IIT JEE , AIEEE Forum - Physics , Mechanics - Easy one:::::::Mechanics:::::::::Position vector:::::::

LAMPARD

The position vector of a particle of mass 1kg moving in space is given by r=(2

2t² i+3

3t³ j+(1/

2)t⁴ j)m.
Then the value of mod(₀

¹ r x (d²r/dt²).dt) is [t is in second].
One method is finding d²r/dt² and calculating cross product but that is obviously too tedious.There has to be a better method,maybe since d²r/dt² is accelaration.
But i cant go further.Can anyone plz try??

LAMPARD

This doesnt seem that difficult...someone,plz try!!!

LAMPARD

Someone,plzzz...

akshay.khare91

since the d^2r /dt^2 is acceleration differentiatethe
given eq. of r wrt to time to get velocity and again differentiate
to get acceleration..

LAMPARD

Dude,i did diff. r to get acceleration.,but then you also have to take cross product of r and acc. then.Finding d²r/dt² itself is a tedious job and then cross product too!!There should be a better method.

akshay.khare91

what is the ans.????

computer001

y dont u do parts once??
rite as d(dr/dt)

karthik2007

LAMPARD, I don't think there is any other way to do this sum.

LAMPARD

Hmm...ok..thnx guys...
@computer001
How to do integration by parts when there is cross product in the integral??

computer001

oh sry din notice the cross.pdt symbol...thot it was normal multiplcation..
but ca n write as mod inside and then integrate...??

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