euree posted!!!!!!

JEE Main 2015 Questions › Mechanics questions › euree posted!!!!!!

#Physics #Mechanics

UP 0 DOWN 0 1 5
Post on FacebookPost anonymously?

5 Answers

1

hacker ·2010-02-09 23:13:19

as friction is absent!\so,

mgl/2[1-sin@]=1/2I(ic)w²

I(ic)=ml²/12+ml²cot²@/4

>w=âˆš12g(1-sin@)/l(1+3cot²@)

UP 0 DOWN 0
Post on FacebookPost anonymously?

1

hacker ·2010-02-09 23:13:54

@euree point out mistake in above solution y i m not getting same answer as you!!!!!!

UP 0 DOWN 0
Post on FacebookPost anonymously?

1

rahul nair ·2010-02-10 03:20:26

u have neglected the Normal reactions exerted by the wall an the ground on the rod......

UP 0 DOWN 0
Post on FacebookPost anonymously?

1

hacker ·2010-02-10 03:55:07

@rahul[i think u can see i have considered earth and rod as a system so i don't think that error is coz of normal reaction[3][3]]

plz post ur solution!

UP 0 DOWN 0
Post on FacebookPost anonymously?

1

rahul nair ·2010-02-10 05:08:29

Eqns for com of rod:
(ax & ay are horizontal & vertical component of acc. respectively)

1.N1=max

2.N2-mg=may

Torque abt COM

3.
(N2cos@-N1sin@)l/2=ml\alpha²/12

x=lcos@/2, y=lsin@/2

Diff. twice

4.ax=l(\alphacos@-w²sin@)/2

5.ay=-l(\alphacos@+w²sin@)/2

solving these eqns,\alpha=3gcos@/2l

&use \alpha=w(dw/d\theta)

so w=âˆš3g(1-sin@)/l

UP 0 DOWN 0
Post on FacebookPost anonymously?

Your Answer

Basic Integral Calculus Operators Symbols Matrix Cases

√ √ || {} [] () → ~ ^ x x x → → ln log lg lim lim cos sin tan cot arcsin arccos arctan

∫ ∫ ∫ ∫ ∫ ∬ ∬ ∬ ∬ ∬ ∭ ∭ ∭ ∭ ∭ ⨌ ⨌ ⨌ ⨌ ⨌

∑ ∑ ∑ ∑ ∑ ∮ ∮ ∮ ∮ ∮ ∏ ∏ ∏ ∏ ∏ ∐ ∐ ∐ ∐ ∐

+ - × ÷ ± ∓ = ∪ ∩ ≠ ∼ ≈ ∅ ∀ ∃ ∈ ∋ ∝ ≤ ≥ ≃ ≅ ≡ < > ≪ ≫ ⊂ ∉ ⊃ ⊆ ⊇ ⇒ ⇐ ⇔ ⇒ ⇐ ⇔ ⇑ ⇓ ⇕ → ← ↔ → ← ↔ ↑ ↓ ↕ ↖ ↗ ↘ ↙ ↪ ↩ ⇀ ↼ ⇁ ↽

° ∂ ∞ α β γ δ θ λ μ ν ξ π ρ σ φ ω ε ƒ κ τ υ Γ Δ Λ Σ Π Ω