At time t, ModifyingAbove r = 7.60t2 i - (3.30t + 3.80t2) j gives the position of a 3.0 kg particle relative to the origin of an xy coordinate system ( ModifyingAbove r With right-arrow is in meters and t is in seconds). (a) Find the torque acting on the particle relative to the origin at the moment 6.27 s (b) Is the magnitude of the particle’s angular momentum relative to the origin increasing, decreasing, or unchanging?

Question

At time t,  ModifyingAbove r = 7.60t2 i - (3.30t + 3.80t2) j gives the position of a 3.0 kg particle relative to the origin of an xy coordinate system ( ModifyingAbove r With right-arrow is in meters and t is in seconds). (a) Find the torque acting on the particle relative to the origin at the moment 6.27 s (b) Is the magnitude of the particle’s angular momentum relative to the origin increasing, decreasing, or unchanging?

irishboy123 · Answer

"ModifyingAbove" ?!?!  what is that?

@martaa
if you are online we can work this, but maybe your previous question first...

anonymous · Answer

oops sorry it's supposed to be  r = 7.60t^(2) i - (3.30t + 3.80t^(2)) j

irishboy123 · Answer

did you mail this one? http://openstudy.com/study#/updates/565a3003e4b0959c2b14707d

irishboy123 · Answer

*Nail*, not mail, lol

anonymous · Answer

yes :)

irishboy123 · Answer

good.

so now onto this one?

anonymous · Answer

sorry about the modying above part i just copied it , just ignore it :))

anonymous · Answer

yeah :)

anonymous · Answer

i'm just gonna edit the question @IrishBoy123  :)

irishboy123 · Answer

cool

anonymous · Answer

At time t, r = 7.60t^(2) i - (3.30t + 3.80t^(2)) j gives the position of a 3.0 kg particle relative to the origin of an xy coordinate system (  r is in meters and t is in seconds). (a) Find the torque acting on the particle relative to the origin at the moment 6.27 s (b) Is the magnitude of the particle’s angular momentum relative to the origin increasing, decreasing, or unchanging?

irishboy123 · Answer

marta

what are you currently learning?!?!

you know $\vec r$.  that is given to us.

so you can diffeentiate to get $\vec v$

and you can apply $\vec L = m (\vec r \times \vec v)$

anonymous · Answer

oh i see :x thank you, i'm stuyding for my exit exam for physics :)

irishboy123 · Answer

cool!  when are the exams?

anonymous · Answer

next week ... :)))

raffle_snaffle · Answer

What is L=m(rxv) ?

raffle_snaffle · Answer

angular momentum?

irishboy123 · Answer

it is the vector product that gives you the angular momentum.....which is a vector also.....

irishboy123 · Answer

@martaamador62 

good luck!!!

feel free to post your doubts here😞

anonymous · Answer

thank you @IrishBoy123  :)