When the centripetal force on an orbiting body disappears (e.g. if it the body is a ball and the force was exerted by a string and the string rips, or, more unrealistically, if the body is the earth and the sun suddenly disappeared), the body continues in linear motion. How is angular momentum conserved here?
35957
I am not a physicist (math/comp-sci) but I understand that Angular Momentum is supposed to be conserved.  I find this confusing because there seems to be many simple, common cases where a restrained, revolving body, when released seems to lose it's angular momentum.  For example, let's say that a hammer thrower is revolving a 10kg hammer around herself at a 1 meter distance from their combined center of mass every second.  Taking the hammer as a point-mass, then before release, the hammer and the thrower each have an angular momentum of      $l\omega = (mr^2)*(2\Pi) = (10*1*1)*(2*3.14..) =~ 62.8 kg\text{-}m^2/s$   Now, after the hammer is released, the thrower still has her same angular momentum (also 62.8), but the hammer no longer seems to have any.  Yes, I know that the hammer is still rotating (spinning) as it did when it was revolving around the thrower, so that angular momentum seems to be conserved, but the angular momentum from revolution (orbiting) seems to be gone.  AFAIK, it's not transferred to it's spin angular momentum (observation seems to bear this out).  Nor is it transferred to the thrower.  Besides having no mechanism for this (she let go of it), if it did, we would have noticed hammer throwers being knocked over by having their angular momentum suddenly doubled.  So where does it go?  Or is it not actually conserved in this case?
44410
A small body of mass $m$ slides down from the top of a hemisphere of radius $r$. There is no friction between the surface of the block and the hemisphere. The height at which the body loses contact with the surface of the sphere is? This is how I understood the problem: First of all, the mass does not lose contact with the surface of the sphere at all, considering it is undergoing centripetal acceleration, where the centripetal force is provided by the component of the mass's weight towards the centre of the hemisphere ($mgcos\theta$, where $\theta$ is the angle between the vector of the mass's weight and its component acting towards the centre). Therefore, since only $mgcos\theta$ is responsible for the centripetal force, I can form a relationship like this:  $mgcos\theta\ =\ \large \frac{mv^2}{r}$ $ v\ =\ \sqrt{rgcos\theta}$ Taking 'h' as the height of the mass from the base of the hemisphere. $cos\theta\ =\ \large \frac{h}{r}$ Then the velocity of the mass becomes: $v\ =\ \sqrt{gh}$ The component of the mass's weight along the centre disappears only when $\theta$ becomes $90$ degrees. At this point, it leaves the surface of the hemisphere. Now, the energy of the mass at the topmost point is: $P.E\ =\ mgr$  As the body slides over the hemisphere's surface, it has a tangential velocity given by the expression I had just previously derived. So by the conservation of total mechanical energy of the body, its energy at any other point on the hemisphere is: $T.E\ =\ mgh + \frac{1}{2}mv^2$ $P.E\ =\ T.E$ $gr\ =\ gh + \frac{v^2}{2}$ $gr\ =\ gh + \frac{gh}{2}$ $h\ =\ \frac{2}{3}r$ But this would mean that the body does indeed leave the surface of the hemisphere. It just doesn't add up. Can someone please explain if my approach and assumptions are valid and how I got to this completely contradictory answer?
57534
I read in  in this site that the KE a free-falling ball acquires is not originated by the attracting body but that energy was actually stored in the ball when it had been lifted to the height it dropped from.  In this way, it was said, gravity is subject to the conservation of energy principle and cannot change the total energy of an object.  Consider now the maneuver known as a  (also gravity assist) used by space probes such as the . A space probe approaches a planet with velocity $v$, slingshots around, and ends up with velocity $v+2U$, where $U$ is the velocity of the planet.  Consider the energy of the probe. Before, it was $E_i=\frac{1}{2} mv^2$ and after it is $E_f=\frac{1}{2} m(v+2U)^2$. It looks like $E_f$ is much bigger than $E_i$ - but where did the additional energy come from?   Is this not a violation of the conservation of energy principle?
42196
When applying a force outside of the center of mass of the body, the body will get both linear and angular momentum. Right?  Does the linear velocity from this force equal to the linear velocity from the same force, which is applied on the center of mass?  $$v_{\text{center}} = v_{\text{non-center}}\quad\text{? (no vectors)}$$  If it does: How is it that the energy applied to the system is equal? In one we have both rotation and linear movement, and in the other just linear movement.
84535
The Runge-Lenz vector is an "extra" conserved quantity for Keplerian $\frac{1}{r}$ potentials, which is in addition to the usual energy and angular momentum conservation present in all central force potentials.    I suppose it is not a coincidence that $\frac{1}{r}$ potentials have this extra conserved quantity and are also one of the two central force potentials that have closed orbits.  Indeed, one may think of the RL vector as an expression for the orientation and eccentricity of the elliptical orbit, which is conserved iff the orbit closes on itself.    This then leads me to think that there must be an analog to the RL vector in an $r^2$ potential, which is the other central force potnetial that has closed orbits.  Surely, we can define a vector pointing along the direction of the orbit's major axis, that has magnitude proprtional to the eccentricity, and this will be conserved.  Is there any way to write down such a vector in terms of dynamical variables, and thus obtain an $r^2$ analog for the RL vector?
8756
I am in search of a simplified version of the derivation of  (both translational and rotational) for a rigid body (3D block) that has a body fixed frame and where the center of mass of the body is not at the center of gravity.  I can find elementary derivations for the same system when the center of mass is at the center of gravity, but not for my system in question.    I am using the derivation as background research for a rotordynamics project that I'm working on.  Any help and or references would be greatly appreciated!  :)
57828
I am trying to understand conservation linear momentum with in angular momentum  Herewith i just explain what i am  looking for.    Imagine an object of mass(M), velocity(V), is undergoing circular motion    with radius(R)     Angular momentum L = RMV  (R-radius M-mass- V- tangential velocity)  Mass is constant here  Since L is conserved   so when R is decreased tangential velocity(V)     has to be increased to keep angular momentum constant.  Does this mean linear momentum not conserved when radius of rotation changes?  When linear momentum is not conserved then what is the force that increases the velocity of object when R decreases.  Could anyone clarify this. or could any one say is my basic understanding about momentum is wrong. for easy understanding i am only considered magnitude of momentum's not cross product of vectors
33949
Suppose we are given a stick and a stone tied to the stick by a string. Now if we rotate the stone around the stick the stone rises in height (see picture below). My question is which force accounts for this rise in height?     According to me (please correct me if I am wrong) this happens due to centrifugal force. The centrifugal force is directed along the string outwards, so we can resolve it into two components, one is the horizontal component and the other is the vertical component. If the mass of the stone is m, its velocity is v, the length of the string is r, and the angle the string makes with the horizontal is θ, then the total centrifugal force is mv^2/r, the horizontal component of the centrifugal force is mv^2/r * cos(θ), and its vertical component is mv^2/r * sin(θ) (see picture below). Can anyone tell me if I am correct? The directions of the forces are given in the picture, but I am not sure whether the directions are right or not. Also note that in the second picture for sake of clarity I have not drawn the direction of rotation of the stone.
123572
Problem #25 from the 2009 $F=ma$ contest:     Two discs are mounted on thin, lightweight rods oriented through their centers and normal to the discs. These axles are constrained to be vertical at all times, and the discs can pivot frictionlessly on the rods. The discs have identical thickness and are made of the same material, but have differing radii $r_1$ and $r_2$. The discs are given angular velocities of magnitudes $\omega_1$ and $\omega_2$, respectively, and brought into contact at their edges. After the discs interact via friction it is found that both discs come exactly to a halt. Which of the following must hold? [...] Ignore effects associated with the vertical rods.   Diagram here:     Probably conservation of angular momentum is useful here, but I am not clear how it can be applied. There are two distinct axes of rotation, one for each disc, so we cannot talk about conservation of angular momentum about one axis. If the two disks rotated about the same axes then the solution is straightforward, but this is not the case.
105256
this is a question I had when I was a kid. I'm a bit ashamed because I think I am missing out on something very obvious since I have the same question despite almost being an engineer now!  From Newtonian physics, I understand how although the gravitational force pulls the planet towards a star, and the planet 'falls' towards the star, due the angular momentum, it also moves laterally. In case of a planetary orbit, this is just enough to keep it moving around it in an elliptical orbit. (With speed and radius being such that angular momentum and energy is conserved)  But let's consider a single planet and star system where the planet moves around the star in a circular path for simplicity. If gravitational force provides centripetal force, what can account for the centrifugal force?  I have a feeling it is related to some inaccurate view I have, of centrifugal force. The  and  talk about two concepts: fictitious and reactive centrifugal force. But it would be safe to assume the star(EDIT: sorry, I meant to choose a frame of reference such that the centripetal force on the planet has to be cancelled by a centrifugal force on the planet in the direction: star to planet) as a stationary frame of reference for the purpose of this question, right? Which means that it is not a fictitious force, right? Or is it not considered a centrifugal force if you take the star as the reference frame?
57823
I don't understand how the centripetal force, which always points to the center of our circular motion can cause this scenario:  We have a big stone which spins very fast, so fast that a part breaks down, because of the centrifugal force (this is at least how my text books describes it).   My problem: the centrifugal force does not really exist (we only use it in accelerated frames of reference, so the newton laws still work there), so if we're in a laboratory frame of reference, which force would "pull" that piece of the stone to the outside of the circle, when we only have the centripetal force (as mentioned pointing to the center of the circular motion...)?   (Please don't try to explain it in an accelerated frame of reference, because there I understand it, but I don't understand it in a laboratory frame of reference)
44284
Let's assume that there is nothing in the universe except Earth. If the Earth rotates on its axis as it does, then would we experience the effects of rotational motion like centrifugal force and Coriolis force?  The meaning of my question is: is Earth rotating relative to space?
14453
If some object is kept in a radial groove made on a disk and the disk is rotated with say constant angular velocity.   Now as we know that the object moves outwards ie away from the centre of the disk. I want to know which force is it which throws the object outwards.   As centrifugal force is a pseudo force and hence when the observer is in ground frame it shouldn't be present so which force exactly pulls it out.
105979
Suppose There's a particle of mass m moving with speed $v_{0}$, at certain moment a force starts acting on it (centripetal force) and it start uniform circular motion  Now every other reference states that centripetal force only brings about change only in direction  But plugging-in things in the equations paint a different picture  suppose  normal/centripetal acceleration = $A$  then $|v| = \sqrt{v_{0}^2+\frac{(A t)^2}{1}}$  we can certainly see that factor of $t$ will bring change in magnitude of velocity   Secondly, why the velocity is always tangential to the circle, but how can we mathematically show such? Intuitively for me if there's constant acceleration toward a certain then at some point of time, the body should 'slack' and fall toward the direction of acceleration   Here' a picture for a reference:
114407
A thought experiment has been bothering me. Assume an enclosed box free-falling in a constant gravity field. The box contains a person and a bowling ball at the top end. The person throws the bowling ball down (in the direction the system is falling/the direction gravity is pulling). Obviously, being a closed system, forces are all internal, so there's no net effect in that sense. But, the center of mass has shifted downward some amount more than what is due to gravity.  The question, then, is this: What does someone outside the box observe? Part of me says they don't see anything, the box appears to fall under the influence of gravity with no visible change. The other part says there's a period corresponding to the flight of the bowling ball where the box falls faster than can be accounted for by gravity.   A follow-up question that I believe is closely related is the following: Imagine jumping off a high-dive board in a cannonball position. Shortly before hitting the water, you extend your legs downward. Do you hit the water faster, slower, or with the same velocity as someone who kept their legs extended throughout the jump?
24267
From Goldstein:     ... The mechanics of the particle is contained in Newton's second law of motion, which states that there exist frames of reference in which the motion of the particle is described by the differential equation $$\mathbf{F} = \frac{d\mathbf{p}}{dt} \equiv \dot{\mathbf{p}},$$ or $$\mathbf{F} = \frac{d}{dt}\left(m\mathbf{v}\right).$$ In most instances, the mass of the particle is constant and [the last equation] reduces to $$\mathbf{F} = m\frac{d\mathbf{v}}{dt} = m\mathbf{a}\textrm{...}$$   What Goldstein is saying troubles me for it implies that $\mathbf{F} = \dot{\mathbf{p}}$ works for a particle of time-varying mass and that contradicts Ján Lalinský's answer to this question: .  What's going on here?  Edit: Some people think that a particle that loses mass that isn't going anywhere--it simply disappears---isn't a useful fiction to solve problems. Consider a ball that is emitting mass isotropically whilst being pushed by some force. In analyzing the motion of this ball there is no need to consider the fact that material is indeed being emitted, all that is needed is the fact that the ball is losing mass. Here we can just use $\mathbf{F} = m\mathbf{a}$ instead of $\mathbf{F} = \dot{\mathbf{p}}$, no?
43205
I'm a physics tutor for 11th and 12th grade students. I taught my students that momentum is conserved only when the net external force on the system is zero. But in the books we are following, momentum was conserved in the following cases in spite of presence of an external force:   A body (mass m1) is dropped from a tower (height h) and another body (mass m2) is projected vertically upward with initial velocity u. Find the velocities just after collision. (Gravitational force is the external force) Two bodies projected towards each other on a rough table surface, with all required parameters given asked to find the velocities after collision. (frictional force is the external force) Body explodes in mid air. Momentum is conserved before and after explosion. (Again gravitational force is the external force)   I explained to my students in all these cases though an external force is present it has negligible effect. But I'm not satisfied with my answer.  EDIT: Thank you for the comments and answer. It is clear that contribution of external force to change in momentum is negligible. Can someone do the calculation of taking change in momentum into consideration and showing that it is negligible?