Seven identical coins are rigidly arranged on | vedclass

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

Question

(C) Let $A$ be the centre of the central coin. The moment of inertia of the central coin about an axis through $A$ is $I_{central} = \frac{m r^2}{2}$.

Vedclass · Accepted Answer

$\frac{111}{2}$

Vedclass · Answer

(C) Let $A$ be the centre of the central coin. The moment of inertia of the central coin about an axis through $A$ is $I_{central} = \frac{m r^2}{2}$.For the six surrounding coins,the distance of their centres from $A$ is $2r$. Using the parallel axis theorem,the moment of inertia of each surrounding coin about the axis through $A$ is $I_{surrounding} = \frac{m r^2}{2} + m(2r)^2 = \frac{m r^2}{2} + 4mr^2 = \frac{9}{2} mr^2$.The total moment of inertia of the system about the axis through $A$ is $I_A = I_{central} + 6 	imes I_{surrounding} = \frac{m r^2}{2} + 6 	imes \frac{9}{2} mr^2 = \frac{m r^2}{2} + 27 mr^2 = \frac{55}{2} mr^2$.Now,we need the moment of inertia about point $P$,which is at a distance $d = 2r$ from $A$. Using the parallel axis theorem,$I_P = I_A + M_{total} 	imes d^2$,where $M_{total} = 7m$.$I_P = \frac{55}{2} mr^2 + (7m)(2r)^2 = \frac{55}{2} mr^2 + 28 mr^2 = \frac{55 + 56}{2} mr^2 = \frac{111}{2} mr^2$.

Similar Questions

Four holes of radius $R$ are cut from a thin square plate of side $4R$ and mass $M$. The moment of inertia of the remaining portion about the $z-$ axis is

The moment of inertia of a uniform hollow hemisphere about the given axes $I_A$ and $I_B$ is:

Four solid spheres,each of diameter $\sqrt{5} \ cm$ and mass $0.5 \ kg$,are placed with their centers at the corners of a square of side $4 \ cm$. The moment of inertia of the system about the diagonal of the square is $N \times 10^{-4} \ kg \cdot m^2$. Find $N$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module