Question

A uniform rod of length l is acted upon by a force F in a gravity-free region, as shown in the figure. If the area of cross-section of the rod is A and it's Young's modulus is Y, then the elastic potential energy stored in the rod due to elongation is U=F2lnAY. Then find the value of n?

Answer 1

6

Answer 2

The problem describes a uniform rod of length $l$ acted upon by a force $F$ in a gravity-free region. The figure shows a single force $F$ applied at one end of the rod. In a gravity-free region, if a single force $F$ is applied to a free body of mass $m$, the body accelerates with $a = F/m$. In this case, the tension within the rod is not uniform; it varies along the length.

Let's assume the force $F$ is applied at one end, say the end at $x=l$, and the other end is at $x=0$. Let the rod have mass $m$. The acceleration of the rod is $a = F/m$. Consider a cross-section at a distance $x$ from the free end (at $x=0$, where no force is applied). The part of the rod from $0$ to $x$ has mass $m(x/l)$. This part is accelerated by the tension $T(x)$ at the cross-section at $x$.

Using Newton's second law for the segment from $0$ to $x$: $T(x) = (\text{mass of segment}) \times (\text{acceleration})$ $T(x) = \left(\frac{m}{l} x\right) a = \left(\frac{m}{l} x\right) \left(\frac{F}{m}\right) = \frac{Fx}{l}$. The tension varies linearly from $0$ at the free end ($x=0$) to $F$ at the end where the force is applied ($x=l$).

The strain at a distance $x$ from the free end is given by Hooke's law: $\epsilon(x) = \frac{\text{Stress}(x)}{Y} = \frac{T(x)/A}{Y} = \frac{Fx/l}{AY} = \frac{Fx}{lAY}$.

The elastic potential energy stored in a small element of length $dx$ at position $x$ is given by: $dU = \frac{1}{2} \times \text{Stress}(x) \times \text{Strain}(x) \times \text{Volume of element}$ $dU = \frac{1}{2} \times \frac{T(x)}{A} \times \epsilon(x) \times (A dx)$ $dU = \frac{1}{2} T(x) \epsilon(x) dx$ Substitute the expressions for $T(x)$ and $\epsilon(x)$: $dU = \frac{1}{2} \left(\frac{Fx}{l}\right) \left(\frac{Fx}{lAY}\right) dx = \frac{1}{2} \frac{F^2 x^2}{l^2 AY} dx$.

To find the total elastic potential energy stored in the rod, integrate $dU$ over the entire length of the rod from $x=0$ to $x=l$: $U = \int_0^l dU = \int_0^l \frac{F^2 x^2}{2 l^2 AY} dx$. Since $F$, $l$, $A$, and $Y$ are constants with respect to the integration variable $x$: $U = \frac{F^2}{2 l^2 AY} \int_0^l x^2 dx$. Evaluate the integral: $\int_0^l x^2 dx = \left[\frac{x^3}{3}\right]_0^l = \frac{l^3}{3} - \frac{0^3}{3} = \frac{l^3}{3}$.

Substitute the result of the integral back into the expression for $U$: $U = \frac{F^2}{2 l^2 AY} \left(\frac{l^3}{3}\right) = \frac{F^2 l^3}{6 l^2 AY} = \frac{F^2 l}{6AY}$.

The elastic potential energy stored in the rod is $U = \frac{F^2 l}{6AY}$. The question gives the formula for the elastic potential energy stored in the rod as $U = \frac{F^2 l}{nAY}$. Comparing the calculated formula with the given formula: $\frac{F^2 l}{6AY} = \frac{F^2 l}{nAY}$. By comparing the denominators, we can see that $n=6$.

The final answer is $\boxed{6}$.

Explanation of the solution:

1. Determine the tension distribution along the rod when a force F is applied at one end in a gravity-free region. The tension varies linearly from 0 at the free end to F at the end where the force is applied. $T(x) = Fx/l$, where x is the distance from the free end.
2. Calculate the strain at each point along the rod using Hooke's Law: $\epsilon(x) = T(x)/(AY) = Fx/(lAY)$.
3. Calculate the elastic potential energy stored in an infinitesimal element of length $dx$ at position $x$: $dU = \frac{1}{2} T(x) \epsilon(x) dx = \frac{1}{2} \frac{F^2 x^2}{l^2 AY} dx$.
4. Integrate the potential energy density over the length of the rod to find the total potential energy: $U = \int_0^l dU = \int_0^l \frac{F^2 x^2}{2 l^2 AY} dx = \frac{F^2}{2 l^2 AY} \int_0^l x^2 dx = \frac{F^2}{2 l^2 AY} \frac{l^3}{3} = \frac{F^2 l}{6AY}$.
5. Compare the calculated potential energy formula $U = \frac{F^2 l}{6AY}$ with the given formula $U = \frac{F^2 l}{nAY}$ to find the value of $n$. The comparison yields $n=6$.