Question

For a uniform disc of mass m and radius r, the moment of inertia of the disc about a tangent in its plane is....

Answer 1

Answer:

the moment of inertia of the disc about a tangent in its plane is

$I = \frac{5}{4}mR^2$

Explanation:

As we know that the moment of inertia of disc about an axis passing through its center and perpendicular to its plane is given as

$I = \frac{1}{2}mR^2$

now by Perpendicular axis theorem we have

$I = I_x + I_y$

$I_x = I_y$

so we have

$I_x = \frac{1}{4}mR^2$

so above is the moment of inertia about an axis passing through the center and lying in the plane of the disc

Now for tangential axis we can use parallel axis theorem

$I = \frac{1}{4} mR^2 + mR^2$

$I = \frac{5}{4}mR^2$

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Topic : Moment of inertia

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