in the equation, i = [E1 + E2 +------+En] - [E1 + E2 +------+E N-n] / R1 + R2 + ------- + RN
here E = αR
Hence,
i = [αR1 + αR2 +-------+ αRn] - [αR1 + αR2 +-------+ αR N-n] / [αR1 + αR2 +-------+ αRN]
typical to solve this, so If I assume all the resistances identical, then
i =...
Hi friend I am Stuck in a problem. Please help me in solving this. Thank you all in advance.
The problem is as follows.
https://fbcdn-sphotos-a-a.akamaihd.net/hphotos-ak-xpf1/t1.0-9/10270329_1576533655907072_4352617381833809587_n.jpg...
That means when I was conserving linear momentum, it was giving the answer as v'(hoop + bullet) = v/2, which was wrong but giving the correct answer(coincidence). I took conservation of linear momentum in wrong manner. This is the best approach.
What do you say Tanya?
I got it haruspex, I will take care for all these. But If in any problem collision takes place like in Prob. 4 and porb 7, we take it as abrupt change or what should we think?
Now I got where I was doing mistake.
conserving linear momentum,
m(v) = 2m v' => v' = v/2
conserving angular momentum about centre of the loop,
mv(R/2) = mR2 + m(R2)/4(loop) + {mR2/4(particle)}. ω + (m). (v/2).(R/2) - (m). (v/2).(R/2)(particle +loop com)
it gives the answer, ω = v/3R
Dear haruspex, Please make me clear by connecting all these:
https://fbcdn-sphotos-d-a.akamaihd.net/hphotos-ak-frc3/q71/s720x720/1511150_1462825700611202_1609703582_n.jpg
So, will this be the right approach?
conserving linear momentum,
m(v) = 2m v' => v' = v/2
conserving angular momentum about centre of the loop,
mv(R) = mR2(loop) + {mR2(particle)}. ω + (2m). (v/2).(R/2)(particle +loop com)
it gives the answer, ω = v/4R
So, will this be the right approach?
conserving linear momentum,
m(v) = 2m v' => v' = v/2
conserving angular momentum about centre of the loop,
mv(R) = mR2(loop) + {mR2(particle)}. ω + (2m). (v/2).(R/2)(particle +loop com)
it gives the answer, ω = v/4R
Well the confusion which is arising is that I got the function
3Rf = F (R - 2x)
x lies between 0 to R
when x = 0, f = F/3
when x = R, f = - f/3
when x = R/2, f = 0
using the condition of pure rolling, I used a = Rα for the bottom point.
Please see,
x = Lcosθ
dx/dt = v = L(-sinθ. dθ/dt)
v = L(-sinθ. ω)
dv/dt = a = -L(sinθ. α + ω. cosθ . ω)
a = -L(sinθ. α + ω2. cosθ )
now how to replace a ?
https://fbcdn-sphotos-c-a.akamaihd.net/hphotos-ak-frc3/1471232_1462515570642215_877595177_n.jpg
I think when the rod is at the position AB making angle 450 with horizontal the instantaneous axis would pass through point P and when the rod will slide the new position would be A'B' and...