Question

A thin wire is bent in form of a helix of radius R and height H. The pitch of helix is H/2 and mass per unit length of wire is $\lambda$. Moment of inertia of wire about the axis of helix is:

• A. $\lambda HR^2$
• B. $2\lambda R^2\sqrt{H^2 + 16\pi^2R^2}$
• C. $\lambda R^2\sqrt{H^2 + 16\pi^2R^2}$
• D. None of these

Answer 1

Answer: (C)

$\lambda R^2\sqrt{H^2 + 16\pi^2R^2}$

Answer 2

The length of a helix with radius $R$, height $H$, and pitch $P=H/2$ is $L = \sqrt{H^2 + (2\pi R \times N)^2}$ where $N$ is the number of turns. The number of turns is $N = H/P = H/(H/2) = 2$. The length element $ds = \sqrt{R^2 + (dz/dt)^2} dt$. With $z(t) = \frac{P}{2\pi}t = \frac{H}{4\pi}t$, we get $dz/dt = H/(4\pi)$. The total length for $t \in [0, 4\pi]$ is $L = \int_0^{4\pi} \sqrt{R^2 + (H/4\pi)^2} dt = 4\pi \sqrt{R^2 + H^2/(16\pi^2)} = \sqrt{16\pi^2 R^2 + H^2}$. The total mass is $M = \lambda L = \lambda \sqrt{H^2 + 16\pi^2 R^2}$. Since all mass elements are at a distance $R$ from the axis, the moment of inertia is $I = MR^2 = \lambda R^2 \sqrt{H^2 + 16\pi^2 R^2}$.