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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.
Solution
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.
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