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 - y plane as shown in the figure. At a time t \lt v_0\sin \theta \over g, the angular momentum of the particle is ................, where \widehat i,\widehat j and \widehat k are unit vectors along x,y and z -axis respectively.

- 1 \over 2mg\,v_0t^2\cos \,\theta \widehat k

- mg\,v_0t^2\cos \theta \widehat j

1 \over 2mg\,v_0t^2\cos \theta \widehat i

Rotational Motion PYQ — JEE Main 2010 · Shift 0 · Q62 with Solution

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Correct answer: C

Position and velocity of the projectile

The particle is projected from the origin with initial velocity $v_0$ at angle $\theta$ with the $x$ -axis.

So, at time $t$ :

$x = v_0 \cos\theta\, t$ $y = v_0 \sin\theta\, t - \frac{1}{2}gt^2$

Velocity components are:

$v_x = v_0\cos\theta$ $v_y = v_0\sin\theta - gt$

Linear momentum

The linear momentum is

$\vec p = m\vec v = m(v_x\hat i + v_y\hat j)$

Hence,

$\vec p = m\left(v_0\cos\theta\,\hat i + (v_0\sin\theta - gt)\hat j\right)$

Angular momentum about the origin

Angular momentum is

$\vec L = \vec r \times \vec p$

where

$\vec r = x\hat i + y\hat j$

Thus,

\begin{vmatrix} \hat i & \hat j & \hat k \\ x & y & 0 \\ mv_x & mv_y & 0 \end{vmatrix}

Only the $\hat k$ -component survives:

$\vec L = (x\,mv_y - y\,mv_x)\hat k$

Substitute values:

$L_z = m\left[x(v_0\sin\theta - gt) - y(v_0\cos\theta)\right]$

Now put

\qquad y = v_0\sin\theta\, t - \frac{1}{2}gt^2$$ So, $$L_z = m\left[v_0\cos\theta\, t (v_0\sin\theta - gt) - \left(v_0\sin\theta\, t - \frac{1}{2}gt^2\right)v_0\cos\theta\right]$$ Expand: $$L_z = m\left[v_0^2\sin\theta\cos\theta\, t - gv_0\cos\theta\, t^2 - v_0^2\sin\theta\cos\theta\, t + \frac{1}{2}gv_0\cos\theta\, t^2\right]$$ The first and third terms cancel: $$L_z = m\left[-gv_0\cos\theta\, t^2 + \frac{1}{2}gv_0\cos\theta\, t^2\right]$$ $$L_z = -\frac{1}{2}mgv_0\cos\theta\, t^2$$ Therefore, $$\boxed{\vec L = -\frac{1}{2}mgv_0 t^2\cos\theta\,\hat k}$$ 4. **Checking options** - **A:** Along $\hat j$, incorrect direction. - **B:** Along $\hat k$ but proportional to $t$, incorrect. - **C:** $-\dfrac{1}{2}mgv_0 t^2\cos\theta\,\hat k$, correct. - **D:** Along $\hat i$, incorrect direction. Hence the correct option is: $$\boxed{\text{C}}$$

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View written solutionFree

Correct answer: C

Position and velocity of the projectile

The particle is projected from the origin with initial velocity $v_0$ at angle $\theta$ with the $x$ -axis.

So, at time $t$ :

$x = v_0 \cos\theta\, t$ $y = v_0 \sin\theta\, t - \frac{1}{2}gt^2$

Velocity components are:

$v_x = v_0\cos\theta$ $v_y = v_0\sin\theta - gt$

Linear momentum

The linear momentum is

$\vec p = m\vec v = m(v_x\hat i + v_y\hat j)$

Hence,

$\vec p = m\left(v_0\cos\theta\,\hat i + (v_0\sin\theta - gt)\hat j\right)$

Angular momentum about the origin

Angular momentum is

$\vec L = \vec r \times \vec p$

where

$\vec r = x\hat i + y\hat j$

Thus,

\begin{vmatrix} \hat i & \hat j & \hat k \\ x & y & 0 \\ mv_x & mv_y & 0 \end{vmatrix}

Only the $\hat k$ -component survives:

$\vec L = (x\,mv_y - y\,mv_x)\hat k$

Substitute values:

$L_z = m\left[x(v_0\sin\theta - gt) - y(v_0\cos\theta)\right]$

Now put

\qquad y = v_0\sin\theta\, t - \frac{1}{2}gt^2$$ So, $$L_z = m\left[v_0\cos\theta\, t (v_0\sin\theta - gt) - \left(v_0\sin\theta\, t - \frac{1}{2}gt^2\right)v_0\cos\theta\right]$$ Expand: $$L_z = m\left[v_0^2\sin\theta\cos\theta\, t - gv_0\cos\theta\, t^2 - v_0^2\sin\theta\cos\theta\, t + \frac{1}{2}gv_0\cos\theta\, t^2\right]$$ The first and third terms cancel: $$L_z = m\left[-gv_0\cos\theta\, t^2 + \frac{1}{2}gv_0\cos\theta\, t^2\right]$$ $$L_z = -\frac{1}{2}mgv_0\cos\theta\, t^2$$ Therefore, $$\boxed{\vec L = -\frac{1}{2}mgv_0 t^2\cos\theta\,\hat k}$$ 4. **Checking options** - **A:** Along $\hat j$, incorrect direction. - **B:** Along $\hat k$ but proportional to $t$, incorrect. - **C:** $-\dfrac{1}{2}mgv_0 t^2\cos\theta\,\hat k$, correct. - **D:** Along $\hat i$, incorrect direction. Hence the correct option is: $$\boxed{\text{C}}$$

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