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I am trying to understand torque and angular momentum. I faced the following problems but couldn't find an answer in my textbook or internet:

  1. why is torque equal to vector product of force and position vector (radius). Why isn't it the addition of centripetal force and the acting force on the object in radius. What does this denote?

  2. And the same holds for angular momentum why is the radius in the equation and why does an object move faster as it comes near the center( where does the inertia opposing the change come from. Is it due to the object tendency to maintain rotation in the originally larger radius?)

Further explanation:

Suppose a force $F_1$ is acting on a rod at a distance $x$ from the hinge , now if i am trying to find the force $F_2$ which would give the same effect on the rod as $F_1$, i would follow the following procedure (neglecting the knowledge of torque):
$a = \alpha r$

$F_1 = m\alpha r_1$

Since the rod is virtually one piece ($\alpha$ is constant)

$F_2 = m \alpha r_2$

Dividing two equations:

$F_2 r_1 = F_1 r_2$

but according to torque:

$F_1 r_1 = F_2r_2$

Refering back to question (1.), why do we need the torque equation to express the tendency to rotate and why do i need more force as i push the rod near its hinge? Is it only due to our experimental results (like how archimedes discovered lever laws)?

another situation:

I know the mathematical derivation of $r^2 \omega = rv = constant$ in a single frame of refrence (when circumferential acceleration $a_\theta$ is zero), but i want to understand it intuitively (without mathematics) because this is the part that leads to angular momentum derivation. where is the stored velocity coming from (which is changed when velocity changes) .When talking about linear momentum. we think of it as a representation for what is the amount of velocity that would be given to a body when it collides with another one.

For the mathematical derivation , it is in lecture 15 of MIT course

Note: I read several answer in the site but none of them addressed the deep meaning of these vectors and quantities.

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2 Answers 2

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Torque is nothing but moment of force, i.e

$\vec{τ} = \vec{r} × \vec{F}$

That is not centripetal force, centripetal force is the force which keeps the object in circular motion and which is always directed radially inwards. And torque is something which is present when there is $Angular$ $Acceleration$ in rotational motion.

Moment of something is defined as measure of a force's tendency to cause a body to rotate about a specific point or axis "or" a moment is an expression involving the product of a distance and physical quantity and that's why we multiply the term "$r$" with it. let's say if we want to study about moment of momentum then we define it as:-

$\vec{L} = \vec{r} × \vec{P}$

Which is also known as angular momentum.

The reason for "why does an object move faster as it comes near the center" is that according to the relation

$V = ωr$

$\frac{V}{r} = ω$

As we can see that from here the angular velocity and the radius is inversely proportional to each other and if we decrease the radius then the angular velocity of the particle increases and that is the reason as the angular velocity increase with decrease in radius.

Edit :- as torque is the moment of force it can be represented as shown above i.e:-

$\vec{τ} = \vec{r} × \vec{F}$

Let's understand this with an example, let the torque be some constant $\vec{τ_0}$ till the end.

And let's consider a rod with some length $l_1$ which is attached to a frictionless hinge (since it may provide an external torque other than the torque which we are providing to the system if it is present).

And exert a force equal to some constant $F_1$ on then end of the rod and mutually perpendicular to the axis of rotation and the rod as well (to keep it simple for further elaboration and your understanding, although you can put the force in any direction you want) hence, if it is perpendicular to the rod By solving the above equation for the torque we get:

$\vec{τ_0} = l_1 F_1 sin \theta \hat{n}$

Note : $\hat{n}$ is the direction perpendicular to the force and the radius vector.

In this case as I already mentioned that the radius vector and force vector are perpendicular to eachother then the theta must be $90°$ so:-

$\vec{τ_0} = l_1 F_1 sin 90° \hat{n}$

Then the torque on the system will come out to be:-

$\vec{τ_0} = l_1 F_1 \hat{n}$

And now let a consider a length equal to $l_2$ Where $l_2 < l_1$

Following the same procedure for $l_2$ as for the length $l_1$ and now for this case exerting the force equal to $F_2$ at the length $l_2$ you will get:

$\vec{τ_0} = l_2 F_2 \hat{n}$

$\hat{n}$ will be in the same direction as before.

Now from the general for the torque i.e:-

$\vec{τ_0} = l F sin \theta \hat{n}$

And as I said the $\vec{τ_0}$ is a constant quantity by comparing the magnitude we get:-

$τ_0 = lF$

$\frac{τ_0}{l} = F$

From this equation we can see that

$\frac{1}{l} ∝ F$

As radial length increases to keep the torque constant the force you will exert must be less or if you apply the force near to the hinge then the force is going to become greater, in mathematical way we can express it as:-

Comparing the torque equations for $l_1$ and $l_2$ (or in other language taking ratio of both the equations)we get:-

$l_1 F_1 = l_2 F_2$

$\frac{l_2}{l_1} = \frac{F_1}{F_2}$

As we know $l_1 > l_2$ so the answer will come out to be a number less than $1$ and if the ratio of $l_1$ and $l_2$ is less than $1$ then the ratio for $F_1$ and $F_2$ will also have to be less than $1$ as well, according to the above equation that means:

$F_2 > F_1$

Hence we've proved it.

Hope it helps.

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  • $\begingroup$ In the part you are explaining why speed increases. I meant linear not angular since angular momentum is conserved and equal to $mvr$ linear velocity should increase too when it moves closer to center $\endgroup$
    – mohamed
    Commented Jan 28, 2022 at 19:15
  • $\begingroup$ No, that completely depends upon the situation what you are thinking about everything matters not just angular momentum! If you see the centripetal force $F = mv^2/r$ and the angular momentum $L = mvr$, relation $v = \omega r$ these relations are different and we are able to conclude the actual reason, as we can see in these situations the tangential velocity is either directly proportional or inversely proportional so that completely depends upon situation we can't merely say that angular momentum is conserved, momentum can also vary when there is torque in the system because........ $\endgroup$ Commented Jan 29, 2022 at 2:43
  • $\begingroup$ ......Rate of change of momentum with respect to time is known as torque! But I took a basic one that's $v = \omega r$ so it depends upon the situation what's actually happening....edit the question if you want and describe what type of situation you are dealing with.. either momentum is conserved or not please specify if you want full answer to your question. $\endgroup$ Commented Jan 29, 2022 at 2:47
  • $\begingroup$ I clarified the question and sorry for being late $\endgroup$
    – mohamed
    Commented Jan 31, 2022 at 13:34
  • $\begingroup$ @mohamed Okay I've edited the answer I think it'll be more clear to you now and answered some of your new questions as well but for the case of tangential velocity you still haven't clarified that if the angular momentum is conserved or not, same case with the centripetal force if it is constant or not, so for this reason I'm going to keep that part same as it was before. $\endgroup$ Commented Feb 2, 2022 at 7:18
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Consider a thin mass-less rod with a friction-less axle at one end. A point mass (m) is attached to the rod at a distance (r) from the axle, and a force (F) is applied perpendicular to the rod (and the axle) at a distance (R) from the axle. In a time (t), the rod rotates from a rest position through an angle (θ). The work done by the force on the rod is transmitted to the mass: W = F(Rθ) = (ma)(rθ) = m(rα)(rθ). Dividing by θ gives: W/ θ = FR = (m$r^2$)α. Which says: The work done per unit angle of rotation = Torque = Iα. Where (I) is the rotational inertia. (This result can be extended to any number of forces and point masses within a rigid system.) Note that if we multiply by (t): (FR)t = Iαt = Iω = (m$r^2$)(v/r) = mvr. (Torque times time = change in angular momentum.) The axle (or hinge) does apply forces which constrain the motion of the rod to rotation but it does no work.

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  • $\begingroup$ I understand this way of deriving torque : work done is conserved and so, $F . 2 \pi r$ is conserved and thus $F R = f r$. but +1, for the way of linking it to work done per angle. This made me think so if Force increases when we get closer to the hinge, then does this mean the hinge resists the force (has a reaction force on the rod) and if so how could someone calculate the reaction force as a function of r? And also does this mean the first part of my question where i get $F_1 r_2=F_2 r_2$ is due to the NET force and not the total force used ( Work done gets the total force x distance ) $\endgroup$
    – mohamed
    Commented Jan 31, 2022 at 17:49
  • $\begingroup$ And another thing if $ F.r = 0 $ and $F.r = mr^2 \alpha$ . Since $ mr^2 != 0 $, $\alpha = 0 $, so $ \delta \omega / \delta t = 0$ and $\omega$ is constant ->$ v/r = constant$ but according to angular momentum : $ v . r = constant$ $\endgroup$
    – mohamed
    Commented Jan 31, 2022 at 18:30
  • $\begingroup$ The reaction force from the axle acts at the axle. It does not move and does no work. The torque (which does the work) is not F.r. It can be put in vector form as R x F, with the resultant vector (as well as the angular acceleration vector) parallel to the axle. In this case, the angular momentum vector can be expressed as r x mv. $\endgroup$
    – R.W. Bird
    Commented Feb 1, 2022 at 13:45
  • $\begingroup$ I didn't mean dot product (my mistake). I meant if you are applying a force perpendicular to the rod and tangential to the circle then the rod should move in a straight line but instead different parts of it move at different speeds ( that's what I mean by reaction force of the hinge (I think it is the constraint of rod and not the hinge)) as well as its direction changes due to centripetal force. $\endgroup$
    – mohamed
    Commented Feb 1, 2022 at 19:43
  • $\begingroup$ Note the correction to my answer. $\endgroup$
    – R.W. Bird
    Commented Feb 2, 2022 at 14:17

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