A uniform bar of length 6l and mass 8m lies o | vedclass

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

Question

(C) Let the center of the bar be the origin $c$. The moment of inertia of the bar about $c$ is $I_{bar} = \frac{(8m)(6l)^2}{12} = 24ml^2$.After the co

Vedclass · Accepted Answer

$\frac{3mv^2}{5}$

Vedclass · Answer

(C) Let the center of the bar be the origin $c$. The moment of inertia of the bar about $c$ is $I_{bar} = \frac{(8m)(6l)^2}{12} = 24ml^2$.After the collision,the masses $m$ and $2m$ stick to the bar at distances $2l$ and $l$ from the center respectively.The total moment of inertia of the system about $c$ is $I = I_{bar} + m(2l)^2 + (2m)(l)^2 = 24ml^2 + 4ml^2 + 2ml^2 = 30ml^2$.The angular momentum of the system about $c$ before the collision is $L = (2m)(v)(l) + (m)(2v)(2l) = 2mvl + 4mvl = 6mvl$.Since there are no external torques,angular momentum is conserved: $L = I\omega$.$6mvl = (30ml^2)\omega \Rightarrow \omega = \frac{6mvl}{30ml^2} = \frac{v}{5l}$.The total energy about the center of mass is the rotational kinetic energy: $E = \frac{1}{2}I\omega^2$.$E = \frac{1}{2}(30ml^2)\left(\frac{v}{5l}ight)^2 = 15ml^2 	imes \frac{v^2}{25l^2} = \frac{15}{25}mv^2 = \frac{3mv^2}{5}$.

Similar Questions

$A$ thin uniform rod of length $L$ and mass $M$ is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is $\omega$. Its centre of mass rises to a maximum height of (where $g$ is the acceleration due to gravity):

$A$ disc is rotating with angular velocity $\vec{\omega}$. $A$ force $\vec{F}$ acts at a point whose position vector with respect to the axis of rotation is $\vec{r}$. The power associated with the torque due to the force is given by ..........

$A$ man, sitting firmly on a rotating stool, has his arms stretched. If he folds his arms, the work done by the man is:

$A$ thin rod of length $l$ and mass $m$ oscillates in a vertical plane about a horizontal axis passing through one of its ends. If the maximum angular velocity of the rod is $\omega$,what is the maximum height reached by its center of mass?

$A$ uniform rod of length $L$ is free to rotate in a vertical plane about a fixed horizontal axis through $B$. The rod begins rotating from rest from its unstable equilibrium position. When it has turned through an angle $\theta$,its angular velocity $\omega$ is given as

Vedclass Test Series

Exam Paper Generator

Online Exam Module