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![[Post New]](/templates/default/images/icon_minipost_new.gif) 6 Feb 2008 14:50:37 IST
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a particle is moving along a horizontal circle such that its total power varies with time as p = k2rt2 , r = radius of circle , find total acc. of particle as a function of time?
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I REALLY DON'T KNOW IF THIS IS CORRECT BUT HAVE A LOOK AT IT P=F x v WHERE P IS THE POWER, F IS THE FORCE AND v IS THE VELOCITY BECAUSE THE PARTICLE IS MOVING IN A CIRCULAR PATH AT ANY TIME t F=mv^2/r P=Fv=mv^3/r=(k^2)r(t^2) NOW REARRANGE THE TERMS WRITING v IN TERMS OF THE k, r,t and m NOW DEFFRENTIATE BOTH SIDES WITH RESPECT TO t dv/dt=a=ans
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FAILURE, THE FIRST STEP TO SUCCESS |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 6 Feb 2008 15:46:08 IST
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what is da final ans i want to confrm
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 6 Feb 2008 16:01:26 IST
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but mv^2/r is only the normal accn what about tangential accn
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 6 Feb 2008 18:11:57 IST
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Isn't the mass of the body given? Anyway:
At t = 0, P = 0, note, hence we can write
Pt = 1/2mv2 at any instant (WE theorem)
Or k2rt3 = 1/2mv2
or mv2/r = 2k2t3
Also, we have v = 2k2rt3/m
Find dv/dt.
Then use maresultant = (mv2/r)2 + (mdv/dt)2
thus find a
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Will nip in at times to solve problems :)
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 6 Feb 2008 18:26:37 IST
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The answer should be a=k t[4/9k2t5 + 3/2r]
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 6 Feb 2008 18:36:13 IST
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Since the particle moves in the horizontal plane, the work W done by the external forces is completely converted to the K.E. (E) of the particle. Hence P = dW/dt = dE/Dt Thus dE/dt = k2rt2 => E = k2rt3/3 (we'll ignore the constant) Hence 1/2mv2 = k2rt3/3. Thus mv2/r = 2k2t3/3 ...... a1 = centripetal acceleration v =  2k 2rt 3/m and hence dv/dt = a 2. = tangential acceleration. The resultant a =  a 12+a 22
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Time wounds all heels |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 8 Feb 2008 10:05:38 IST
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karthik is right acc to me
in circular motion , we knw that
P = Ftangential . V
k2rt2 = m dv/dt.V
integrate both sides
V comes out to be (2k2rt2/3m)1/2
anet = ( acentrepital2 + atangential2)1/2
= [ (V2/r)2 + (dv/dt)2]1/2
so u can solve
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 8 Feb 2008 12:26:50 IST
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P=Ft * v [Ft=tangential force] k^2*rt^2=mvdv/dt therefore v^2=2k^2*rt^3/3m
accn tangential =dv/dt=underroot(3k^2rt/2m) accn centripetal=v^2/r=(2*k^2*t^3)/3m a total=underroot[(3ksquare*r*t/2m)+ (4k^4*t^6/9m^2)] =underroot[ksquare*t/m * (3r/2 + 4ksquare*t^5/9m)]
i am almost sure the answer is wrong and hence would hope for some help!!
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Impossible To be Impossible is Impossible |
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