Question

A particle of mass m is projected with initial speed u at an angle of $45$ degree with the horizontal. What is the torque of the force on the object about the point of projection at the highest point?

Answer 1

First we have to draw a rough diagram representing all the parameters which are required to solve this problem. Then we have to find the time required to reach the maximum height. Using that time we have to find the horizontal distance covered by the particle. Now the force acting on the particle is the gravitational force and it is perpendicular to the horizontal distance. With the help of this we can find the torque.

Complete step by step answer:
As per the statement given we have a particle of mass m projected with initial speed u at an angle of $45$ degree with the horizontal.
We need to find the torque of the force on the object about the point of projection at the highest point.

So the vertical and the horizontal components of the initial speed which is projected at an angle of $45$ degree with the horizontal be ${u_x} = u\cos 45^\circ$ and ${u_y} = u\sin 45^\circ$ respectively.
Let the mass of the particle be m.
The time taken by the particle to reach at the maximum height be t. Then using the equation of motion we will get,
$v = u + at$
Where,
Final velocity of the particle is equal to $v = 0m{s^{ - 1}}$
Initial velocity of the particle is equal to $u = u\sin 45^\circ$ here we are using the velocity along y direction.
Acceleration of the particle due to gravity is equal to $a = - gm{s^{ - 2}}$.
Now putting the respective value we will get,
$0 = u\sin 45^\circ - gt$
Rearranging the above equation we will get the time as,
$t = \dfrac{{u\sin 45^\circ }}{g}$
Now the horizontal distance during the time to reach the maximum height be r. Then using the equation of motion we will get,
$s = ut + \dfrac{1}{2}a{t^2}$
Where,
Initial velocity of the particle is equal to $u = u\cos 45^\circ$ here we are using the velocity along x direction.
Time taken to cover that distance is equal to $t = \dfrac{{u\sin 45^\circ }}{g}$.
Acceleration of the particle in horizontal direction which is equal to $a = 0$.
On putting the values we will get,
$r = u\cos 45^\circ \times \dfrac{{u\sin 45^\circ }}{g}$
$\Rightarrow r = \dfrac{{{u^2}\sin 45^\circ \cos 45^\circ }}{g}$
Multiplying and dividing with two we will get,
$r = \dfrac{{{u^2}2\sin 45^\circ \cos 45^\circ }}{{2g}} = \dfrac{{{u^2}\sin 2\left( {45^\circ } \right)}}{{2g}} = \dfrac{{{u^2}\sin 90^\circ }}{{2g}}$
Hence the horizontal distance will be,
$r = \dfrac{{{u^2}}}{{2g}}$
This horizontal distance r represents the distance of vertically downward gravitational force mg on the particles that is the horizontal distance and the force mg are perpendicular to each other.
Hence the magnitude of the torque on the particle at this position is product of the force and the horizontal distance along it which is $r$.
Mathematically,
$\tau = F \cdot {r_ \bot }$
Where,
Torque is equal to $\tau$.
Gravitational force on the particle be $F = mg$.
The horizontal distance of the particle be $r = \dfrac{{{u^2}}}{{2g}}$
Now on putting the known values we will get,
$\tau = mg \cdot \dfrac{{{u^2}}}{{2g}}$
Cancelling the common terms we will get,
$\tau = \dfrac{1}{2}m{u^2}$

Note: Here that word toques in physics is simply the tendency of a force to turn or twist an object. Actually the real formula for torque is $\tau = Fr\sin \theta$ but as the force and the horizontal distance of the above solution are perpendicular to each other the angle between them becomes $90^\circ$. Also remember that Torque is vector quantity.