A thin uniform rod of mass M and length L is | vedclass

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

Question

(D) The rod is pivoted at a distance $\frac{L}{3}$ from the lower end. The center of mass of the rod is at a distance $\frac{L}{2}$ from the lower end

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The rod is pivoted at a distance $\frac{L}{3}$ from the lower end. The center of mass of the rod is at a distance $\frac{L}{2}$ from the lower end.Initially, the center of mass is at a height $h_i = \frac{L}{2} - \frac{L}{3} = \frac{L}{6}$ above the pivot point.When the rod hits the table, the center of mass is at the same level as the pivot point, so the final height $h_f = 0$.The change in potential energy is $\Delta PE = Mg(h_f - h_i) = -Mg\frac{L}{6}$.By the principle of conservation of energy, the loss in potential energy equals the gain in rotational kinetic energy: $Mg\frac{L}{6} = \frac{1}{2} I \omega^2$.The moment of inertia $I$ about the pivot point (which is at distance $\frac{L}{3}$ from the end) is calculated using the parallel axis theorem: $I = I_{cm} + M d^2 = \frac{ML^2}{12} + M(\frac{L}{2} - \frac{L}{3})^2 = \frac{ML^2}{12} + M(\frac{L}{6})^2 = \frac{ML^2}{12} + \frac{ML^2}{36} = \frac{3ML^2 + ML^2}{36} = \frac{4ML^2}{36} = \frac{ML^2}{9}$.Substituting $I$ into the energy equation: $Mg\frac{L}{6} = \frac{1}{2} (\frac{ML^2}{9}) \omega^2$.$Mg\frac{L}{6} = \frac{ML^2}{18} \omega^2$.$\omega^2 = \frac{MgL}{6} \cdot \frac{18}{ML^2} = \frac{3g}{L}$.$\omega = \sqrt{\frac{3g}{L}}$.

Similar Questions

The instantaneous angular position of a rotating wheel is given by the formula $\theta (t) = 2t^3 - 6t^2$. At what time will the torque on this wheel be zero? $t = \dots \text{s}$

$A$ disc of radius $0.4$ m and mass $1$ kg rotates about an axis passing through its centre and perpendicular to its plane. The angular acceleration is $10$ rad/s$^2$. The tangential force applied to the rim of the disc is (in $N$)

In the following figure,$r_1 = 5 \, cm$ and $r_2 = 30 \, cm$. If the moment of inertia of the wheel is $5100 \, kg \cdot m^2$,then its angular acceleration will be:

$A$ disc of mass $M \ kg$ and radius $R \ m$ is rotating at an angular speed of $\omega \ rad/s$ when the motor is switched off. Neglecting the friction at the axle,the force that must be applied tangentially to the wheel to bring it to rest in time $t$ is ...............

$A$ uniform rod $AB$ of length $l$ and mass $m$ is free to rotate about point $A.$ The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about $A$ is $ml^2/3$,the initial angular acceleration of the rod will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module