A non-uniform rod OA of linear mass density l | vedclass

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

Question

(C) The center of mass $X_{cm}$ is given by:$X_{cm} = \frac{\int x dm}{\int dm} = \frac{\int_{0}^{L} x (\lambda_0 x) dx}{\int_{0}^{L} (\lambda_0 x) dx

Vedclass · Accepted Answer

$\frac{4g}{3L}$

Vedclass · Answer

(C) The center of mass $X_{cm}$ is given by:$X_{cm} = \frac{\int x dm}{\int dm} = \frac{\int_{0}^{L} x (\lambda_0 x) dx}{\int_{0}^{L} (\lambda_0 x) dx} = \frac{\lambda_0 [x^3/3]_0^L}{\lambda_0 [x^2/2]_0^L} = \frac{L^3/3}{L^2/2} = \frac{2L}{3}$.The total mass $M$ of the rod is:$M = \int_{0}^{L} \lambda_0 x dx = \frac{\lambda_0 L^2}{2}$.The moment of inertia $I$ about the hinge $O$ is:$I = \int x^2 dm = \int_{0}^{L} x^2 (\lambda_0 x) dx = \lambda_0 \int_{0}^{L} x^3 dx = \frac{\lambda_0 L^4}{4}$.Just after the string is cut,the torque $	au$ about the hinge $O$ is due to gravity acting at the center of mass:$	au = Mg 	imes X_{cm} = \left( \frac{\lambda_0 L^2}{2} ight) g \left( \frac{2L}{3} ight) = \frac{\lambda_0 g L^3}{3}$.Using the relation $	au = I \alpha$:$\frac{\lambda_0 g L^3}{3} = \left( \frac{\lambda_0 L^4}{4} ight) \alpha$.Solving for $\alpha$:$\alpha = \frac{\lambda_0 g L^3}{3} 	imes \frac{4}{\lambda_0 L^4} = \frac{4g}{3L}$.

$A$ non-uniform rod $OA$ of linear mass density $\lambda = \lambda_0 x$ (where $\lambda_0$ is a constant) is suspended from the ceiling with a hinge joint at $O$ and a light string as shown in the figure. Find the angular acceleration of the rod just after the string is cut.

Similar Questions

$A$ disc of mass $10 \ kg$ and radius $0.1 \ m$ is rotating at $120 \ rpm$. $A$ retarding torque brings it to rest in $10 \ s$. If the same torque is due to a force applied tangentially on the rim of the disc, then the magnitude of the force is: (in $\pi \ N$)

$A$ disc is rotating with an angular velocity $\omega_0$. $A$ constant retarding torque is applied on it to stop the disc. The angular velocity becomes $\frac{\omega_0}{2}$ after $n$ rotations. How many more rotations will it make before coming to rest?

$A$ circular disc of diameter $0.8 \ m$ and mass $4 \ kg$ is rolling on a smooth horizontal plane. If $2.56 \ N \ m$ torque is acting on the disc,then its angular acceleration is

Which of the following relations is incorrect?

Vedclass Test Series

Exam Paper Generator

Online Exam Module