Question

The moment of inertia of sphere is 40kg-m2 about the diameter. Determine the moment of inertia about any tangent .​

Answer 1

Answer:

(1) Given,

Moment of inertia of the sphere about its diametre = (2/5)mR²

Use, parallel axis theorem ,

Moment of inertia of the sphere about tangent = I + mR²

= (2/5)mR² + mR²

= (7/5)mR²

(B) given,

Moment of inertia of disc of mass m and radius R about any of its diametre = mR²/4

See the figure ,

Moment of inertia about diametre = Ix = Iy= (1/4)mR²

Use , perpendicular axis theorem ,

We know, Iz = Ix + Iy

Where Iz is moment of inertia about perpendicular axis of plane of disc .

Iz = (1/4)mR² + (1/4)mR²

= (1/2)mR²

Now, moment of inertia of disc about passing through a point of its edge

______________________________

Use , parallel axis theorem ,

I = Iz + mR²

= (1/2) mR² + mR²

= (3/2)mR²

Explanation:

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Answer 2

$\huge\underline\blue{\sf Answer:}$

$\large\red{\boxed{\sf I=140kg-{m}^{2}}}$

$\huge\underline\blue{\sf Solution:}$

$\large\underline\pink{\sf Given: }$

• Inertia (I) of sphere = $\sf{40\:kg-{m}^{2}}$

$\large\underline\pink{\sf To\:Find: }$

• Moment of inertia about any tangent = ?

━━━━━━━━━━━━━━━━━━━━━━━━━━

We know that Inertia for sphere is :

$\large{♡}$$\large{\boxed{\sf I={\frac{2}{5}}M{R}^{2}}}$

On Putting value of inertia ,

$\large\implies{\sf 40={\frac{2}{5}}M{R}^{2}}$

$\large\implies{\sf M{R}^{2}={\frac{40×5}{2}}}$

$\large\implies{\sf M{R}^{2}=100 }$

By theorem of parallel axes :

$\large{♡}$$\large{\boxed{\sf I=I_{CM}+M{R}^{2}}}$

$\large\implies{\sf {\frac{2}{5}}M{R}^{2}+M{R}^{2}}$

$\large\implies{\sf {\frac{7}{5}}M{R}^{2}}$

On putting value of $\sf{M{R}^{2}}$

$\large\implies{\sf {\frac{7}{5}}×100 }$

$\huge\red{♡}$$\large\red{\boxed{\sf I=140kg-{m}^{2}}}$