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Suppose we have a tyre hanging in mid air and has a metal rod in the middle of the tyre. If we apply a force at the end of the tyre then a Torque will be produced. I heard from someone that if we apply force at the end of it(the tyre) then there will be an increase in angular momentum in direction perpendicular to both the force and the moment arm.

So, my question is: What does the term "there will be an increase in angular momentum in directon perpendicular to both the force and the moment arm" means and what will it be beneficial of?

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Torque gives the rate of change of angular momentum, so the benefit is, once you understand how to compute torque (look that up, it's pretty simple in special cases like tires), and once you understand what angular momentum is (look that up too, it's also pretty simple in those types of special cases), you can use one to get the rate of change of the other. That's the game we play when trying to understand rotational motion, very analogous to how we use force to be the rate of change of momentum when we want to understand linear motion.

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There are a few things to consider here.

Fist, the fact that torque equals the rate of change of angular momentum. If angular momentum lies along the x-axis, then torque about the x-axis changes the magnitude of angular momentum, and torque about the y or z axis change the direction of angular momentum (but not its magnitude). The change in direction is towards the direction of the torque.

So what direction is the torque applied?

The vector equation for torque as a result of a force $\boldsymbol{F}$ applied at a distant location $\boldsymbol{r}$ is $$ \boldsymbol{\tau} = \boldsymbol{r} \times \boldsymbol{F} $$ where $\times$ is the vector cross product. A property of the cross product is that the results is mutually perpendicular to both $\boldsymbol{r}$ and $\boldsymbol{F}$.

In summary, Torque is perpendicular to the force and the moment arm, and angular momentum changes direction towards the direction of torque.

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Torque $\tau$ causes angular momentum $L$ in a linear manner, $\tau=dL/dt$, so torque and angular momentum are in the same direction (lets call them parallel). Torque itself comes from a force $F$ acting at a distance $r$ from the rotation centre:

$$\tau=r\times F$$

This is a cross-product. A cross-product mathematically gives a resulting vector which is perpendicular to both of the involved vectors. So, the resulting $\tau$ vector is perpendicular to both the $F$ vector and the $r$ vector. This is simply a mathematical quirk due to the cross-product operation.

Now, it is not that simple to talk about the "direction" of a rotational parameter. What is the direction of angular velocity or angular momentum or angular acceleration? You would maybe say clockwise or counter-clockwise, but that is a curved direction and not easily expressed mathematically.

Instead, it is easier mathematically to talk about the axis, about which the rotational parameter acts. When a torque turns a wheel, it is turning about an axis through its centre. So, if you can define that axis, then you have a hold on the direction, mathematically (which can be described with coordinates and such - much easier to work with).

The torque vector found in the cross-product thus points along this rotation axis. And then we use the right-hand-rule to say that if the axis points outwards (out of the screen/paper), then the actual direction is counter-clockwise and vice versa.

This is why you often in technical literature will hear the description of "a direction along an axis", even though we are talking about a rotational parameter. Saying "directed along an axis" actually rather means "turning about that axis".

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protected by ACuriousMind Apr 9 at 16:34

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