The particle of mass m is moving in time t on a trajectory given by- r - 10 - t2 - - 5 - t-5 -Where - and - are dimensional constants-The angular momentum of the particle becomes the same- as it was for t - 0- at time t - seconds-

R = 10 α t^2 î + 5 β (t 5) ĵ v =d rdt= 20 α t î + 5 βĵ L = m (r×v) = m [10α t^2 î + 5 β (t 5)ĵ] × [20 α t î + 5 βĵ] L= m [50 αβ t^2 k̂ 100 αβ t (t 5) k̂] At ...

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Standard XII

Physics

Conservative and Non Conservative Forces

The particle ...

Question

The particle of mass $m$ is moving in time $t$ on a trajectory given by, $\to r = 10 α t^{2}^i + 5 β (t - 5)^j$
Where $α$ and $β$ are dimensional constants.
The angular momentum of the particle becomes the same, as it was for $t = 0$ , at time $t =$ seconds.

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Solution

$\to r = 10 α t^{2}^i + 5 β (t - 5)^j$
$\to v = \frac{d \to r}{d t} = 20 α t^i + 5 β^j$
$\to L = m (\to r \times \to v)$
$= m [10 α t^{2}^i + 5 β (t - 5)^j] \times [20 α t^i + 5 β^j]$
$\to L = m [50 α β t^{2}^k - 100 α β t (t - 5)^k]$
At $t = 0, \to L = 0$
$\Rightarrow 50 α β t^{2} - 100 α β t (t - 5) = 0$
$t - 2 (t - 5) = 0$
$t = 10 sec$

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The particle of mass m is moving in time t on a trajectory given by, →r=10 αt2^i+5β(t−5)^j Where α and β are dimensional constants. The angular momentum of the particle becomes the same, as it was for t=0, at time t= seconds.

→r=10αt2^i+5β(t−5)^j →v=d→rdt=20αt^i+5β^j →L=m(→r×→v) =m[10αt2^i+5β(t−5)^j]×[20αt^i+5β^j] →L=m[50αβt2^k−100αβ t(t−5)^k] At t=0,→L=0 ⇒ 50αβt2−100αβ t(t−5)=0 t−2(t−5)=0 t=10 sec

The particle of mass $m$ is moving in time $t$ on a trajectory given by, $\to r = 10 α t^{2}^i + 5 β (t - 5)^j$
Where $α$ and $β$ are dimensional constants.
The angular momentum of the particle becomes the same, as it was for $t = 0$ , at time $t =$ seconds.