Angular Definition and 999 Threads

A disc initially has angular velocities as shown It's angular momentum along the y-axis initially is ##L_s## I tried to find its angular momentum and ended up with this:##L=I_{x} \omega_{x}+I_{y} w_{y}+I_{z} z_{z}##The z component of angular momentum is thus ##L_{z}=I_{z} \omega_{z}## However...

Here's the problem setup, my student and I are stuck. A disk is rotating at constant angular velocity ω, and we are watching a point on the rim, parameterized by the angular position θ, move. Because we are observing the motion from an inclination angle Ψ, we do not always observe the...

From a previous post about the Relationship between the angular and 3D power spectra , I have got a demonstration making the link between the Angular power spectrum ##C_{\ell}## and the 3D Matter power spectrum ##P(k)## : 1) For example, I have the following demonstration, ## C_{\ell}\left(z...

I assumed the angular velocity of the center of mass of the two discs about z axis to be w1 note that angular velocity of center of mass of both discs and center of anyone disc about z axis is same, you can verify that if you want, me after verifying it will use it to decrease the length of the...

Hi, I need to come up with a math model for a digital ignition system. I've been thinking about it and I think that I need to measure 2 things to be able to calculate when I have to start charging the coil. They are the angular velocity and the acceleration but how can I do it? the idea is to...

Any spinning item, proton, electron, even planet, has angular momentum that creates force. How can an electron exist in a random orbital cloud around a spinning proton if it has an angular momentum and requires force to alter from any circular orbital plane (like a planet orbiting a star)?

What we know: The ball is dropped at the tip A with some speed ##v_0## and rebounds with speed ##v##. This collision produces an angular impulse, changing the angular momentum of the bar with the flywheels. Solution inspired by an answer provided by @TSny in the similar question. Angular...

Torque appiled by smaller disc = mga emf of disc due to B = Bwr2/2 Current I = Bwr2/2R force = IBr = Bwr3/2r torque = rF = Bwr4/2r mga = Bwr4/2r

If the crawling insect were stationary at a certain instant of time, then it would have the same angular velocity as that of disk, which is w in a clockwise direction. But now it's velocity at any instant is the vector sum of velocity due to rotation and the velocity it crawls at. My attempt is...

I calculate as follow and get a correct answer, but I wonder why the weight of the ladder 6 kg is not included in the mass (m) in the numerator. w= √(mgd/I) = √ { （42*10*1)/ [(1/12)(6)(2^2)+42*1] } = √ (420/44) = 3.06

ω^2 - (ωo)^2 = 2 (-630) (0.32) = -403.2 This is what I have now and I stuck here.

53 rpm equals 5.55 rad/sec multiply 5.55 by 2pi to get angular velocity of 34.8717 Is the answer 34.8717? What should I have done to more accurately solve the problem with a better understanding? What other steps should I take when solving similar problems? and lastly, Is the mass relevant...

Hi, I wanted to have a precision about a question that has been post on this relation between P(k) and C_l The author writes the ##C_\ell## like this : $$C_\ell(z,z') = \int_0^\infty dkk^2 j_\ell(kz)j_\ell(kz')P(k)$$ I don't undertstand the meaning of ##z## and ##z'## : these are not...

Find the probability distributions of the orbital angular momentum variables ##L^{2}## and ##L_{z}## for the following orbital state functions: ##\Psi(x) = f(r) sin(\theta) cos(\theta)## ##\Psi(x) = f(r) cos^{2}(\theta)##I am aware that the prob. distribution of an observable is ##|<a_{n} |...

In the context of Survey of Dark energy stage IV, I need to evaluate the error on a new observable called "O" which is equal to : \begin{equation} O=\left(\frac{C_{\ell, \mathrm{gal}, \mathrm{sp}}^{\prime}}{C_{\ell, \mathrm{gal}, \mathrm{ph}}^{\prime}}\right)=\left(\frac{b_{s p}}{b_{p...

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[Mentor Note -- Specialized question moved to the general technical forums] Homework Statement:: To show that ##J = Ma## for the charged Kerr metric [Wald Ch. 11 Pr. 6] Relevant Equations:: \begin{align*} \mathrm{d}s^2 = &- \left( \frac{\Delta - a^2 \sin^2{\theta}}{\Sigma}\right) \mathrm{d}t^2...

So I tried the problem and it’s different from the solution. I’m confused on why my attempt didn’t work, is it because the wheel is undergoing general planar motion? I tried to just apply Newton’s 2nd law to find the acceleration of the centre and then use that to find angular acceleration. The...

So for this question I got the right angular velocity. But I don’t get the same velocity for point A. I don’t understand why it’s cos30, problem asked for V_a when theta = 45 so I used cos45. I have my work below.

I am very confused when textbooks say the direction of Angular velocity is perpendicular ot radius and theta for that matter direction is in perpendicular direction. I know this comes from cross product rule but what is the meaning of Angular velocity and Angular momentum directing in upward...

I know the torque will be zero when the deflection is zero and will be maximum when the angular displacement is θ0 but how to determine the exact shape of the graph? Thanks

Angular velocity ω is by definition the runned angle dθ per time dt elapsed: ω=dθ/dt. If the time elapsed in the center of the Earth is dt, the dilated time elapsed on satellite is dt′. What is the satellite's angular velocity? Is it dθ/dt or dθ/dt′?

I calculated as attached and got it right. However, I just wonder why we can't use conservation of energy as the question has already specified 'frictionless', meaning no energy loss and energy distributed to the rotation only.

To show that when ##[J^2, H]=0 ## the propagator vanishes unless ##j_1 = j_2## , I did (##\hbar =1##) $$ K(j_1, m_1, j_2 m_2; t) = [jm, e^{-iHt}]= e^{iHt} (e^{iHt} jm e^{-iHt}) - e^{-iHt} jm $$ $$ = e^{iHt}[jm_H - jm] $$ So we have $$ \langle j_1 m_1 | [jm, e^{-iHt} ] | j_2 m_2 \rangle $$ $$ =...

I calculate in this way : Angular Momentum = I W = [ ( 1/12 ML^2 + m(L/2)^2 ] (V/ L/2) = [ 1/12 ML^2 + 1/4 mL^2 ] 2V/L = 2VL/4 [ M/3 + M] but can not find a matching answer. Why?

mball = 2 kg, mputty = 0.05 kg, L = 0.5 m, v = 3m/s a) Moment of inertia : I = (2mball + mputty ). ¼ L^2 = 0.253125 kg.m^2 Linitial = Lfinal => mputty. v. r = I.ω => ω = (4.mputty.v.r) / I = 0.148 rad/s b) K initial = 1/2 m v^2 = 0.225 J K final = 1/2 Iω^2 = 2.85.10^(-3) J => Kfinal /...

In quantum mechanics hydrogen atom problem ##L=\sqrt{l(l+1)}\hbar##. What that means physically when ##L=0##?

Hi guys, I don't really know how to solve this problem. The point is finding ##\omega## when ##m_2## passes from ##m_1##'s original position. Ideally, I'm thinking about some conservation of energy/momentum to apply here, but I'm quite confused. Any hint?

Hello! If we have a transition between 2 ro-vibrational levels of the same electronic state of a diatomic molecule the selection rules require for the changes in the rotational quantum number J that ##\Delta J = \pm 1##. Why can't we have ##\Delta J = 0##? The photon carries one unit of angular...

Question 1: I believe that the ratio would be b. 8:1 because by combining the formula for kinetic energy and momentum the expression Ek=p^2/2m can be obtained. Thus, for a body of mass 2kg with twice the momentum: Ek=2^2/2*2=1 For a body of mass 4kg with half the momentum: Ek=1^2/2*4=1/8...

1) Applying conservation of linear momentum: $$m.u = M.V + m.v$$ where ##V## is final linear speed of the rod $$V=\frac{m.u-m.v}{M}$$2) Applying formula of circular motion: $$V=\omega . r$$ $$\omega = \frac{\left(\frac{mu-mv}{M} \right)}{\frac{1}{2}L-x}$$ Is this correct?And can this be...

The answer here is A What i did is getting the area as follows, 2×4×1/2 +3×-6×1/2 +4×-6 = -29 and then use this Δω=ωf-ωi -29=ωf-5 ωf=24 but there is no such choice.

Hello! I just started reading some molecular physics and I am a bit confused about the electron angular momentum in diatomic molecules. Let's say we have just 2 protons and an electron for simplicity and we are in the Born-Oppenheimer approximation, so we assume that the nuclei are fixed in...

Take for example earth. Earth has angular momentum about its own axis. However, if we ignore the orbital portion, the angular momentum of the Earth relative to the sun's axis is the same. Another example is the spinning bike wheel/person holding it in a chair. It has angular momentum about its...

So clearly the easiest way to relate the angular speed to the linear speed would be to start from ##\tan(θ) = x/h## and take a time derivative of both sides. However, it also shouldn't be difficult to find the angular speed geometrically. Using the diagram below one can see that: ##sin(dθ) =...

i feel like subbing the numbers into the equation isn't enough because of the second minima and maxima thing? not sure what to do... would appreciate any help.

Unfortunately, I couldn't arrive to the correct answer ($$=0.28mL^2 \omega^2$$ ) and will be happy to understand what am I doing wrong. **My attempt:** Using $$ E_k = \frac{1}{2} I \omega^2 $$ I obtain that the difference I need to calculate is $$ \frac{1}{2} (2mL^2)(0.8\omega)^2 +...

In the classic example of a person holding a spinning bike wheel, as they flip the wheel over, angular momentum is conserved by the person/chair spinning with 2x the angular momentum of the initial wheel. Not questioning that. However, I thought ang momentum is always conserved about a...

Under the topic of simple harmonic motion comes the composition of two SHM's with the same angular frequency, different phase constants, and amplitudes in the same directions and in perpendicular directions. composition of SHM's in same direction: say a particle undergoes two SHM's described by...

Good day here is the exerciceThe only velocity I do have is the velocity v os the center of pulley 5, I tried to find the center of instantaneous velocity to find the angular velocity of pulley 5 but I couldn't, any hint would be highly appreciated! Best regards!

L = mvr = mr (dr/dt) = 2m*r*(dr/dt)/2 = 2m*(dA/dt) So, A = (L/2m)T so, ## L = \frac{2 \pi a b m}{T}## Now, ##T^2 = \frac{4 \pi^2}{GM} a^3## So from all these, I get ##L = \sqrt{ \frac{GM m^2 b^2}{a}}## But answer given is ##L = \sqrt{ \frac{2GM m^2 ab}{a+b}}## (This, they have derived from...

I know that in QM, there is LS coupling. So the interaction is there. But is such an interaction possible in macroscopic objects like a planet?

The "egg" initially spun around axis 1 with at ##\omega_s##. After being disturbed, it has started to possesses angular velocities along 2 and 3. The question is to find the rotational speed of ##\vec \omega=\vec\omega_1+\vec\omega_2+\vec\omega_3## to a fixed observer. It is calculated that...

Question 7.6 Official solution It seems that the solution uses the conservation of angular momentum to solve this question (τ=0). But the problem is that the frame is set on the centre of mass of the guy, which is non inertial. I would like to know why it is correct to do it this way. My...

List of relevant equations: Angular Momentum = L (vector) = r(vector) x p(vector) Angular velocity of rotating object = w(vector), direction found using right hand rule. Torque = T(vector) = dL(vector)/dt I have a few questions about torque and angular momentum direction and...

Question: If we place the frame of reference on an accelerating point, does the total rotational momentum still remain the same? I attempted to solve this question by manipulating the equations as shown below. $$\text{Define that }\vec r_i=\vec R+\vec r_i'\text{, where r is the position vector...

First I calculated the momentum of m1. Since m2 was at rest after the collision, all its momentum was transferred, so m1 has a momentum of 158 i hat. L=r x p, so its 916 k hat. This would also be the change in L because it was initially 0 when m1 had no velocity, so I know this is the net...

First I found the moment of inertia, I=1.8(5.5^2+3.9^2+4.9^2) =125.046 Then I tried to find the rotation rate using the equation L=rotation rate*I rotation rate=3773/125.046=30.173 But the answer is suppose to be 21.263?

I know how to get to the answer but that's what is confusing me. To find final velocity I multiply the acceleration by the time the object fell. Then multiply the velocity by the mass to get momentum. Now the angular momentum is r x p. Since the initial angular momentum was 0, this was also...

So I first tried to find L using torque, Torque=d/dt*L, and took the integral of this. Ended up with 23.28484t Now I square the equation L=rotation rate*I to get L^2=rotation acceleration *I^2 Angular acceleration=L^2/I^2 I feel like I am doing something wrong though, this doesn't give the...