Question

A rod of mass m, uniform cross sectional area A and length L is accelerated by applying force F as shown in figure on a smooth surface. If Young’s modulus of elasticity of the material of rod is Y. (Consider x as measured from the right end)

Elastic potential energy stored in the rod is

• A. F^2L / AY
• B. F^2L / 2AY
• C. F^2L / 3AY
• D. F^2L / 6AY

Answer 1

Answer: (D)

F^2L / 6AY

Answer 2

Solution:

1. Find the tension distribution

Consider an element of the rod from a point $x$ (measured from the right end) to the left end. The mass of this segment is

$$dm = \frac{m}{L}(L-x) \,.$$

Since the entire rod is accelerating with acceleration $a=\frac{F}{m}$ (because the force $F$ is only applied at one end), the force needed to accelerate the segment is

$$dF = dm \cdot a = \frac{F}{m}\cdot \frac{m}{L}(L-x) = \frac{F}{L}(L-x)\,.$$

Thus, the internal tension at the section located at $x$ is

$$T(x)=\frac{F}{L}(L-x)\,.$$

2. Determine strain and elastic energy density

The strain at $x$ is given by Hooke’s law as

$$\epsilon(x)=\frac{T(x)}{AY}=\frac{F(L-x)}{AYL}\,.$$

The elastic energy density (energy per unit volume) in the rod is

$$u(x)=\frac{1}{2}Y\,\epsilon(x)^2\,.$$

3. Integrate to find the total elastic potential energy (U)

The total energy stored in the rod is obtained by integrating the energy density over the entire volume of the rod:

$$U=\int_0^L u(x)\,A\,dx = \frac{1}{2}YA\int_0^L \left(\frac{F(L-x)}{AYL}\right)^2 dx \,.$$

Simplify the integrand:

$$\left(\frac{F(L-x)}{AYL}\right)^2 = \frac{F^2(L-x)^2}{A^2Y^2L^2}\,.$$

Thus,

$$U=\frac{1}{2}YA\cdot \frac{F^2}{A^2Y^2L^2}\int_0^L (L-x)^2\,dx = \frac{F^2}{2AYL^2}\int_0^L (L-x)^2\,dx \,.$$

Evaluate the integral by letting $u=L-x$; when $x=0$, $u=L$ and when $x=L$, $u=0$. Then,

$$\int_0^L (L-x)^2\,dx = \int_0^L u^2\,du = \frac{L^3}{3}\,.$$

Therefore,

$$U=\frac{F^2}{2AYL^2}\cdot \frac{L^3}{3} = \frac{F^2L}{6AY}\,.$$

Final Answer:

The elastic potential energy stored in the rod is

$$\boxed{\frac{F^2 L}{6AY}}\,.$$