A particle of mass m is projected from origin O with speed u at an angle with positive x-axis. Find the angular momentum of particle at any time t about O before the particle strikes the ground again.
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Let particte is at Pat any instant, and its position vector is r. Then, $\text{r}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}$ Where, $\text{x}=\text{u}\cos\theta\text{ t}$ and $\text{y}=\text{u}\sin\theta\text{ t}-\frac{1}{2}\text{gt}^2$ and if v = velocity vector of particle at that instant. Then, $\text{r}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}$ Where, $\text{x}=\text{u}\cos\theta\text{ t}$ and $\text{y}=\text{u}\sin\theta\text{ t}-\frac{1}{2}\text{gt}^2$ d if v = velocity vector of particle at that instant.
Then, $\text{v}=\text{v}_{\text{x}}\hat{\text{i}}+\text{v}_{\text{y}}\hat{\text{j}}$ Where, $\text{v}_{\text{x}}=\text{u}\cos\theta\text{ t}$ and $\text{v}_{\text{y}}=\text{u}\sin\theta\text{ t}-\text{gt}$ So, $\text{r}=\text{u}\cos\theta\hat{\text{i}}+\Big((\text{u}\sin\theta)\text{t}-\frac{1}{2}\text{gt}^2\Big)\hat{\text{j}}$and $\text{v}=\text{u}\cos\theta\hat{\text{i}}+(\text{u}\sin\theta-\text{gt})\hat{\text{j}}$
So, L = angular momentum of particle $=\text{m}(\text{r}\times\text{v})$ $=\text{m}\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\\text{u}\cos\theta&\Big(\text{u}\sin\theta-\frac{1}{2}\text{gt}\Big)\text{t}&0\\\text{u}\cos\theta&(\text{u}\sin\theta-\text{gt})&0\end{vmatrix}$ $=-\frac{1}{2}\text{m}(\text{u}\cos\theta)\text{gt}^2.\hat{\text{k}}$ So, angular momentum is along z-axis.
Then, $\text{v}=\text{v}_{\text{x}}\hat{\text{i}}+\text{v}_{\text{y}}\hat{\text{j}}$ Where, $\text{v}_{\text{x}}=\text{u}\cos\theta\text{ t}$ and $\text{v}_{\text{y}}=\text{u}\sin\theta\text{ t}-\text{gt}$ So, $\text{r}=\text{u}\cos\theta\hat{\text{i}}+\Big((\text{u}\sin\theta)\text{t}-\frac{1}{2}\text{gt}^2\Big)\hat{\text{j}}$and $\text{v}=\text{u}\cos\theta\hat{\text{i}}+(\text{u}\sin\theta-\text{gt})\hat{\text{j}}$
So, L = angular momentum of particle $=\text{m}(\text{r}\times\text{v})$ $=\text{m}\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\\text{u}\cos\theta&\Big(\text{u}\sin\theta-\frac{1}{2}\text{gt}\Big)\text{t}&0\\\text{u}\cos\theta&(\text{u}\sin\theta-\text{gt})&0\end{vmatrix}$ $=-\frac{1}{2}\text{m}(\text{u}\cos\theta)\text{gt}^2.\hat{\text{k}}$ So, angular momentum is along z-axis.
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