A uniform rod, of mass m, length l and r radius of cross section, is rotated about an axis passing through one of its ends and perpendicular to its length with constant angular velocity ω in horizontal plane. If Y is the Young's modulus of the material of rod, the increase in its length due to rotation of rod is

Option 1 -

(mω²l²)/(πr²Y)

Option 2 -

(mω²l²)/(2πr²Y)

Option 3 -

(mω²l²)/(3πr²Y)

Option 4 -

(2mω²l²)/(πr²Y)

0 2 Views | Posted 4 weeks ago
Asked by Shiksha User

  • 1 Answer

  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    4 weeks ago
    Correct Option - 3


    Detailed Solution:

    dm = (m/L)dx
    ∴ T = (mω²/2L) (L² - x²)
    ∴ ΔL = ∫? (mω²/2Lπr²Y) (L² - x²)dx
    = ΔL = mω²L²/3πr²Y

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V
Vishal Baghel

If μ  is Poisson’s ratio,

Y = 3K (1 - 2 μ )      ……… (1)

and Y = 2 n ( 1 + μ )  ……… (2)

With the help of equations (1) and (2), we can write

  3 Y = 1 η + 1 3 k K = η Y 9 η 3 Y

V
Vishal Baghel

Initially S? L = 2m
S? L = √2² + (3/2)²
S? L = 5/2 = 2.5 m
? x = S? L - S? L = 0.5 m
So since λ = 1 m. ∴? x = λ/2
So white listener moves away from S? Then? x (= S? L − S? L) increases and hence, at? x = λ first maxima will appear.? x = λ = S? L − S? L.
1 = d - 2 ⇒ d = 3 m.

V
Vishal Baghel

Loss in elastic potential energy = Gain in KE
½ (YA/L)x² = ½mv²
0.5 × (0.5×10? × 10? / 0.1) × (0.04)² = 20×10? ³ v²
0.5 × (5×10²) × 1.6×10? ³ = 20×10? ³ v²
0.4 = 20×10? ³ v²
v² = 20 => v = √20 ≈ 4.47 m/s
(Re-checking calculations)
0.5 * ( (0.5e9 * 1e-6) / 0.1) * (0.04)^2 = 0.5 * (5e2) * 1.6e-3 = 4.
0.5 * 20e-3 * v^2 = 10e-3 v^2
4 = 10e-3 v^2
v^2 = 400 => v = 20 m/s

V
Vishal Baghel

As we know that

Y = F L A ? L          

? L = 0 . 0 4 m = F L A Y . . . . . . . . . . . . . . . ( i )           

If length and diameter both are doubled

? L ' = F . 2 L 4 A . Y = F L 2 Y = 0 . 0 2 m = 2 c m       

V
Vishal Baghel

Energy density (u) = 1 2 × σ × ε = 1 2 × Y ε × ε = 1 2 ε 2 Y

e n e r g y s t o r e d / m 2 = 1 2 ε 2 Y A

Strain ( ε ) = Δ l l 0 = α Δ T = 1 0 5 × 1 0 = 1 0 4

e n e r g y s t o r e d / m 2 = 1 2 × 1 0 8 × 1 0 1 1 × 1 0 2 = 5

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