A hollow cylinder has mass $M$ outside radius $R_{2}$ and inside radius $R_{1}$ .Its moments of inertia about an axis parallel to its symmetry axis and tangent to the outer surface in equal to:

\frac{M}{2} (R_{2}^{2} + R_{1}^{2})

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{M}{2} (R_{2}^{2} - R_{1}^{2})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{M}{4} (R_{2} + R_{1})^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{M}{2} (3 R_{2}^{2} + R_{1}^{2})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\frac{M}{2} (R_{2}^{2} + R_{1}^{2})$
Given, $M a s s = m, r a d i u s, R_{1}, R_{2}$

By the principle of superposition $I = I_{o u t} - I_{i n}$

Now the mass,

$M_{o u t} = \frac{m}{π (R_{2}^{2} - R_{1}^{2})} \times π R_{2}^{2}$

$M_{i n} = \frac{m}{π (R_{2}^{2} - R_{1}^{2})} \times π R_{1}^{2}$

Now, $I = \frac{M_{o u t} R_{2}^{2}}{2} - \frac{M_{i n} R_{1}^{2}}{2} = \frac{\frac{m}{R_{2}^{2} - R_{1}^{2}} \times R_{2}^{4}}{2} - \frac{\frac{m}{R_{2}^{2} - R_{1}^{2}} \times R_{1}^{4}}{2}$

$= \frac{m}{R_{2}^{2} - R_{1}^{2}} 2 (R_{2}^{4} - R_{1}^{4}) \Rightarrow I = \frac{m (R_{2}^{2} + R_{1}^{2})}{2}$

Suggest Corrections

Similar questions

Q. A hollow cylinder has mass

M

outside radius

R_{2}

and inside radius

R_{1}

. Its moment of inertia about an axis parallel to its symmetry axis and tangential to the outer surface is equal to

Q. A hollow cylinder has mass M. outside radius

R_{2}

and inside radius

R_{1}

Its moment of inertia about an or axis parallel to its symmetry axis and tangential to tho outer surface is equal to :

Charge ‘q’ is uniformly distributed over the surface of an annular-non-conducting disc of inner radius $R_{1}$ and outer radius $R_{2}$ . The disc is made to rotate about an axis passing through its centre and perpendicular to its plane with a constant frequency v (rotations per second). Magnetic moment of the disc can be expressed as :

Q. The moment of inertia of a hollow cylinder of mass

M

and inner radius

R_{1}

and outer radius

R_{2}

about its central axis is

Calculate the moment of Inertia of an annular disc of inner radius $R_{1}$ and outer radius $R_{2}$ about an axis perpendicular to the plane in which the disc lies and passing through $O$ .

Join BYJU'S Learning Program

A hollow cylinder has mass M outside radius R2 and inside radius R1.Its moments of inertia about an axis parallel to its symmetry axis and tangent to the outer surface in equal to:

The correct option is A M2(R22+R21)Given, Mass=m,radius,R1,R2By the principle of superposition I=Iout−IinNow the mass,Mout=mπ(R22−R21)×πR22Min=mπ(R22−R21)×πR21Now, I=MoutR222−MinR212=mR22−R21×R422−mR22−R21×R412=mR22−R212(R42−R41)⇒I=m(R22+R21)2

A hollow cylinder has mass $M$ outside radius $R_{2}$ and inside radius $R_{1}$ .Its moments of inertia about an axis parallel to its symmetry axis and tangent to the outer surface in equal to: