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

The position vector of a $1\,kg$ object is $\overrightarrow{r} = (3\hat{i} - \hat{j})\,m$ and its velocity is $\overrightarrow{v} = (3\hat{j} + \hat{k})\,m/s$. The magnitude of its angular momentum is $\sqrt{x}\,N\cdot m\cdot s$,where $x$ is:

$A$ body of mass $2 \ kg$ is moving in a circular path of radius $2 \ m$ with uniform speed. If the centripetal force acting on it is $100 \ N$,then its angular momentum is ....... $J \cdot s$.

$A$ particle of mass $m$ is projected from a point $P$ on the ground with an initial velocity $v_0$ at an angle of $45^{\circ}$ with the horizontal at $t = 0$. Find the magnitude of the angular momentum of the particle at time $t = \frac{v_0}{g}$.

$A$ particle moves along a circular path with decreasing speed. Hence,

Which graph represents the relationship between angular momentum $L$ and angular velocity $\omega$ for a rigid body rotating about a fixed axis?

Vedclass Test Series

Exam Paper Generator

Online Exam Module