JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2021
- Hard
- Numerical
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 \({t}=\) .....\(seconds.\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(10\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{ r }=10 \alpha t ^{2} \hat{ i }+5 \beta( t -5) \hat{ j }\)
\(\overrightarrow{ v }=20 \alpha t \hat{ i }+5 \beta \hat{ j }\)
\(\overrightarrow{ L }= m (\overrightarrow{ r } \times \overrightarrow{ v })\)
\(=m\left[10 \alpha t ^{2} \hat{ i }+5 \beta( t -5) \hat{ j }\right] \times[20 \alpha t \hat{ i }+5 \beta \hat{ j }]\)
\(\overrightarrow{ L }= m \left[50 \alpha \beta t ^{2} \hat{ k }-100 \alpha \beta\left( t ^{2}-5 t \right) \hat{ k }\right]\)
At \(t =0, \overrightarrow{ L }=\overrightarrow{0}\)
\(50 \alpha \beta t ^{2}-100 \alpha \beta\left( t ^{2}-5 t \right)=0\)
\(t -2( t -5)=0\)
\(t=10 \sec\)
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