A particle of mass 'm' is moving in time 't' on a trajectory given by r? = 10αt² i? + 5β(t-5) j?. Where α and β are dimensional constants. The angular momentum of the particle become the same as it was for t = 0 at time t = ______ seconds.

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Raj Pandey | Contributor-Level 9

5 months ago

r? = 10αt²î + 5β (t-5)?
v? = dr? /dt = 20αtî + 5β?
As L? = m (r? * v? )
So, at t=0, L=0
given L is same at t=t as at t=0
⇒ r? * v? = 0
⇒ (10αt²î + 5β (t-5)? ) * (20αtî + 5β? ) = 0
⇒ 50αβt² (î*? ) + 100αβt (t-5) (? *î) = 0
⇒ 50αβt² k? - 100αβt (t-5) k? = 0
⇒ 50t² - 100t (t-5) = 0
⇒ 50t² - 100t² + 500t = 0
⇒ -

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