion.
Not sure what to do here, except using the conversation of angular momentum. Even then, is angular momentum conserved in this case even after attaching an external object here? Else, what laws can I use to solve this problem?
Using conversation of angular momentum:
$$\dfrac...
Hello all! Hope everyone's been doing well!
My question relates to the nebular theory of solar system formation. It is generally accepted that via the nebular hypothesis, matter in a nebula contracts on its own gravity and begins to spin, but I'm having trouble understanding why it must begin...
initial total KE= (1/2)(0.6kg)(8m/s)^2 = 19.2J
(0.6kg)(8m/s) = (0.6kg+1.8kg)(vf)
vf= 2m/s
final KE= (1/2)(0.6kg+1.8kg)(2m/s)^2 = 4.8J
I tried to use linear speed=angular speed * radius : thus
2m/s= angular speed * (3.3m/2)
angular speed= 1.2 rad/s
Apparently that is wrong.
Motivation: In my thermodynamics + statistical physics class, we derived the equipartition theorem for ideal gasses using Boltzmann factors, dividing the phase space of a gas particle in position+momentum space into units of size x*p=h based on the quantum nature of the space of states that are...
In calculating the matrix elements for the raising operator L(+) with l = 1 and m = -1, 0, 1 each of my elements conforms to a diagonal shifted over one column with values [(2)^1/2]hbar on that diagonal, except for the element, L(+)|0,-1>, where I have a problem.
This should be value...
As an analogue, if 5J of work is done on an object then the linear KE might increase by 2J and the angular by 3J, so the work is divided between the linear and rotational forms.
Now suppose there is a sphere sliding on a frictionless surface. If an impulse of magnitude 1Ns is applied to the...
ω(10)=(1.3)∗(1.0−e^(−10/22) )= 0.475 rad/s
0.475 rad/s=0 +α(10second)
α=0.0475 rad/s^2
∫ω(t)=Θ =1.3t + 28.6e^(-t/22) | (t=10s, t=0)
total angle by which the wheel rotates over this period of t=10 seconds = 2.55 rad
Θ= 2(pi)(8m)= 1.3t + 28.6e^(-t/22)
0=1.3t + 28.6e^(-t/22) - 2(pi)(8m)
t=34...
1) the motion equations for ##m_2## are: $$T-m_2 g=0 \rightarrow T=m_2 g$$
##m_1##: $$T=m_1\frac{v^2}{r_0} \rightarrow \vec {v_0}=\sqrt{\frac{r_0 g m_2}{m_1}}\hat{\theta}$$
2) This is where I am stuck, first I wrote ##m_2## motion equation just like before, but in polar coordinates...
Summary: An explanation into the mechanisms that make irregular shaped meteorites, including some up one tenth the size of the moon, seemingly vanish, leaving symmetrical craters with surprisingly shallow floors.
We’re going to look at the initial conditions of the moons formation from the...
I wrote Newton's equations for each body (I took ##x## as the axis aligned with the tension)
##m_1##:
##x)f*_1 -T_1+T_2=0##
Where ##f*_1=\omega ^2 r_1##
##m_2##
##x)f*_2 -T_2=0##
##x)f*_2=T_2##
Where ##f*_2=\omega ^2 r_2##
I wrote that ##T_2=1100 N## and solved for ##\omega##, and I got...
Relevant Equations:
Angular momentum density stored in an electromagnetic field: $$\vec{l}_{em} = \epsilon_0[\vec{r} \times (\vec{E} \times \vec{B})]$$
Electric field of an electric charge: $$\frac{q_e}{4\pi\epsilon_0}\frac{r - r'}{|r - r'|^3}$$
Magnetic field of a magnetic charge...
A) The slider experiments three forces, all of them are on the ##x## axis (considering ##x## axis as the axis aligned with the arm): Normal force (exerted by the support), elastic force and centrifugal force, which is ##m.(\omega^2 r)##
Elastic force is equal to
##Fe=-k \delta =-2 (R-2R)=2R##...
So first of all, I will have to find the centre of mass.
##X_{cm} = \frac{1}{M}\int x dm##
likewise for Y.
##M## =150kg
From the above-given points, I can find the equation of a line to be
## y =-\frac{3}{4}x +3## .
Area density = ##150kg/(0.5*4*3) = 25kg/m^2##
##X_{cm} = \frac{1}{M}\int x dm...
Hi,
Since this is a question about COAM (Conservation of Angular Momentum), I will assume I can leave out the part on translation and just use the formula below:
##Initial Angular Momentum= Final Angular Momentum##
whereby ##I = \frac {1}{12}ML^2## (of rod)
So,
##\frac {1}{12}ML^2(1.5)=\frac...
I have a disc. The center of the disc is its center of mass and the motion of the disc is purely rotational (no translation). What is the angular velocity in the center of the rotating disc?
Hi everyone. Do correct me if I am thinking wrongly.
So to find angular velocity, won't I just have to integrate angular acc = 2t, which means angular velocity = t^2? Hence, won't the answer be 3^2=9?
The answer seems to be 5.43 :/
Thanks
I think the answer is ##\frac{mV}{M}## but I am not sure. Won't the cylinder tries to rotate due to the collision at one end? Is this anything related to Angular Momentum?
The Answers given were,
Hi guys, I'm trying to understand between gyroscope angular displacement and euler angles?
for example { Δx = Δx + h * Rx * SCx);} this is gyroscope output about anguler displacement.This value can be used to determine angle that
device created.Why we should euler angles to fly.(I know...
I suppose that the principle of conservation of angular momentum holds also for a cloud of particles weekly interacting at low pressure, density and temperature. And it should be still applicable when the particles or the atoms would start condensing and forming fusion products or simply solid...
There is a cornering maneuver in rallying called the "Scandinavian flick" or the "pendulum turn". It involves steering away from the corner before actually steering into the corner. This creates a pendulum effect which makes the car turn more sharply into the corner.
Sorry for the poor video...
I believe I know that when an object, in terms of linear motion, accelerates, it is being resisted by inertia, thus creating so called fictitious forces. Now, that said, how does angular acceleration affect spinning objects like say, a gymnast, when they spin around the axis of rotation? Do they...
I'm reading through "University Physics 14th edition" by Young and Freeman. Section 10.5 on angular momentum for a rigid body around a fixed axis of rotation is derived as L = Iω. However, it shows that this is only the case for the fixed axis of rotation being an axis of symmetry.
In section...
Problem Statement: How to calculate minumum angular velocity of a mass on a spinning plate
Relevant Equations: f=mrw^2
Hi, here's the question:
a) A rough horizontal plate rotates with a constant angular velocity of w about a fixed vertical axis. A particle of mass m lies on the plate at a...
Problem Statement: The uniform 4-kg cylinder A, of radius r = 150 mm, has an angular velocity w = 50 rad/s when it is brought into contact with an identical cylinder B which is at rest. The coefficient of kinetic friction at the contact point D is uk. After a period of slipping, the cylinders...
In this video, around 2:28 He explains Earth maintain its same angular momentum even after sun disappears. I didn't get it.
How Earth maintain its same angular momentum even after sun disappears?
Hello to all good people of physics forums. I just wanted to ask, whether the angular and linear (orbital) speed in perihelion of eliptical orbit are related the same way as in circular orbit (v = rw). If we take a look at the angular momentum (in polar coordinates) of reduced body moving in...
a) Kepler's first law states that a planet like Earth displays an elliptical orbit with the sun in focus. Using M = dL/dt, prove that a planet cannot leave its plane of orbit. Note: M here is an externally applied torque that the sun exerts on the planet.
diagram of the situation described
b)...
Both point A and B are moving in parallel in same direction, therefore rod is not rotating at this instance and angular acceleration is 0. Question states angular velocity is equal to zero.
Plugging into Angular Acceleration"AB" = r*sqr(angular velocity"AB") = 0.26m* sqr(0) = 0
[Moderator's...
3. Find the hamilton equations
4. using 3. prove the the angular momentum in the z axis ##L_z=m(x\dot y-xy\dot)## is preserved.
I got in ##3##:
How can I prove 4?
What I know is the following:
The total angular momentum of the nucleus is just the total sum of the angular momentum of each nucleon.
If the nucleons are even the total angular momentum in the ground state will simply be ##0+##.
If the odd number of nucleons is close to one of the magic...
I write Conservation of Energy:
Potential Energy loss(change):
U = m g ##\Delta##h = m g (R+r) (1-cos##\alpha##)
kinetic Energy gain(change):
K = (##\frac {m v^2} 2## + ##\frac {I \omega^2} 2##) + (##\frac {M v_2^2} 2## + ##\frac {I_2 \omega_2^2} 2##)
U = K
m g (R+r) (1-cos##\alpha##) =...
Hi, a newbie to the site and hoping someone can help. Its been a long time since I studied physics or math at school.
I am having some difficulty calculating the angular velocity of the arm trhough a partial movement (phase). The data set I have separates the angular velocity into the three...
(i) A rotating machine shaft turns through 2100 revolutions in 3 minutes. Determine the average angular velocity, w, of the machine shaft in rad/s.
the answer for this is
2100 x 2 pie = 4200 pie radian
3 mins = 180 seconds
4200 / 180 = 73.3 rad/s
(ii) Determine the area and arc length of the...
I was wondering why in the video the moment of inertia for the clay ball (upon collision) was simply 1ml^2. That is the constant for a hollow cylinder. The problem specifies that the object is a ball, so the cylinder classification makes no sense, and also I'm pretty sure clay is rather dense...
So, I was reading my textbook in the section regarding net torque, and they gave an example of a seesaw with one person at each end, and they said that there is a net external torque due to the force of gravity on each person. I completely understand that; however, when I was reading another...
I had a question about the equation (1/2)mv^2...
Why is the velocity squared? Why not simply (1/2)mv? Does it have anything to do with the intrinsic angular momentum ie does the intrinsic angular momentum change in anyway as velocity increases in a particular reference frame leading to the...
I don't have too much of a clue of how to begin the problem.
I first wrote the angular moementum of the system of particles: →M=∑mi(→ri×→vi)M→=∑mi(r→i×v→i). Then I know that the angular momentum from of the moving reference frame would have the velocity as the sum of the velocity of the frame...
n is the principal quantum number.
l is the angular momentum quantum number.
ml is the magnetic quantum number.
The possible values of l are 2, 3, and 4. I'm not sure if l can be equal to 4.
On the answer key, it shows l = 2, 3.
The speed of the sphere after the impact will be the same since the collision is elastic and the kinetic energy remains the same. So the change of momentum will be given by the cosine law right? What bothers me is the second question about the force that acts on the sphere (which can be given by...
PART B ONLY:
The cylinder undergoes torque when the mass m2 is removed:
τNET = Iα = FNETr
= 45α = FT(0.5)
FT = 90α, therefore msystem = 90 kg
After this step, I am not sure what to do.
τ = ΣF = Fg - Ft
= ma = (20 kg)(9.8 m/s2) - (90α)
= 196 -...