Search results

1. Current Electricity Problem

Thank you friends, I have got the answer. I was doing a silly mistake. Special Thanks To NascentOxygen and ehild
2. Current Electricity Problem

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 =...
3. Current Electricity Problem

NascentOxygen: Please tell me how we'll find the current in this circuit?
4. Current Electricity Problem

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...
5. Rotational Motion Problem - 7

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?
6. Rotational Motion Problem - 2

Yes, I got it. Thank you all for the help.
7. Rotational Motion Problem - 13

From now on I will always try to take a fixed point as the reference, then I watch over the quick approach. Thanks a lot for the help.
8. Rotational Motion Problem - 2

The post which I wrote, Because during pure rolling contact point does not slip. Hence friction is zero.. Is this statement correct?
9. Rotational Motion Problem - 2

I got it. That means Any of the choice could be there. Hence Can't be interpreted?
10. Rotational Motion Problem - 7

But this will give the answer ω = v/ 4r, and answer is v/ 3r
11. Rotational Motion Problem - 13

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?
12. Rotational Motion Problem - 2

Last one when x = R/2, f = 0, Because during pure rolling contact point does not slip. Hence friction is zero.
13. Rotational Motion Problem - 7

I conserve A.m. about the center of mass of the system.
14. Rotational Motion Problem - 14

https://fbcdn-sphotos-f-a.akamaihd.net/hphotos-ak-prn2/1476572_1462877957272643_1028005643_n.jpg https://fbcdn-sphotos-g-a.akamaihd.net/hphotos-ak-prn1/q71/s720x720/936643_1462877960605976_1351577648_n.jpg
15. Rotational Motion Problem - 7

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
16. Rotational Motion Problem - 13

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
17. Rotational Motion Problem - 16

Will the torque due to friction act as impulsive torque here or the torque due to reaction force?
18. Rotational Motion Problem - 13

Yes I solved both the two approaches and both are giving the same answer α = g/8R
19. Rotational Motion Problem - 7

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
20. Rotational Motion Problem - 7

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
21. Rotational Motion Problem - 7

Yes it will be at the distance R/2 from the centre of the loop.
22. Rotational Motion Problem - 2

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.
23. Rotational Motion Problem - 1

Thank you Tanya I got the answer.

25. Rotational Motion Problem - 7

I think that point would be point C, the centre of line AB. And if i'll conserve angular momentum then post 1 again comes in front of me.
26. Rotational Motion Problem - 1

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 ?
27. Rotational Motion Problem - 1

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...