Ncert Exercises & Solutions

Q.26. A steel rod of length 2l, cross-sectional area A and mass M is set rotating in a horizontal plane about an axis passing through the centre. If Y is Young's modulus for steel, find the extension in the length of the rod. (assume the rod is uniform)

Hint: The centrifugal force in the rod produces strain in the rod.

Step 1: Find the tension in the rod.

Consider an element of width dr at r as shown in the diagram.

Let T(r) and T(r+dr) be the tensions at r and r+dr respectively.

Net centrifugal force on the element

= ω^{2} r d m

(where

ω

is the angular velocity of the rod)

= ω^{2} r μ d r

(∵ μ = mass/length)

\Rightarrow

T (r) - T (r + d r) = μ ω^{2} r d r

\Rightarrow

- d T = μ ω^{2} r d r

[

∵

Tension and centrifugal forces are opposite]

∴

- \int_{T = 0}^{T} d T = \int_{r = l}^{r = r} μ ω^{2} r d r

[∵ T = 0 at r = l]

\Rightarrow

T (r) = \frac{μ ω^{2}}{2} (l^{2} - r^{2})

Step 2: Find the strain produced in the rod.

Let the increase in length of the element dr be

Δ r

So, Young's modulus,

Y = \frac{Stress}{Strain} = \frac{T (r) / A}{\frac{Δ r}{d r}}

∴ \frac{Δ r}{d r} = \frac{T (r)}{Y A} = \frac{μ ω^{2}}{2 Y A} (l^{2} - r^{2})

∴ Δ r = \frac{1}{Y A} \frac{μ ω^{2}}{2} (l^{2} - r^{2}) d r

∴

∆

= change in length in right part =

\frac{1}{Y A} \frac{μ ω^{2}}{2} \int_{0}^{l} (l^{2} - r^{2}) d r

= (\frac{1}{Y A}) \frac{μ ω^{2}}{2} [l^{3} - \frac{l^{3}}{3}] = \frac{1}{3 Y A} μ ω^{2} l^{2}

∴

Total change in length

= 2 Δ = \frac{2}{3 Y A} μ ω^{2} l^{2}

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