A thin circular ring of mass M and radius r i | vedclass

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

Question

(A) The initial angular momentum of the ring is $L = I\omega = Mr^2\omega$.When four particles of mass $m$ are placed on the ring,the new moment of in

Vedclass · Accepted Answer

$\frac{M\omega}{M + 4m}$

Vedclass · Answer

(A) The initial angular momentum of the ring is $L = I\omega = Mr^2\omega$.When four particles of mass $m$ are placed on the ring,the new moment of inertia of the system becomes $I' = Mr^2 + 4(mr^2) = (M + 4m)r^2$.Since there is no external torque acting on the system,the angular momentum remains conserved.Therefore,$L_{initial} = L_{final}$.$Mr^2\omega = (M + 4m)r^2\omega'$.Solving for the new angular velocity $\omega'$,we get $\omega' = \frac{Mr^2\omega}{(M + 4m)r^2} = \frac{M\omega}{M + 4m}$.

$A$ thin circular ring of mass $M$ and radius $r$ is rotating about its axis with a constant angular velocity $\omega$. Four particles,each of mass $m$,are gently placed on the ring at opposite points of two perpendicular diameters. The new angular velocity of the ring will be:

Similar Questions

Which one of the following characters is not typical of the class Mammalia?

$\begin{aligned} & f(x, y)=2(x-y)^2-x^4-y^4 \\ & \left|\left(f_{x x} f_{y y}-f_{x y}^2\right)\right|_{(0,0)} \end{aligned}$

The bond order of individual carbon-carbon bonds in benzene is

The value of $k$,if $(1,2)$ and $(k,-1)$ are conjugate points with respect to the ellipse $2x^2 + 3y^2 = 6$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module