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 point of projection at the highest point?

Answer 1

Given that a particle of mass m is projected with initial speed u at an angle of 45 degree with the horizontal.

So the vertical and horizontal components of velocity of projection are `usin45^@ and u cos45^@` respectively.

If the time to reach at maximum height be `t` then

`0=usin45^@-gxxt`

`=>t=u/gsin45^@`,

Where g is the acceleration due to gravity.

Horizontal displacement during this time will be
`r=ucos45^@xxt=ucos45^@xxu/gsin45^@`
`=u^2/(2g)`

This `r` represents the distance of vertically downward gravitational force `mg` on the particle.

Hence the the magnitude of torque on the particle at this position will be

`tau=r*mg=u^2/(2g)*mg=1/2m u^2`