A thin wire of length \(l\) having a linear density \(\rho\) is bent into a circular loop with \(C\) as its…
- \(\frac{5}{16} \frac{\rho l^3}{\pi^3}\)
- \(\frac{1}{16} \frac{\rho l^3}{\pi^3}\)
- \(\frac{1}{8} \frac{\rho /^3}{\pi^3}\)
- \(\frac{3 \rho /^3}{8} \frac{\pi^2}{}\)
Solution
Here, circular loop of the wire with centre, C. \(R\) is the radius of circular loop of wire. \(\therefore\) Moment of inertia axis passing through the centre and perpendicular to the plane \(A B\). \(I_{A B}=M R^2\)...(i) \(\therefore\) Moment of inertia axis passing through the centre of circular loop about its diameter , \(I_{\mathrm{CM}}=\frac{1}{2} M R^2\)...(ii) Now, the total moment of inertia for circular loop of wire. If \(R\) is the distance between the axis, \(I_{\mathrm{CM}}\) and \(I_{A B}\) are the respective moments of inertia about these axies, Then, \(\quad I=I_{\mathrm{CM}}+I_{A B}\) From Eqs. (i) and (ii), we get \(I=\frac{1}{2} M R^2+M R^2 \Rightarrow I=\frac{3}{2} M R^2\)...(iii) We know that, Mass \(=\) length \(\times\) density Mass of the wire, \(M=\rho l\)...(iv) Thus, the length of the wire, \(l=\) circumference of the circular loop of wire \(\therefore \quad l=2 \pi R \text { or } R=\frac{l}{2 \pi}\)...(v) Now, from Eq. (iv) and (v) putting the value of \(M\) and \(R\) in Eq. (iii), we get \(I=\frac{3}{2}(\rho l) \times\left(\frac{l}{2 \pi}\right)^2 \Rightarrow I=\frac{3 \rho l^3}{8 \pi^2}\) So, the moment of inertia of a circular loop of wire is, \(I=\frac{3 \rho l^3}{8 \pi^2}\).
Asked in: AP EAMCET 2019 (22 Apr Shift 1)