A small particle of mass m is projected at an angle θ with the x -axis with an initial velocity v₀ in the x-y plane as shown in the figure. At a time t < (v₀sin θ)/(g), the angular momentum of the particle is................, where i, j and k are unit vectors along x, y and z -axis respectively. (a) -mg v₀t²cos (b) mg v₀tcos (c) -1/2mg v₀t²cos (d) 1/2mg v₀t²cos

Answer: (c). x=v₀cos θ t, y=v₀sin θ t-12gt²; vx=v₀cos θ, vy=v₀sin θ-gt. Lz=m(xvy-yvx)=mv₀cos θ [t(v₀sin θ-gt)-(v₀sin θ t-12gt²)] Lz=-12mgv₀t²cos θ So the answer is (c) -12mgv₀t²cos θ k.

A small particle of mass m is projected at an angle θ with the x… | JEE Main 2010 (AIEEE)

Q84JEE Main 2010 · AIEEEMedium

Angular Momentum & its Conservation

A small particle of mass $m$ is projected at an angle $\theta$ with the $x$ -axis with an initial velocity $v_0$ in the $x\text{-}y$ plane as shown in the figure. At a time $t < \frac{v_0\sin\theta}{g},$ the angular momentum of the particle is ................,

where $\hat{i}, \hat{j}$ and $\hat{k}$ are unit vectors along $x, y$ and $z$ -axis respectively.

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Answer: (c)

$x=v_0\cos\theta\,t$ , $y=v_0\sin\theta\,t-\frac12gt^2$ ; $v_x=v_0\cos\theta$ , $v_y=v_0\sin\theta-gt$ .
$L_z=m(xv_y-yv_x)=mv_0\cos\theta\,[t(v_0\sin\theta-gt)-(v_0\sin\theta\,t-\frac12gt^2)]$
$L_z=-\frac12mgv_0t^2\cos\theta$
So the answer is (c) $-\frac12mgv_0t^2\cos\theta\,\hat k$ .

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Angular Momentum & its Conservation question from JEE Main 2010 (AIEEE)

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