Suppose we have a weightless, rigid rod fixed at one end, but free to swing at the other, where there is a mass $m$ attached.    If we want to determine the tangential acceleration of the mass using linear mechanics, we can use the formula,  $F = ma$  Thus,  $a = \frac{F}{m}$  Thus provided the force is always acting perpendicular to R, the mass will accelerate tangentially at $a$.    Now if instead we analyse the problem using torque, we can arrive at the equation,  $\frac{dL}{dt} = \tau =R\times F = RF$  Therefore, $\Delta L = RFt$ where t is the time the force is acting on the mass (again assuming the force is always applied in a direction perpendicular to R).  Since $L = mRv$, where $v$ is the tangential velocity of the mass, we have,  $mRv = RFt$  Therefore,  $\frac{v}{t} = \frac{F}{m}$  $a = \frac{F}{m}$  which is exactly as predicted from the linear analysis.    So in this very simplistic scenario, the acceleration of the body could be derived using either linear mechanics or utilising the concepts of torque and angular momentum.  Given this, in general when does the application of linear momentum start to break down when describing rigid body physics problems, and the concepts of torque and angular momentum become necessary to arrive at a solution?
123462
For a long time I have wondered if there is a way to show that the rotational analogs of Newton's Laws are a direct consequence of just those laws, or are we adding more to them?  I understand mathematically that we take Newton's second law and do "r cross both sides," but that has always struck me as using more than just the 2nd law.  Here it matters where forces on the body are applied, but I don't see where Newton's Laws talk about points where forces are applied to a body. For translational motion it doesn't matter.  So are we adding a sort of "extra law" when we do this for rotation?  The only thing I can think of would be to "disassemble" an extended body into differential point masses and work with internal constraint forces that make the body rigid. If every mass is just a point mass it would get around the issue I'm having.  I've never seen any discussion along those lines, though.
21597
Assume a point-mass $m$ is travelling in a straight line, and a force $F$ will act on $m$ (in the same direction as $m$'s velocity) over a constant distance $d$; why doesn't $m$'s velocity matter to the calculation of work done on $m$ by $F$? Work is defined such that, in this example, the work done by $F$ on $m$ is equal to $Fd$, but it seems that if $m$ were moving slower, it would spend more time in the field, allowing $F$ more time to act on $m$, thereby doing more work. In fact, if $m$'s velocity were very great, it would hardly spend time in $F$'s field at all (so very little work done). Maybe I misunderstand work; can someone address this confusion of mine?
58409
I'm trying to simulate a wheel as it hits the ground.  Problem 1  Suppose a disc is dropped from a height. It has initial velocity of $-x,-y$ caused by throwing and gravity. It has no initial angular velocity. When it hits the ground it should have some rotation resulting from the collision.   How to calculate force that caused torque for that rotation?   Problem 2  The same disc is dropped from height. It doesn't have velocity on the side direction. It already spinning fast. When it hits the ground the spinning should cause some translation.   How to calculate force that caused this linear acceleration? Some rotation must be lost in the collision, how to calculate that?   To simplify assume static $friction = 1$ and $g = -10$.
10573
I have been looking for this for quite some time now. A simple pendulum behaves in SHM. Let's put that pendulum in an upward accelerating elevator. The component of the force that acts in SHM $(\text{mg}\sin\theta)$ still stays the same in my head.  However, websites and books tell me to use $m(g+a)\sin\theta$ where $a$ is the acceleration of the elevator.  I tried to look up Free Body Diagrams, but I can't find any for the case of accelerating frames. Can someone explicitly prove this without using the flimsy argument of "Think it's a noninertial frame with a new effective g"?
39677
I know that :  Position $$r:\mathbb{R} \to\mathbb{R^3} \ \ \ \ \ ;  \overrightarrow{r}(t)$$  Velocity $$v :\mathbb{R} \to\mathbb{R^3} \ \ \ \ \ ;  \overrightarrow{v}(t)$$  Acceleration $$a:\mathbb{R} \to\mathbb{R^3} \ \ \ \ \ ;  \overrightarrow{a}(t)$$  Now i want to know that , what is function of force $F$ and work $W$ ?
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
117026
Lets imagine a binary system of two astronauts in space connected to one another via light rope.  The rope is taut and they're spinning round and round with their axis of rotation being the the axis perpendicular to the their centre of mass.  Now, my question is this. Lets say they each let go of the rope; they move off at tangents.  Is angular momentum conserved ? And what is their subsequent motion ?  My hypothesis : They will move off at constant velocity at a tangent to their circular motion. I believe that because they now under the effects of no external force by the rotational analogue of newton's second law (dL/dt = T) their angular momentum will be of equal magnitude but opposite sign.  Does this make sense ?
118793
Consider a particle that is described by $x(t)=(2.0-0.1t) \cos(0,5t)$ and $y(t)=(2.0-0.1t) \sin(0,5t)$ t in seconds and x,y in meters.    In previous subquestions we were asked to determine an expression for the module of the velocity vector and the tangent acceleration. The expressions obtained are:  $v=\sqrt{k^2 + w^2(r_0-kt)^2}$  $a_t=-\frac{kw^2(r_0-kt)}{\sqrt{k^2+w^2(r_0-kt)^2}}$  with $k=0.1$ and $w=0.5$ and $r_0=2.0$.  Now my question is in the next question, to find an expression with the normal acceleration.   My attempt was to take   $a_c = \frac{v^2}{R}$  $R = \sqrt{x^2+y^2}= 2.0-0.1t$  $a_c=\frac{k^2}{r_0-kt} + w^2$  The answer given by my textbook is however:  $a_c= 4k^2w^2 + w^4(r_0-kt)^2-\frac{k^2w^4(r_0-kt)^2}{{k^2+w^2(r_0-kt)^2}}$  I don't understand how this formula was derived. It seems like the third term of the sum is $-a_t^2$ but I can't understand the other two terms and what variables ares involved. And also I want to know why my answer is wrong or if it is equivalent to the one given, I can't understand that. Thanks!
63588
Here is a classic problem:     A 60kg boy and a 40kg girl are sitting in front of each other on skateboards on a frictionless horizontal surface. The boy pushes the girl and the girl gains a velocity of 0.3m/s during the 0.5s the boys hand was in contact with her. What is the boy's resulting velocity?   Considering Newton's Third Law, the best way is to find the total force applied and rearrange the formula to find the boy's speed.  Newton's Law Method:  $F = m*a$  $a=\frac{0.3m/s}{0.5s}$  $F = 40kg * 0.6m/s^2$  $F = 24N$  Therefore the boy's speed will be:   $\frac{24N}{60kg} = a$  $a = 0.4m/s^2$  $v = 0.4m/s^2 * 0.5s$  $v = 0.2m/s$  Kinetic Energy Method:  $E_{k}=\frac{1}{2}*m*v^2$ the girl's kinetic energy  $E_{k}=\frac{1}{2}*40kg*(0.3m/s)^2$  $E_{k}=1.8J$  Since every action has an equal and opposite reaction the boy should have the same amount of kinetic energy.  $E_{k}=\frac{1}{2}*60kg*(0.2m/s)^2$  $E_{k}=1.2J$  As you can see, using Newton's Third Law, the kinetic energy was not the same. So what's going on here? I thought that in this case they should both have the same amount of kinetic energy due to Newton's Law. I must be missing something since there is more energy going in one direction than the other after meeting. My book used the first method as the answer and, well, it feels wrong.
24170
This may seem like a slightly trite question, but it is one that has long intrigued me.  Since I formally learned classical (Newtonian) mechanics, it has often struck me that angular momentum (and generally rotational dynamics) can be fully derived from normal (linear) momentum and dynamics. Simply by considering circular motion of a point mass and introducing new quantities, it seems one can describe and explain angular momentum fully without any new postulates. In this sense, I am lead to believe only ordinary momentum and dynamics are fundamental to mechanics, with rotational stuff effectively being a corollary.  Then at a later point I learned quantum mechanics. Alright, so orbital angular momentum does not really disturb my picture of the origin/fundamentality, but when we consider the concept of spin, this introduces a problem in this proposed (philosophical) understanding. Spin is apparently intrinsic angular momentum; that is, it applies to a point particle. Something can possess angular momentum that is not actually moving/rotating - a concept that does not exist in classical mechanics! Does this imply that angular momentum is in fact a fundamental quantity, intrinsic to the universe in some sense?  It somewhat bothers me that that fundamental particles such as electrons and quarks can possess their own angular momentum (spin), when otherwise angular momentum/rotational dynamics would fall out quite naturally from normal (linear) mechanics. There are of course some fringe theories that propose that even these so-called fundamental particles are composite, but at the moment physicists widely accept the concept of intrinsic angular momentum. In any case, can this dilemma be resolved, or do we simply have to extend our framework of fundamental quantities?
54091
I struggle to find a satisfying reason and formulation, why for a point mass,  that is fixed in a rotational system and the radius is changed (like pulling on a string with a small ball rotating around it), the speed of this point mass would increase due to the conservation of angular momentum.  that follows a trajectory with decreasing radius (like a Roller Coaster on a Spiral shaped rail), the speed of this point mass would not increase, since conservation of angular momentum is not applicable in this case. (I already started doubting, if it really does not increase).   The only reason I found so far is that in the second case, the rail acts as an external force to the mass, while in the first case the radius change is an internal change of the system that comes along with putting energy into the system by working against the centrifugal forces. While I expect this is a somewhat correct answer, I still find this is a little bit hand waving argument. I expect that I just miss a proper clean definition of external force vs internal force. I also constructed the following thought experiment for myself, that involves an external rail, but I would expect to react like the first case: When a point mass rotates on a circular stretchable rail and the radius of that rail would be changed by applying some force against the centrifugal force of the point mass, I'd also expect the velocity of the point mass increases, since some force applied over a distance (energy change) is involved. But I doubt that the applied force over a distance is the underlying fundamental reason and I expect that there is another more satisfying fundamental difference between the fixed predetermined trajectory and the thought experiment / the first case mentioned in the beginning.
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).
6494
I know this question is asked here a lot, but I just had to ask this to finalise the concept.  When a system lets say a rod of length $L$ and mass $M$ is rotating with angular speed $omega_1$ its initial angular momentum is $L1 = (1/12)ML^2\omega_1$ and its initial kinetic energy is $KE = (1/24)ML^2{\omega_1}^2$.  Now after some time the rod is folded in half its angular momentum kept conserved i.e. without applying any external force or torque, its new angular velocity becomes $\omega_2 = 4\omega_1$ and its new kinetic energy becomes $KE_2 = (1/6)ML^2{\omega_1}^2$.  This is 4 times the original kinetic energy when no external force works, since the rod is folded, you can even say melted and formed into a smaller and denser rod, it has not undergone compression/expansion of any sort but still there is change in kinetic energy.   The most sense I could make out of this was that all the particles while rotating felt a centripetal force and the particles of half of the rod under this force went in its direction and did some work which appears as the change in kinetic energy. I have written the same and a proof as an answer .  Now if I am write in my concept where did this energy come from, tension was providing the centripetal force but no work was done against tension as the rod was folded in half not compressed. If I am wrong then where did the energy come from?  Extra: I also tried the analogy of this question in translatory motion, suppose there is body of mass $m$ moving with velocity $v$  suddenly, its mass becomes $m/2$ then its velocity becomes $2v$ and its KE becomes $4KE_1$ there is no need to explain the energy conservation here since mass suddenly does not disappear into thin air, however moment of inertia can be changed and hence the question.  Addendum : Since folding the rod seems to bring about unnecessary questions about ways of folding, you can imagine that if rod was melted and formed into a longer rod all the while the system was rotating and angular momentum conserved, then new length becomes $2L$ new angular velocity becomes ${\omega_1}/4$ and new KE becomes $(1/96)ML^2{\omega_1}^2$. This time the energy becomes ${1/4}^{th}$ of the initial, where did this energy go to ? Certainly movement against centripetal force takes place, but since there is no extension in existing rod, energy can not be stored as spring energy in it, or so I think.
53653
A mass is attached to a rope, and put into a circular motion. If I pull the string from the center, the tangential speed of the mass will increase (by conservation of angular momentum).   I am applying a force only in the radial direction, so how can the tangential velocity increase if there is no tangential force?
18903
Assume there is a rigid body in deep space with mass $m$ and moment of inertia $I$. A force that varies with time, $F(t)$, is applied to the body off-center at a distance $r$ from its center of mass. How do I calculate the instantaneous acceleration, rotational acceleration, and trajectory of this object, assuming it starts from rest?
65676
Consider a cylinder that rolls without sliding on an inclined plane. If it's placed at the top of the plane, with its center of mass at a height $h$ from the bottom, it will have a potential energy $mgh$ (considering the bottom of the plane as the zero point). Then, the cylinder will start rolling down the plane, as its weight does work on it, causing its potential energy to be transformed into rotational and translational kinetic energy. I'm confused about the origin of the rotational kinetic energy of the cylinder. Since the static friction and the normal force are applied at the point of contact with the plane, neither of them can do work on the cylinder (which is why its mechanical energy is conserved), since the cylinder is, by constraint, not sliding. This seems to imply that the cylinder's rotational kinetic energy comes from the work done by its weight. However, I don't understand how this happens, since a rigid body's weight can be seen as being applied on its center of mass, which means it can exert no torque on it. Instead, the only torque exerted with respect to its center of mass comes from static friction. Moreover, if the inclined plane was frictionless, there would be no torque exerted on the cylinder, so it would not rotate. Yet, I don't see how this situation could be distinguished from the previous one by only looking at the cylinder's mechanical energy (since the normal force would still be doing no work). So, my question is: how is gravitational potential energy converted into rotational kinetic energy, and what is friction's role on this?
38481
In my book, the following are described to be the steps that one must follow if he wishes to find the direction of static friction force acting on an object:  &quot;(1) Draw the F.B.D(free body diagram) of the object in question with respect to the other object on which it is kept. (2) Include the pseudo force also, if the contact surface is accelerating. (3) Decide the direction of the resultant force and resolve this force into two components; one along the surface of contact and the other along the normal to this surface. (4) The direction of static friction is opposite to the component of the result force along the contact surface.&quot;  Firstly, are these steps correct? Will I always obtain the correct answer by following these steps? Secondly, consider the following problem:  In questions like these, when I try to find the resultant force and then assign the direction to static friction, I find that my answer is incorrect! Are there some other steps to follow when I know that the object is definitely accelerating in one direction or another? Edit 1: I approached the above question by doing exactly what my book suggests; first, I identified all the forces acting on the smaller block, being the normal force applied by the block A and its weight mg downward. Adding these two forces, I get another force that actually acts on the body like so:  I wonder if I'm not supposed to take either of the contact forces into consideration when drawing the F.B.D to figure out the direction of static friction acting on the body. Also, are there other steps for finding the direction of kinetic friction? Edit 2: From Judge's answer, I have been able to figure out my mistake, it being that I completely missed step (3) and didn't resolve the resultant force into components! Sheesh. Sorry! Much thanks in advance! :) Regards.