Let θ be the orientation (angle) of a body (such as a cat), and let ω be its angular velocity.  It is well-known that θ can change even when the body is not rotating, using the conservation of angular momentum; that is, even when ω = d&#xfeff;θ/d&#xfeff;t = 0. That's how cats land on their feet so well.  But how can θ possibly ever change, when its derivative is zero?! What's wrong with the math?
101945
To explain why a falling cat can turn by 180 degree without external torque and without violation of the conservation of angular momentum, one usually models the cat as two cylinders as in     This may explain the turn. However I often heard contrary to this that she can rotate her body simply because she rotates her tail very fast into the opposite direction (and essentially keeps the rest of the body rigid).  So, what effect does the tail have in reality? Is there any detailed model, which takes the tail rotation into account and calculates how large its effect is?
59500
Consider this equation :   $$\sqrt{\left( \frac{dy\cdot u\,dt}{L}\right)^2+(dy)^2}=v\,dt,$$  where $t$ varies from $0$ to $T$ , and $y$ varies from $0$ to $L$.  Now how to proceed ?   This equation arises out of following problem :   A cat sitting in a field suddenly sees a standing dog. To save its life, the cat runs away in a straight line with speed $u$. Without any delay, the dog starts with running with constant speed $v&gt;u$ to catch the cat. Initially, $v$ is perpendicular to $u$ and $L$ is the initial separation between the two. If the dog always changes its direction so that it is always heading directly at the cat, find the time the dog takes to catch the cat in terms of $v, u$ and $L$.  See my solution below :   Let initially dog be at $D$ and cat at $C$ and after time $dt$ they are at $D'$ and $C'$ respectively. Dog velocity is always pointing towards cat.  Let $DA = dy, \;AD' = dx$  Let $CC'=udt,\;DD' = vdt$ as interval is very small so $DD'$ can be taken straight line.  Also we have $\frac{DA}{DC}= \frac{AD'}{ CC'}$ using triangle property.  $\frac{dy}{L}= \frac{dx}{udt}\\ dx = \frac{dy.udt}{L}$  $\sqrt{(dx)^2 + (dy)^2} = DD' = vdt \\ \sqrt{(\frac{dy.udt}{L})^2 + (dy)^2} = vdt $  Here $t$ varies from $0-T$, and $y$ varies from $0-L$. Now how to proceed?
70006
We all know $a = \frac{\mathrm{d}v}{\mathrm{d}t}$. A little application of the chain rule leads to the relation $$a=v\frac{\mathrm{d}v}{\mathrm{d}x}$$ But the above equation shows that $a=0$ whenever $v=0$. And this must be wrong, as when we throw something vertically from the earth surface it stops and then returns due to gravity. There is always an acceleration in the downward direction.  I have come up with two possible ways to solve this problem:   We know that $v = \frac{\mathrm{d}x}{\mathrm{d}t}$, so whenever $v=0$ then $\mathrm{d}x=0$ (i.e. no displacement in that infinitely small time). Also, $$a = \frac{\mathrm{d}v}{\mathrm{d}t} = \frac{\mathrm{d}v}{\mathrm{d}x}\frac{\mathrm{d}x}{\mathrm{d}t}$$ only if $\mathrm{d}x$ is not equal to $0$, as multiplying both numerator and denominator with $0$ will make it $0/0$ (undefined). Or multiplying by $0/0$ is not equivalent to multiplying by $1$.  Now since $\mathrm{d}x=0$ whenever $v=0$, therefore we can not write $a = v\frac{\mathrm{d}v}{\mathrm{d}x}$ when $v=0$. If we plot a $v$-$x$ curve for the motion, whenever $v=0$, as explained, $\mathrm{d}x=0$. Therefore, at that instant $\frac{\mathrm{d}v}{\mathrm{d}x}$ (the slope of the $v$-$x$ curve) will be not defined (as $\mathrm{d}x=0$), therefore the equation $a = v\frac{\mathrm{d}v}{\mathrm{d}x}$ will be undefined and we cannot determine acceleration at that instant with the $v$-$x$ curve.   Now that actually means that by plotting $v$-$x$ curve we lose information about acceleration of the particle when $v=0$.   I want a method to find the acceleration at that point using $v$-$x$ curve. Also a explanation for this defect of the $v$-$x$ curve.  Also I would like to add that so far no physics book (that I had read) has explained this before writing this relation. Also they do not mention that this will not work for $v=0$.     I would like to add the situation where I observed this: when you throw something vertically upward with velocity such that it reaches to a height 10 then the v-x curve will be-   the slope of above curve is clearly not defined at x=10(or the instant where velocity is 0) But we know that there was a constant acceleration throughout the flight. Then how can we find acceleration at that instant??? Also assume that we only have v-x curve. Or if we cannot then why??
88036
I have already read        and many others   But I still have the following problem:   On one hand I've been told that if pure rotation takes place then the object goes on forever. This must imply no frictional force in any form acting on the body. On the second hand I've been told that during pure rotation(no external forces like wind) on a plane surface friction acts opposite to velocity of center of mass (accompanied by change of normal) and slows it down.     Both seem logical enough: The point of contact doesn't slip therefore no friction is required; So no retarding force and it must go on. Things don't go on forever; Friction must be present to slow it down.    So, which hand is correct? Or is there a third hand altogether?
115444
While reading Chapter 1 of an astrodynamics textbook, I came across the statement:  $$\mathbf{v}\cdot \mathbf{{\dot{v}}}=v{\dot{v}}$$  In other words, the dot product of velocity and the time-rate-of-change of velocity is simply equal to the product of the magnitudes of the velocity and acceleration vectors.  I think I see how this works mathematically:  $$\frac{1}{2}\frac{d}{dt}\left ( \mathbf{v}\cdot \mathbf{v} \right )=\frac{1}{2}\left [\mathbf{v}\cdot \frac{d\mathbf{v}}{dt}  +\mathbf{v}\cdot \frac{d\mathbf{v}}{dt}\right]=\mathbf{v}\cdot \frac{d\mathbf{v}}{dt}=\mathbf{v}\cdot \mathbf{\dot{v}}$$  Also, since a vector is always parallel to itself,  $$\mathbf{v}\cdot \mathbf{v}=vv=v^{2}$$  So,  $$\frac{1}{2}\frac{d}{dt}\left ( \mathbf{v}\cdot \mathbf{v} \right )=\frac{1}{2}\frac{d}{dt}\left (v^{2}  \right )=\frac{1}{2}\left (2v  \right )\frac{dv}{dt}=v\frac{dv}{dt}=v\dot{v}$$  My hangup is this: are a vector and the first time derivative of that same vector always parallel (and in the same direction)?  It would seem so, since the (not shown) cosine of theta term equals 1.  Seems strange to me, because an object can be traveling in a certain direction while experiencing acceleration in another direction, such as the parabolic trajectory of a projectile near the Earth's surface (ignoring drag).  Velocity will be tangent to the arc, while the only acceleration is straight down (due to gravity) and never actually parallel to velocity, so long as there is a horizontal component to the trajectory.  Am I missing somehting very obvious here?  Thank you for your time and consideration.  Citation: See page 15 (sheet 30 of 470) section 1.4.1.2 of .     Some very helpful people left great answers, thank you. More webmining also yielded , which provided yet another good answer to the question. Seems like that one statement is responsible for a fair bit of consternation.
64939
To my knowledge there are three types of acceleration when a body (e.g. a rod) is moving in a circle about an axis. These are:   Angular acceleration : this is the rate of change of angular velocity. Tangential acceleration : this is the linear acceleration of the system in a tangential direction to the circle and equals the radius times the angular acceleration. Radial/centripetal acceleration : this is the linear acceleration of the system that is directed inwards towards the center of the circle.   I also think there are two types of velocity:   Linear velocity: this is its velocity in the tangential direction and is constantly changing Angular velocity or angular frequency: this is the rate of change of angle.   Is the above correct? And have I missed anything?
63135
I have difficulties in understanding, in the rotation of a rigid body, the properties of the component of the angular momentum vector $ \vec {L} $ which is perpendicular to the fixed axis of rotation $ z $. I will call this component $ \vec {L_n } $. Suppose that the angular velocity  $ \Omega $ is constant in direction but can vary in magnitude.    $ | \vec {L_ {n, i}} | = m_i r_i R_i \Omega cos \theta_i \implies | \vec {L_n} | = \Omega \sum m_i r_i R_i cos \theta_i$  Can I therefore say that $ | \vec {L_n} | \propto | \vec {\Omega} | $ (1)?  If so, suppose to apply a torque perpendicular to the $z$ axis and parallel to $ \vec {L_n} $, so that the magnitude of this vector increases. Follows from (1) that there should be an angular acceleration $ \vec {\alpha} $, although we are in the absence of a torque with an axial component.  This would go against the fact that $ I_z \vec {\alpha} = \vec { M_z}$ (Where $I_z $ is the moment of inertia with respect to the $z$ axis and $ M_z $ is the axial component of the exerted torque).
94636
Suppose we have a disk of radius $r$ and mass $m$ travelling at velocity $v$. I want to calculate the instantaneous angular momentum with axis through the edge of the disc (on the circumference).   Angular momentum $= I \omega$. $I = \frac{1}{2}mr^2 + mr^2 = \frac{3}{2}mr^2$ by the parallel axis theorem. $\omega = \frac{v}{r}$. Therefore, angular momentum $= \frac{3mrv}{2}$.  Alternatively, angular momentum $=p\times r= m r \times v = mrv$.  Why do these two methods differ? Which, if any, are correct?
42352
I have been told [Warning: I leave this because it's what I asked and allows to understand the dialogues in the comments, but Azad, whom I thank, has pointed that the formula does not hold in general in the form it is expressed] that the angular momentum of and rigid body with respect to any point $P$ can always be expressed as $$\mathbf{L}_{P}=\mathbf{r}_{cm}\times M\mathbf{v}_{cm}+\big(\sum_im_iR_i^2\big)\boldsymbol{\omega}$$ where $\mathbf{r}_{cm}$ is the position of the centre of mass with respect to $P$, $M$ the mass of the body, $R_i$ the distance of the $i$-th point, having mass $m_i$, composing the body, and $\sum_im_iR_i^2=I$ its moment of inertia with respect to the instantaneous axis of the rotation around the centre of mass of angular velocity $\boldsymbol{\omega}$.  I know that the velocity $\mathbf{v}_i$ of each point $P_i$, having mass $m_i$, of a rigid body of mass $M$ can be see as the sum of a translation velocity of one of its points $C$ plus a rotation velocity around that point: $\mathbf{v}_i=\mathbf{v}_{C}+\boldsymbol{\omega}\times\overrightarrow{CP_i}$. If we chose $C$ as the centre of mass I see that $$\mathbf{L}_{cm}=\sum_i \overrightarrow{CP_i}\times m_i\mathbf{v}_{i}=\sum_i \overrightarrow{CP_i}\times m_i\mathbf{v}_{cm}+\sum_i \overrightarrow{CP_i}\times m_i(\boldsymbol{\omega}\times\overrightarrow{CP_i})$$$$=\sum_i \overrightarrow{CP_i}\times m_i(\boldsymbol{\omega}\times\overrightarrow{CP_i}) $$because, if I am not wrong, $\sum_i \overrightarrow{CP_i}\times m_i\mathbf{v}_C=(\sum_i m_i\overrightarrow{CP_i})\times\mathbf{v}_C=\mathbf{0}$ since $\sum_i m_i\overrightarrow{CP_i}$ is the position of the centre of mass with respect to itself, which is $\mathbf{0}$.  How can it be proved that $\sum_i \overrightarrow{CP_i}\times m_i(\boldsymbol{\omega}\times\overrightarrow{CP_i})=(\sum_im_iR_i^2)\boldsymbol{\omega}$? I have searched a lot on the Internet and on books, but I find nothing. To give some background of mine, I have studied nothing of analytical mechanics. I find the formula very, very interesting both in itself and because, if the moment of inertia does not depend upon time, $\forall t\quad I(t)= I(t_0)$, the above expression can be differentiated to get the formula of the resultant torque with respect to the centre of mass $\sum\boldsymbol{\tau}_{cm}=\frac{d\mathbf{L}_{cm}}{dt}=I\boldsymbol{\alpha}_{cm}$ where $\boldsymbol{\alpha}$ is the angular acceleration around the centre of mass. I heartily thank you for any answer!    Some unfruitful trials: by using the "BAC CAB identity" as suggested by Azad, whom I heartily thank, $\mathbf{a}\times(\mathbf{b}\times\mathbf{c})=(\mathbf{a}\cdot\mathbf{c})\mathbf{b}-(\mathbf{a}\cdot\mathbf{b})\mathbf{c}$, I can see that$$\sum_i \overrightarrow{CP_i}\times m_i(\boldsymbol{\omega}\times\overrightarrow{CP_i})=\sum_im_i\|\overrightarrow{CP_i}\|^2\boldsymbol{\omega}-m_i(\overrightarrow{CP_i}\cdot\boldsymbol{\omega})\overrightarrow{CP_i}$$which, by decomposing every $\overrightarrow{CP_i}$ into an axial component $\mathbf{A}_i$ and a radial component $\mathbf{R}_i$, whose norms respectively are $A_i$ and $R_i$, with $R_i$ as the distance from $i$ to the axis of rotation, becomes $$\sum_im_iR_i^2\boldsymbol{\omega}+\sum_i m_i A_i^2\boldsymbol{\omega}-m_i(\mathbf{A}_i\cdot\boldsymbol{\omega})\overrightarrow{CP_i}$$but I cannot prove that $\sum_i m_i A_i^2\boldsymbol{\omega}-m_i(\mathbf{A}_i\cdot\boldsymbol{\omega})\overrightarrow{CP_i}=\mathbf{0}$.
85569
Why is the radians implicitly cancelled?  Somehow, the feet just trumps the numerator unit. For all other cases, you need to introduce the unit conversion fraction, and cancel explicitly.  Is it because radians and angles have no relevance to linear speed (v), so they are simply discarded?
55609
In a situation where a disk is rolling WITH slipping on the ground i.e velocity of centre of mass is greater than $r\omega$, is angular momentum conserved about a point on the ground.  What confuses me is that friction decelerates the disk to make $v_{com} = r\omega$ and in this process some velocity is lost so according to formula of angular momentum $L=mvr$,about a point on the ground $L$ decreases as $v$ decreases.  But since friction is also acting about a point on the ground, torque about a point on the ground is zero, so then how would angular momentum change.  edit:this question is about applying conservation of angular momentum in rolling with slipping.
67162
The following quote comes from:      Consider an ice skater that has just gone into a spin with arms   stretched out.  If it helps, the skater is wearing lead bracelets on   each wrist.  As the skater spins, the angular momentum and angular   kinetic energy are constant;  friction with the air and with the ice   can be eliminated.  The skater must exert an inward centripetal force   on the bracelets to keep them rotating in the same circle, but this   force does no work, since the bracelets are not changing their radius   of rotation.      Then the skater pulls in his arms and the bracelets, closer to his   axis of rotation.      Several interrelated things happen:           since there is no external torque, the angular momentum of the system stays the same;   Since the various masses of the skater are all moving towards the axis of rotation, the total moment of inertia of the system   decreases;     Combining 1 and 2, the angular velocity of the system increases;  if the moment is halved, the angular velocity doubles   Combining 2 and 3, with some quantitative treatment, the angular kinetic energy increases;  if the moment is halved and the   angular velocity is doubled, the kinetic energy doubles;   The skater does work;  his arms are now exerting an inward force through a distance as his arms draw the bracelets inward;  the work done can be rigorously shown to be identical to the increase in   kinetic energy...      However, it only covers a case when work is done to pull arms inwards. It is understandable that the work would be done in such case.  But what is happening in the opposite case when the skater has their arms inwards and lets them go outwards? S/he doesn't do any work but kinetic energy (and total energy) is lost. Why is it lost?  E.g.  I = 2kg * (1m)^2 w = 3.14/s L = 2kg * m^2 * 3.14/s Ek = 9.869  After shifting 2kg by 0.1m outwards: I = 2kg * (1.1m)^2 = 2.42 kg * m^2 L = 2 * 3.14 = 2.42 * w w = 2.5963 Ek = 8.1566   So, where did 1.7124 J go?  The quoted answer describes the easy case and omits the hard one.
44002
Are angular quantities vector?   ... It is not easy to get used to representing angular quantities as vectors. We instinctively expect that something should be moving along the direction of a vector. That is not the case here. Instead, something(the rigid body) is rotating around the direction of the vector. In the world of pure rotation,a vector defines an axis of rotation, not a direction in which something moves.   This is what my book (authored by Halliday, Walker, Resnick) says. But they mentioned with a caution that      Angular displacements cannot be treated as vectors. ....to be represented as a vector, a quantity must also obey the rules of vector addition..... Angular displacements fail this test.   Why do they fail in this test? What is the reason?
71958
We know that it takes no energy to change the direction of a vector, we know also that it takes no energy to displace a body in motion if a push is applied exactly at a right angle on its center of mass.  Considering that, if that is true, can you explain why a simple change of direction without any increase of speed is considered an acceleration?
52873
thanks for any help.  I'm trying to show that in a 2body problem, angular momentum is conserved given that $\dfrac{dp}{dt}=\dfrac{-GMm(rv)}{r³}$, where p is momentum, t time, G gravitational constant, M mass of 1 object, m mass of the other, (rv) is the vector between them and r is the magnitude of the vector between them.  I've had a lot of attempts but I don't seem to get very far.  Where L is angular momentum, $r_1$ is position vector of mass M and $r_2$ is position vector of mass m, $p_1$ is momentum of mass M and $p_2$ is momentum of mass m.  L=$r_1Xp_1+r_2Xp_2$  If $\dfrac{dL}{dt}=0$ then dL will be conserved over time, applying product rule             $L=\dfrac{dr_1}{dt}Xp_1+r_1X\dfrac{dp_1}{dt}+\dfrac{dr_2}{dt}Xp_2+r_2X\dfrac{dp_2}{dt}$  noticing that $p_1=m\dfrac{dr_1}{dt}$ and $p_2=m\dfrac{dr_2}{dt}$ clearly $p_1$ is parallel to$ \dfrac{dr_1}{dt}$ and $p_2$ is parallel to $\dfrac{dr_2}{dt}$ and hence when cross producted with them, the result is 0.  Giving $L=r_1X\dfrac{dp_1}{dt}+r_2X\dfrac{dp_2}{dt}$  I gather I must now show that either these two terms are 0, or they are equal and     opposite. I can't see any reason why they would be 0, and (slightly guessing) it seems to    make sense to me that they are terms related to how the angular momentum of one body changes    with the other body, and I assume they must therefore be equal and opposite. However when I    do  $r_1X\dfrac{dp_1}{dt}=-r_2X\dfrac{dp_2}{dt}$      I notice $\dfrac{dp_1}{dt}=\dfrac{-GMm(rv)}{r³}$ and $\dfrac{dp_2}{dt}=\dfrac{GMm(rv)}{r³}$ (since (rv)'s direction is flipped)     Therefore $r_1X\dfrac{-GMm(rv)}{r³}=-\dfrac{r_2XGMm(rv)}{r³}$     Defining $w=\dfrac{GMm}{r³}$:  $r_1X(rv)(-w)=r_2X(rv)(-w)$     Scalar products are commutative so the (-w)'s easily cancel giving  $r_1X(rv)=r_2X(rv)$  Which then the only solutions are if all components of $r_1$ are equal to all components of      $r_2$, however angular momentum should clearly be conserved in all positions, not just when the masses are ontop of each other.  Thanks again for any help.
59688
Why is angular momentum conserved when a planet revolves about sun in an elliptical orbit? Why is linear momentum not conserved in this case?  Please use the minimum amount of equations and try to explain using pure physics logic if possible.
93469
Say we have a rod with a fixed axis of rotation. If we give it a push at some point other than the axis, it will start to rotate. Say our force is $F$, then we might write: $\Delta P=F\Delta t$. I’m wondering how to incorporate angular momentum here. We know that for a single particle it holds that $L=r\times p$. And for an extended object with a fixed axis of rotation, we generally might write $L=I\omega$. However, can I write $L=r\times p$ too for the extended object? I would think that I would need to integrate. So we have $p_\alpha=m_\alpha v_\alpha=m_\alpha r_\alpha\omega$, where we consider an infinitesimal mass $\alpha$. So should I write $L=\int r^2\omega\,dm$? I’m confused whether to work with this integral, or just use $L=r\Delta P$ at once.  The reason I'm asking, is because I'm trying to calculate the sweet spot of a baseball bat. So this is my attempt at solving the exercise.