Question

A rod of length L rotate about an axis passing through its centre and normal to its length with an angular velocity \(\omega\). If A is the cross-section and \(D\) is the density of material of rod. Find its rotational $\mathrm{K} . \mathrm{E}$. \(\\{\mathrm{A}\\}(1 / 2) \mathrm{AL}^{3} \mathrm{D} \omega^{2}\) \\{B \(\\}(1 / 6) \mathrm{AL}^{3} \mathrm{D} \omega^{2}\) \(\\{C\\}(1 / 24) A L^{3} D \omega^{2}\) \(\\{\mathrm{D}\\}(1 / 12) \mathrm{AL}^{3} \mathrm{D} \omega^{2}\)

Short answer

The rotational kinetic energy of the rod is \(K.E = \frac{1}{6} A L^3 D \omega^2\).

Step-by-step solution

Find the moment of inertia

To find the moment of inertia \(I\) of the rod, we can use the following formula: \[I = \int_0^L dm \, x^2\] Since the mass is uniformly distributed along the length, we can write \(dm = \frac{M}{L} dx\) , where \(M = A*L*D\) is the total mass of the rod. Now, the moment of inertia is given by: \[I = \int_0^L \frac{M}{L} x^2 dx\]

Calculate the integral

The integral will be: \[I = \frac{M}{L} \int_0^L x^2 dx\] \[I = \frac{M}{L} \left[\frac{x^3}{3}\right]_0^L\] \[I = \frac{M}{L} \left(\frac{L^3}{3}\right)\] \[I = \frac{1}{3} M L^2\] Now, substituting the value of M, we get: \[I = \frac{1}{3} (A * L * D) L^2\] \[I = \frac{1}{3} A L^3 D\]

Find the rotational kinetic energy

The formula for rotational kinetic energy is: \(K.E = \frac{1}{2} I \omega^2\) Now, substituting the value of I, we get: \(K.E = \frac{1}{2} (\frac{1}{3} A L^3 D) \omega^2\) \(K.E = \frac{1}{6} A L^3 D \omega^2\) So, the correct answer is \(B) (1 / 6) AL^{3} D \omega^{2}\).