So, I understand that, when an object undergoes rotational motion, the individual masses undergo/experience various forces and momentums based on their distances from the point of rotation. As such, a mass that is farther away would, if it were to undergo rotational motion, accelerate/experience a greater force than one that is closer. I also understand, generally, how torque is derived from angular momentum, though I don't see why the cross products of the individual linear momentums of the masses with their distances from the point of rotation need to be taken as it seems repetitive. And, obviously it would make sense that if you applied the same mathematical operation to all of the different values (force, momentum, mass)then they would match up. What I'm am confused on is, for an object in free space for example, how do you know that a force applied to it at some point is going to have this exact effect on it? Like, regarding torque, isn't it only representative of the force that an individual mass would have IF it were to accelerate rotationally. But when you apply the force, you are simply accelerating it linearly, and the mass then experiences various internal forces that affect its trajectory from there. I'm confused on how you know how all of these forces/internal forces will play out/interact with each other, obviously they will all cancel out, but how do you know that they will cancel out in such a way that it creates a net torque about the center of mass equal to the force times its distance from the center of mass? (without assuming that the force itself creates a torque, since my confusion is how we know the force creates a torque) I feel like I must be missing something obvious as everyone else I see on similar questions understands this concept fine. Edit: deciding to accept linked answer as it does technically answer the question I asked. Though I still feel that an explanation in terms of Newton's 3 laws for linear motion should be possible, the answers provided have helped me better intuitively understand why linear forces should also exert torques based on other conservation laws which seem to make sense themselves. Edit: deleting question as I don't think it adds any relevance and I have posted the link I found to other relevant questions. Edit: Nvm, I can't delete the question.
120616
I used to think that torque and force were equally “fundamental”. In other words, my understanding was that we usually use Cartesian coordinates in many common problems because it is a convenient system, so as a result instantaneous forces which act in straight lines seem “easier” mathematically but torques require some extra “baggage”. This baggage includes typically teaching that torque is defined in terms of force.  But if say we happened to choose polar coordinates for the problem the situation would appear the other way around. So it would be arbitrary if we chose to define forces in terms of torques instead.  But later on I learned that angular momentum is conserved independently of regular linear momentum (IIRC). Given the definitions of force &amp; torque as derivatives of momentum, this makes it seem much less certain that one should define either torque in terms of force or vice versa — it gives the impression they are more distinct than it first seemed.  That said, as far as I know a lot of physics is about defining &amp; describing “fundamental forces” — not “fundamental torques”.   So is choosing to use either force or torque as the basis of laws &amp; problems arbitrary? Or is there an actual fundamental rationale to when one or the other should be used?
14136
I'm far from being a physics expert and figured this would be a good place to ask a beginner question that has been confusing me for some time.  According to Galileo, two bodies of different masses, dropped from the same height, will touch the floor at the same time in the absence of air resistance.  BUT Newton's second law states that $a = F/m$, with $a$ the acceleration of a particle, $m$ its mass and $F$ the sum of forces applied to it.  I understand that acceleration represents a variation of velocity and velocity represents a variation of position. I don't comprehend why the mass, which is seemingly affecting the acceleration, does not affect the "time of impact".  Can someone explain this to me? I feel pretty dumb right now :)
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?
87430
I was doing calculations with torque and then I came across something very confusing:  I understand that the magnitude of the torque is given by product of the displacement(from the center of rotation) at a point and the force applied at that point. This magnitude perhaps, says something about how "powerful" the rotation is. The question that confuses me, however, is what does the direction of the  tell? The direction is perpendicular both to the force and the displacement. What really is this the direction of?
83769
If I'm not wrong, torque is perpendicular to both the radius and force i. e. It is along the axis of rotation. Questions that arise are- why do we consider the length between the axis/point of rotation while calculating torque? More importantly why is torque a cross product?
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)
51914
&quot;The centre of mass of a system of particles is the point that moves as though (1) all of the system's mass were concentrated there and (2) all external forces were applied there.&quot; (Resnick et al, p.215) &quot;The gravitational force on a body effectively acts at a single point, called the centre of gravity of the body.&quot; (Resnick et al, p.330) According to these concepts: If we have a body large enough that we could not assume that the centre of mass and the centre of gravity doesn't coincide. Then, by definition of the centre of gravity, the gravitational force does not act on the centre of mass. This means that we have an external force which we can't consider to be acting on the centre of mass. Q (1)-Does the centre of mass still move as if the gravitational force were applied there? Q (2)-Aren't this concepts the best definitions of the center of mass and centre of gravity? Q (3)-What would be the axis rotation if I apply a force to one corner of this body? Source: Fundamentals os physics / Robert Resnick, David Halliday, Robert Resnick - 10th edition.
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).
25589
From Wikipedia: The work done by a constant force of magnitude F on a point that moves a displacement d in the direction of the force is the product: $$W = Fd.$$  If I lift some object from a ground, the force to be put in above equation is the gravitational force $mg$.   But while I am moving the object upwards, against the gravity, I must pull the object with greater force than $mg$, right? So the net force is $F_{myforce}-mg$ which results in object being lifted. But this is not the case. Why? Does it matter how strongly I pull the object?  What work will I do if I move this object in absence of gravity (in space)? Will work done be zero or not?
69255
While studying rotational mechanics, I came across a section where it mentioned that angular momentum may not necessarily be parallel to angular velocity. My thoughts were as follows:  Angular momentum ($L$) has the relation $L=I\omega$ where $\omega$ is angular velocity and $I$ is the moment of inertia, so following this relation, it seems they should be in the same direction. Why are they not?
68074
I understand the equations of angular motion and how they relate to one another, however it is not clear to me where the torque equation comes from.   We have $F=ma$. By solving for the angular interpretation of linear acceleration, we have $$a=\frac{dv}{dt}r=\frac{d\omega}{dt}r=\alpha r$$ and thus $F=m\alpha r$.   With $\tau=Fr$, $\tau=m\alpha r^2$. When we have multiple torques we can add them up because $\tau$ is linearly related to force and force is a sum. By defining $I=\sum{m_ir_i^2}$, we have $\tau=I\alpha$  I understand that the concept of inertia is really an abstraction. We see the discrete $\sum{m_ir_i^2}$ or $\int{m_ir_i^2}$ forms frequently when dealing with equations of angular dynamics and thus decide to name the quantity moment of inertia based on the common role it plays in the equations (eg: for kinetic energy)  My question, a clarification   However, I'm unclear about how we know $\tau=Fr$. Is $\overrightarrow{\tau}=rF$ a wholly mathematical corollary of newton's laws of motion, (without creating a tautology), or a later development backed by empirical research.
103728
Consider this situation:  A ball is moving forward and undergoing rotation. Assume that it is not slipping. Eventually, the velocity and rate of rotation of the ball decrease, and it comes to a halt. But if you observe the direction of friction (when the ball is rotating clockwise), you will see that the friction should have provided a clockwise torque to the ball and the angular velocity of the body should have increased. But this doesn't happen. Why?
112481
Observe  case. We can make two choices regarding the axis of rotation. If we choose an axis of rotation that passes through the center of mass $G$ then the only force that would cause the ball to rotate is the friction force and some force $F$ that's acting on the center of mass would just provide translational motion.   However, if we choose the point $P$ as the axis of rotation then it is the opposite. In that case $F$ causes rotation and friction provides translational motion. However, I know that without friction no rolling is possible (the ball would just slide) but how is that possible if the force of friction contributes in no way to rotating while analyzing the instantaneous axis of rotation?
94446
In most of the cases the center of mass is chosen for rigid body motion description, but this is not an obliged choice, since the motion of any point $P$ of the rigid body can be seen as the combination of a traslation of another point $Q$ and a rotation about an axis passing through $Q$. I'm trying to understand what are the main reason why the center of mass is  chosen as the origin for the frame of reference in rigid body dynamics.  I thought about these main reasons:   Center of mass theorem: $\sum F^{(external)}=M a_{CM}$. Nevertheless this concerns traslation motion only and the traslational velocities of the points of the rigid body are all the same, so in principle I could write $\sum F^{(external)}=Ma_{P}$, where $P$ is a random point of the rigid body. Does that makes sense? Koenig theorems (kinetic energy and angular momentum) Parallel axis theorem   The last two, as far as I understood, are valid only if the center of mass frame is used and in no other case. Is it correct?  Are these the main reason why the center of mass frame is used in rigid body dynamics, or are there other reason behind this choice?
47567
I don't understand the fact that let's say hypothetically I was pushing a block on a frictionless surface, and I kept applying the same force on the object how it would keep accelerating! Wouldn't there have to be a certain velocity where it would stop getting faster? If not then what is actually acting on the object to cause it to accelerate infinitely?
60355
"We have focused our discussion on one-dimensional motion. It is natural to assume that for three-dimensional motion, force, like acceleration, behaves like a vector."- (Introduction to Mechanics) Kleppner and Kolenkow  We learn it very early in the course of our study that Force is vector; But, if I were the physicist defining the the Newton's second law (experimentally) and analysing the result F=ma, how would I determine whether Force is vector or scalar(especially in 3-D).   Actually, when I read the aforementioned sentences from the book, I wanted to  know why do the authors expect it to be natural for us to think that in 3-D "Force" behaves like a vector. I know a(acceleration ) is vector and mass a scalar and scalar times vector gives a new vector but is there another explanation for this?
28410
Why is force = mass $\times$ acceleration? I have searched in many sites but didn't actually get at it. Simply I want to know that if a mass in space moves (gains velocity thus further accelerates), how can I think, postulate and further believe that force = multiplication  of mass and its acceleration ?