The moment of inertia of a uniform rod of mas | vedclass

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

Question

(C) Initial angular momentum of the system about the pivot point is due to the bullet: $L_i = Mv(L/2)$.After the collision,the bullet is embedded in t

Vedclass · Accepted Answer

$3v/2L$

Vedclass · Answer

(C) Initial angular momentum of the system about the pivot point is due to the bullet: $L_i = Mv(L/2)$.After the collision,the bullet is embedded in the rod,so the system rotates with angular velocity $\omega$. The final moment of inertia of the system is $I_f = I_{	ext{rod}} + I_{	ext{bullet}} = \frac{ML^2}{12} + M(L/2)^2 = \frac{ML^2}{12} + \frac{ML^2}{4} = \frac{ML^2 + 3ML^2}{12} = \frac{4ML^2}{12} = \frac{ML^2}{3}$.By the principle of conservation of angular momentum,$L_i = L_f$,so $Mv(L/2) = I_f \omega$.Substituting the values: $\frac{MvL}{2} = \left(\frac{ML^2}{3}ight) \omega$.Solving for $\omega$: $\omega = \frac{MvL}{2} \cdot \frac{3}{ML^2} = \frac{3v}{2L}$.

Similar Questions

If a body has a moment of inertia $I$ and an angular velocity $\omega \, rad/s$,then what is the value of its angular momentum $L$?

$A$ uniform rod $AB$ of mass $m$ and length $l$ is at rest on a smooth horizontal surface. An impulse $P$ is applied to the end $B$. The time taken by the rod to turn through a right angle is

$A$ particle of mass $m$ moves along line $PC$ with velocity $v$ as shown in the figure. What is the angular momentum of the particle about point $O$?

When a mass moves in a plane about a fixed point,the direction of its angular momentum is along which of the following?

The position vector of a particle is $\vec{r} = (\hat{i} + 2\hat{j} - \hat{k})$ and its momentum is $\vec{p} = (3\hat{i} + 4\hat{j} - 2\hat{k})$. The angular momentum of this particle is perpendicular to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module