Two half rings are joined as shown in the fig | vedclass

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

Question

(A) The moment of inertia of a complete ring of mass $2M$ and radius $R$ about its diameter is $I = \frac{1}{2}(2M)R^2 = MR^2$.Since the system consis

Vedclass · Accepted Answer

$MR^2$

Vedclass · Answer

(A) The moment of inertia of a complete ring of mass $2M$ and radius $R$ about its diameter is $I = \frac{1}{2}(2M)R^2 = MR^2$.Since the system consists of two half-rings of mass $M$ each,joined such that the axis $XX'$ passes through their common diameter,we can treat the system as a single ring of mass $2M$ and radius $R$ rotating about its diameter.The moment of inertia of a thin ring of mass $m$ and radius $R$ about its diameter is $\frac{1}{2}mR^2$.For the first half-ring of mass $M$,the moment of inertia about the axis $XX'$ is $I_1 = \frac{1}{2}MR^2$.For the second half-ring of mass $M$,the moment of inertia about the axis $XX'$ is $I_2 = \frac{1}{2}MR^2$.Using the principle of superposition,the total moment of inertia of the system is $I = I_1 + I_2 = \frac{1}{2}MR^2 + \frac{1}{2}MR^2 = MR^2$.

Two half rings are joined as shown in the figure. Each half ring has a radius $R$ and mass $M$. The moment of inertia of the system about the axis $XX'$ is

Similar Questions

The moment of inertia of a semicircular ring about an axis passing through the center and perpendicular to the plane of the ring is $\frac{1}{x} MR^2$,where $R$ is the radius and $M$ is the mass of the semicircular ring. The value of $x$ will be $...........$

$A$ uniform square plate $S$ (side $c$) and a uniform rectangular plate $R$ (sides $b$,$a$) $(a > b)$ have identical areas and masses. Based on the figure,which of the following is correct?
$(a)\ I_{xR}/I_{xS} < 1$
$(b)\ I_{yR}/I_{yS} > 1$
$(c)\ I_{zR}/I_{zS} > 1$

Three rods each of length $L$ and mass $M$ are placed along $X$,$Y$,and $Z$-axes in such a way that one end of each rod is at the origin. The moment of inertia of this system about the $Z$-axis is

One circular ring and one circular disc,both having the same mass and radius. The ratio of their moments of inertia about the axis passing through their centers and perpendicular to their planes will be:

The radius of gyration of a uniform rod of length $l$ about an axis passing through one of its ends and perpendicular to its length is

Vedclass Test Series

Exam Paper Generator

Online Exam Module