Question

A particle of mass m is projected with a speed u at an angle θ with the horizontal. Find the torque of the weight of the particle about the point of projection when the particle is at the highest point.

Answer 1

Range of the particle \[= \left( \frac{u^2 \sin2\theta}{g} \right)\]

At the highest point, we have

Total force acting on the particle = mg (downward)

Distance between the line of force and the point of projection

\[\frac{\left(\text{total range}\right)}{2} = \frac{u^2 \sin2\theta}{2g}\],

\[\text{So, }\overrightarrow{\tau}  =  \overrightarrow{F}  \times  d_\perp  = mg \times  u^2 \frac{\sin2\theta}{2g}\] 

\[\overrightarrow{\tau}  = m u^2 \frac{\sin2\theta}{2}\] 

\[= m u^2 \sin\theta\cos\theta\]

Therefore, the direction of torque is perpendicular to the plane of the motion.