A particle of mass $m = 1 k g$ is projected with speed $u = 20 \sqrt{2} m / s$ at angle $θ = 45^{\circ}$ with horizontal. Find the torque (in $N m$ ) of the weight of the particle about the point of projection when the particle is at the highest point.

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Solution

$τ = \frac{m g R}{2} = \frac{(m g) u^{2} sin 2 θ}{2 g}$
$= \frac{m u^{2}}{2}$ (as $θ = 45^{\circ})$
$= \frac{(1) (20 \sqrt{2})^{2}}{2}$
$= 400 N m$ (perpendicular to the plane of motion)

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A particle of mass m=1kg is projected with speed u=20√2m/s at angle θ=45∘ with horizontal. Find the torque (in Nm) of the weight of the particle about the point of projection when the particle is at the highest point.

τ=mgR2=(mg)u2sin2θ2g=mu22 (as θ=45∘)=(1)(20√2)22=400Nm (perpendicular to the plane of motion)

A particle of mass $m = 1 k g$ is projected with speed $u = 20 \sqrt{2} m / s$ at angle $θ = 45^{\circ}$ with horizontal. Find the torque (in $N m$ ) of the weight of the particle about the point of projection when the particle is at the highest point.

$τ = \frac{m g R}{2} = \frac{(m g) u^{2} sin 2 θ}{2 g}$
$= \frac{m u^{2}}{2}$ (as $θ = 45^{\circ})$
$= \frac{(1) (20 \sqrt{2})^{2}}{2}$
$= 400 N m$ (perpendicular to the plane of motion)