Question

A rod of mass M; length L is made of material of Young’s modulus Y. The rod rotates in a horizontal plane about an axis through its one end and perpendicular to length of rod with a constant angular speed ω. The increase in length of rod is

(ρ = density of material of the rod)

(a) \(\frac{1}{2}(\frac{pω^2}{Y})L^2\)

(b) \(\frac{1}{3}(\frac{pω^2}{Y})L^3\)

(c) \(\frac{1}{2}(\frac{pω^2}{Y})L^3\)

(d) \(\frac{2}{3}(\frac{pω^2}{Y})L^3\)

Answer 1

 (b) \(\frac{1}{3}(\frac{pω^2}{Y})L^3\)

Consider a small element of length dx of the rod at a distance x from axis of rotation as shown in Fig. Let T be tension at distance x from axis of rotation. Then

T = Centerifugal force on the element considered = (dm)xω²

dm = mass of the element considered = a(dx)p

a = area of cross–section of rod; p = density of rod.

∴ T = aPω² xdx

The increase in length (dl) of the element considered; in terms of Young’s modulus (Y) is

The total increase in length of rod