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A particle of mass $'{m}'$ is moving in time $'t'$ on a trajectory given by
$\overrightarrow{{r}}=10 \alpha {t}^{2}\, \hat{{i}}+5 \beta({t}-5)\, \hat{{j}}$
Where $\alpha$ and $\beta$ are dimensional constants. The angular momentum of the particle becomes the same as it was for ${t}=0$ at time ${t}=$ .....$seconds.$
$15$
$10$
$20$
$25$
Solution
$\overrightarrow{ r }=10 \alpha t ^{2} \hat{ i }+5 \beta( t -5) \hat{ j }$
$\overrightarrow{ v }=20 \alpha t \hat{ i }+5 \beta \hat{ j }$
$\overrightarrow{ L }= m (\overrightarrow{ r } \times \overrightarrow{ v })$
$=m\left[10 \alpha t ^{2} \hat{ i }+5 \beta( t -5) \hat{ j }\right] \times[20 \alpha t \hat{ i }+5 \beta \hat{ j }]$
$\overrightarrow{ L }= m \left[50 \alpha \beta t ^{2} \hat{ k }-100 \alpha \beta\left( t ^{2}-5 t \right) \hat{ k }\right]$
At $t =0, \overrightarrow{ L }=\overrightarrow{0}$
$50 \alpha \beta t ^{2}-100 \alpha \beta\left( t ^{2}-5 t \right)=0$
$t -2( t -5)=0$
$t=10 \sec$
