Question

A uniform rod of mass M, length $\ell$ and area of cross section A is hanging vertically from ceiling as shown. If Young's modulus of elasticity is Y, the elastic potential energy stored in upper half of the rod, is $\frac{14M^2g^2\ell}{nAY}$, then value of n is _______.

Answer 1

96

Answer 2

For a rod under its own weight, the tension at a distance x from the top is

$$T(x)=\frac{Mg}{\ell}(\ell-x).$$

The elastic energy in an element $dx$ is

$$dU=\frac{T(x)^2}{2AY}\,dx.$$

Thus, the energy in the upper half (from $x=0$ to $x=\ell/2$) is

$$U=\frac{1}{2AY}\frac{M^2g^2}{\ell^2}\int_0^{\ell/2}(\ell-x)^2dx.$$

Substitute $u=\ell-x$ giving limits $u=\ell$ to $u=\frac{\ell}{2}$. Then,

$$\int_0^{\ell/2}(\ell-x)^2dx = \int_{\ell/2}^{\ell}u^2du=\left[\frac{u^3}{3}\right]_{\ell/2}^{\ell}=\frac{\ell^3}{3}-\frac{\ell^3}{24}=\frac{7\ell^3}{24}.$$

Thus,

$$U=\frac{M^2g^2}{2AY\ell^2}\cdot\frac{7\ell^3}{24}=\frac{7M^2g^2\ell}{48AY}.$$

Given that

$$U=\frac{14M^2g^2\ell}{nAY},$$

equate the two expressions:

$$\frac{7}{48}=\frac{14}{n}\quad\Longrightarrow\quad 7n=48\cdot14\quad\Longrightarrow\quad n=\frac{48\cdot14}{7}=96.$$