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 moment of inertia $(I)$ of the system about the axis $YY'$ is the sum of the moments of inertia of the three individual rings.$I = I_1 + I_2 +

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The moment of inertia $(I)$ of the system about the axis $YY'$ is the sum of the moments of inertia of the three individual rings.$I = I_1 + I_2 + I_3$For ring $1$,the axis $YY'$ passes through its diameter. The moment of inertia of a ring about its diameter is $I_1 = \frac{1}{2}MR^2$.For rings $2$ and $3$,the axis $YY'$ is tangent to the ring in its own plane. Using the parallel axis theorem,$I = I_{cm} + Md^2$,where $I_{cm} = \frac{1}{2}MR^2$ and $d = R$. Thus,$I_2 = I_3 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.Adding these together: $I = \frac{1}{2}MR^2 + \frac{3}{2}MR^2 + \frac{3}{2}MR^2 = \frac{7}{2}MR^2$.

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

Similar Questions

Which insecticide is used for termite control in buildings?

The following four wires of length $L$ and radius $r$ are made of the same material. Which of these will have the largest extension,when the same tension is applied?

If the circle $x^2+y^2+8x-4y+c=0$ touches the circle $x^2+y^2+2x+4y-11=0$ externally and cuts the circle $x^2+y^2-6x+8y+k=0$ orthogonally,then $k=$

An electron beam travels with a velocity of $1.6 \times 10^7 \ m/s$ perpendicularly to a magnetic field of intensity $0.1 \ T$. Calculate the radius of the path of the electron beam. (Given: $m_e = 9 \times 10^{-31} \ kg$,$e = 1.6 \times 10^{-19} \ C$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module