A circular plate of mass M and radius R has i | vedclass

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

Question

(A) The mass $M$ of the plate is given by integrating the density over the area: $M = \int_0^R \rho(r) (2\pi r dr) = \int_0^R \rho_0 r (2\pi r dr) = 2

Vedclass · Accepted Answer

$8/5$

Vedclass · Answer

(A) The mass $M$ of the plate is given by integrating the density over the area: $M = \int_0^R \rho(r) (2\pi r dr) = \int_0^R \rho_0 r (2\pi r dr) = 2\pi \rho_0 \int_0^R r^2 dr = \frac{2\pi \rho_0 R^3}{3}$.The moment of inertia about the center of mass $(I_{CM})$ is: $I_{CM} = \int_0^R r^2 dm = \int_0^R r^2 (\rho_0 r \cdot 2\pi r dr) = 2\pi \rho_0 \int_0^R r^4 dr = \frac{2\pi \rho_0 R^5}{5}$.Expressing $I_{CM}$ in terms of $M$: $I_{CM} = (\frac{2\pi \rho_0 R^3}{3}) \cdot \frac{3}{5} R^2 = \frac{3}{5} MR^2$.Using the parallel axis theorem,the moment of inertia about an axis passing through the edge and perpendicular to the plate is: $I = I_{CM} + MR^2 = \frac{3}{5} MR^2 + MR^2 = \frac{8}{5} MR^2$.Comparing this with $I = aMR^2$,we get $a = 8/5$.

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

$A$ solid sphere of radius $R$ has mass $M$. The moment of inertia of the solid sphere about an axis at a distance $\frac{R}{2}$ from the centre is:

The moment of inertia of a circular ring about an axis perpendicular to its plane and passing through its center is $200 \, gm \cdot cm^2$. The moment of inertia about its diameter is ....... $gm \cdot cm^2$.

Four identical thin rods,each of mass $M$ and length $l$,form a square frame. The moment of inertia of this frame about an axis passing through the centre of the square and perpendicular to its plane is

According to the theorem of parallel axes $I = I_g + Md^2$,the graph between $I$ and $d$ will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module