Angular momentum and torque meaning

Question

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.

Answer 1

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.

Answer 2

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.