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: $E = \frac{1}{2}I\omega^2$.$E = \frac{1}{2}(30ml^2)\left(\frac{v}{5l}ight)^2 = 15ml^2 	imes \frac{v^2}{25l^2} = \frac{15}{25}mv^2 = \frac{3mv^2}{5}$.

Column-$I$	Column-$II$
$(1)$ $W = F \Delta x$	$(a)$ $P = \tau \omega$
$(2)$ $P = Fv$	$(b)$ $W = \tau \Delta \theta$
	$(c)$ $L = I \omega$

Similar Questions

$A$ sphere of mass $m$ and moment of inertia $I$ about its center of mass is rolling without slipping down an inclined plane from rest. Which of the following represents its total kinetic energy?

Match the linear motion formulas in Column-$I$ with their corresponding rotational motion formulas in Column-$II$.

Column-$I$ Column-$II$

$(1)$ $W = F \Delta x$ $(a)$ $P = \tau \omega$

$(2)$ $P = Fv$ $(b)$ $W = \tau \Delta \theta$

$(c)$ $L = I \omega$

$A$ 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):

Write the formula of work done by torque in a rotational rigid body about a fixed axis.

Vedclass Test Series

Exam Paper Generator

Online Exam Module