Question

The shape of an elastic rod of length \( l \) is represented by the position vector \( \vec{r}(s) \) corresponding to the arc length \( s \). If the bending energy of the rod is \[ E = K \int_0^l \left( \frac{\partial^2 \vec{r}}{\partial s^2} \right)^2 \, ds, \] what is the dimension of \( K \) in terms of mass \( M \), length \( L \), and time \( T \)?

• A. \( M L T^{-2} \)
• B. \( M L^3 T^{-2} \)
• C. \( M L^2 T^{-2} \)
• D. \( M L^3 T^{-1} \)

Hint:

When performing dimensional analysis, carefully consider the dimensions of each term and use them to solve for unknown constants. In this case, \( K \)'s dimension is derived by balancing the units of the bending energy equation.

Answer 1

The Correct Option is B

Step 1: Analyzing the term \( \frac{\partial^2 \vec{r}}{\partial s^2} \) The position vector \( \vec{r}(s) \) represents the shape of the elastic rod, and \( s \) is the arc length, which has the dimension of length \( L \). The second derivative \( \frac{\partial^2 \vec{r}}{\partial s^2} \) represents the curvature of the rod. The curvature has the dimension of inverse length \( L^{-1} \), because it describes the rate of change of the angle per unit length. Thus, \( \left( \frac{\partial^2 \vec{r}}{\partial s^2} \right)^2 \) will have the dimension of \( L^{-2} \). 
Step 2: Analyzing the integral term \( \int_0^l \left( \frac{\partial^2 \vec{r}}{\partial s^2} \right)^2 \, ds \) The integral is over the length of the rod, \( l \), which has the dimension of length \( L \). Since the integrand has the dimension of \( L^{-2} \), the entire integral will have the dimension of: \[ \left[ \int_0^l \left( \frac{\partial^2 \vec{r}}{\partial s^2} \right)^2 \, ds \right] = L^{-2} \times L = L^{-1}. \] Step 3: Analyzing the bending energy \( E \) The bending energy \( E \) has the dimension of energy, which is \( M L^2 T^{-2} \) (mass × length² × time⁻²). Step 4: Determining the dimension of \( K \) Now, we can equate the dimensions of both sides of the equation for bending energy: \[ E = K \times L^{-1}. \] Substituting the dimensions of \( E \) and \( L^{-1} \): \[ M L^2 T^{-2} = [K] \times L^{-1}. \] Solving for \( [K] \), we get: \[ [K] = M L^3 T^{-2}. \] Thus, the dimension of \( K \) is \( M L^3 T^{-2} \).