Torque, $\tau$ is defined as:

$$\vec{{\tau}} = \vec{r}\times\vec{F}$$

Intuitively, it's the rotational effect of force. However, using the above definition of torque we can calculate the torque due to a force acting on any particle about a reference point.

Consider projectile motion for instance. A particle was projected at an angle $\theta$ with a speed $u$. I calculated the torque due to weight at the highest point of the particle's trajectory about the point of projection to be $-mu^2\sin^2\theta$.

The thing that confuses me is that here there's torque but no rotation. What is the purpose of torque in projectile motion then?

## Best Answer

The equations that you have obtained for the trajectory of a projectile can be obtained by considering the torque which is acting on a mass $m$ and equating the torque to the rate of change of angular momentum.

Consider a mass which is projected at an angle $\theta$ to the horizontal from location $O$ at position $\vec r_{\rm O} = 0 \hat i+0 \hat j$ with velocity $\vec v_0 = \dot x_0 \hat i + \dot y_0 \hat j$ at time $t=0$.

At time $t$ the new position of the mass is $\vec r = x \hat i + y \hat j$ and it has a velocity $\vec v = \dot x \hat i + \dot y \hat j$.

The constant force acting on the mass is $-mg \hat j$.

At time $t$ the torque acting on the mass about location $O$ is

$\tau_{\rm O} = (x \hat i + y \hat j) \times (-mg\hat j) = -mgx\,\hat k$

where $\hat k$ is the unit vector coming out of the screen.

At time $t$ the angular momentum of the mass about location $O$ is

$L_{\rm O} = (x \hat i + y \hat j) \times m(\dot x \hat i + \dot y \hat j) = m(x \dot y - \dot x y)\, \hat z \Rightarrow \dfrac{d\vec L_{\rm O}}{dt} = m(x \ddot y - \ddot xy)\,\hat z$

$\vec \tau_{\rm O} = \dfrac{d\vec L_{\rm O}}{dt} \Rightarrow -gx = x \ddot y - \ddot xy$.

As the last equation must be true for all $x$ and $y$ this means that

$\ddot x =0$ and $\ddot y = -g$

which are the equations that you get by applying Newton's second law to the horizontal and vertical motions of a projectile.

In this case you could say that the torque has caused a change of direction, "rotation", of the velocity vector or you could say that the position vector $\vec r$ has been "rotated" but the key idea is that the torque has changed the angular momentum.