The linear mass density of a thin rod AB of l | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The moment of inertia $I$ is given by $I = \int r^{2} dm = \int_{0}^{L} x^{2} \lambda(x) dx$.Substituting $\lambda(x) = \lambda_{0}(1 + \frac{x}{L

Vedclass · Accepted Answer

$0.388$

Vedclass · Answer

(D) The moment of inertia $I$ is given by $I = \int r^{2} dm = \int_{0}^{L} x^{2} \lambda(x) dx$.Substituting $\lambda(x) = \lambda_{0}(1 + \frac{x}{L})$:$I = \int_{0}^{L} x^{2} \lambda_{0}(1 + \frac{x}{L}) dx = \lambda_{0} \int_{0}^{L} (x^{2} + \frac{x^{3}}{L}) dx$$I = \lambda_{0} [\frac{x^{3}}{3} + \frac{x^{4}}{4L}]_{0}^{L} = \lambda_{0} (\frac{L^{3}}{3} + \frac{L^{3}}{4}) = \frac{7}{12} \lambda_{0} L^{3} \quad ...(i)$Now,calculate the total mass $M$:$M = \int_{0}^{L} \lambda(x) dx = \int_{0}^{L} \lambda_{0}(1 + \frac{x}{L}) dx = \lambda_{0} [x + \frac{x^{2}}{2L}]_{0}^{L} = \lambda_{0} (L + \frac{L}{2}) = \frac{3}{2} \lambda_{0} L$Thus,$\lambda_{0} L = \frac{2}{3} M \quad ...(ii)$Substituting $(ii)$ into $(i)$:$I = \frac{7}{12} (\lambda_{0} L) L^{2} = \frac{7}{12} (\frac{2}{3} M) L^{2} = \frac{14}{36} M L^{2} = \frac{7}{18} M L^{2}$$I \approx 0.388 M L^{2}$.

Similar Questions

Two loops $A$ and $B$ of radii $R_1$ and $R_2$ are made from a uniform wire. If the moment of inertia of $A$ is $I_A$ and that of $B$ is $I_B$,then find the ratio $R_2 / R_1$ given that $I_A / I_B = 27$.

Two rings have their moments of inertia in the ratio $2:1$ and their diameters are in the ratio $2:1$. The ratio of their masses is

$A$ circular disc has radius $R_1$ and thickness $T_1$. Another circular disc made of the same material has radius $R_2$ and thickness $T_2$. If the moment of inertia of both discs are same and $\frac{R_1}{R_2}=2$, then $\frac{T_1}{T_2}=\frac{1}{\alpha}$. The value of $\alpha$ is . . . . . . .

$A$ circular disc is made using iron and aluminum. To maximize the moment of inertia about its geometric axis,which configuration is preferred?

Vedclass Test Series

Exam Paper Generator

Online Exam Module