Three rings each of mass M and radius R are a | vedclass

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

Question

(D) The system consists of three rings. Let the top ring be $1$,and the two bottom rings be $2$ and $3$. The axis $YY'$ passes through the center of r

Vedclass · Accepted Answer

$\frac{7}{2}MR^{2}$

Vedclass · Answer

(D) The system consists of three rings. Let the top ring be $1$,and the two bottom rings be $2$ and $3$. The axis $YY'$ passes through the center of ring $1$ (diameter) and is tangent to rings $2$ and $3$ (parallel to their diameters).$1$. For ring $1$,the axis $YY'$ is its diameter. The moment of inertia is $I_1 = \frac{1}{2}MR^2$.$2$. For rings $2$ and $3$,the axis $YY'$ is parallel to their diameters at a distance $R$. Using the parallel axis theorem,$I = I_{cm} + Md^2$,where $I_{cm} = \frac{1}{2}MR^2$ and $d = R$. Thus,$I_2 = I_3 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.$3$. The total moment of inertia of the system is $I = I_1 + I_2 + I_3 = \frac{1}{2}MR^2 + \frac{3}{2}MR^2 + \frac{3}{2}MR^2 = \frac{7}{2}MR^2$.

Three rings each of mass $M$ and radius $R$ are arranged as shown in the figure. The moment of inertia of the system about the axis $YY'$ is:

Similar Questions

Which of the following is a Dobereiner's triad?

$A$ magnetic needle lying parallel to a magnetic field requires $W$ units of work to turn it through $60^{\circ}$. The torque required to maintain the needle in this position will be

Identify $A$ and $B$ in the following reactions:

Magnetic induction at the centre of a circular loop of area $\pi \ m^2$ is $0.1 \ T$. The magnetic moment of the loop is ( $\mu_0 = \text{permeability of air}$ ).

If $(2+\sin x) \frac{dy}{dx} + (y+1) \cos x = 0$ and $y(0) = 1$,then $y\left(\frac{\pi}{2}\right)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module