Rotational Motion

JEE Main level, numerical answer, medium

211 of 280

Question 211 of 280

A particle of mass mm is moving in time tt on a trajectory given by,
r⃗=10αt2i^+5β(t−5)j^\vec{r}=10\alpha {t}^{2}\hat{i}+5\beta \left(t-5\right)\hat{j}
where α\alpha and β\beta are dimensional constants.
The angular momentum of the particle becomes the same as it was for t=0t=0 at time t=t=_____ seconds.

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Answer: 10

r⃗=10αt2i^+5β(t−5)j^\vec{r}=10\alpha {t}^{2}\hat{i}+5\beta \left(t-5\right)\hat{j}
v⃗=20α+i^+5βj^\vec{v}=20\alpha +\hat{i}+5\beta \hat{j}
L⃗=m(r⃗×v⃗)\vec{L}=m\left(\vec{r}\times \vec{v}\right)
=m[10αt2i^+5β(t−5)j^]×[20αi^+5βj^]=m\left[10\alpha {t}^{2}\hat{i}+5\beta \left(t-5\right)\hat{j}\right]\times \left[20\alpha \hat{i}+5\beta \hat{j}\right]
L⃗=m[50αβt2k^−100αβ(t2−5t)k^]\vec{L}=m\left[50\alpha \beta {t}^{2}\hat{k}-100\alpha \beta \left({t}^{2}-5t\right)\hat{k}\right]
At t=0,L⃗=0⃗t=0,\vec{L}=\vec{0}
50αβt2−100αβ(t2−5t)=050\alpha \beta {t}^{2}-100\alpha \beta \left({t}^{2}-5t\right)=0
t−2(t−5)=0t-2\left(t-5\right)=0
t=10sec⁡t=10\sec

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