Question

iv). In a certain unit, the radius of gyration
of a uniform disc about its central and
transverse axis is
$\sqrt{2.5}$
Its radius of
gyration about a tangent in its plane (in
the same unit) must be

(A)
$\sqrt{5}$

(B)
$2.5$

(C)
$\sqrt[2]{2.5?}$

(D)
$\sqrt{12.5}$
​

Answer 1

Answer:

mnnbh all dg ne liya tha aur group of Narendra singh Negi aur group of Narendra

Explanation:

gghjjb ghb to be the first to comment on this device to moving in this email is strictly prohibited to moving

Answer 2

Answer:

To Find :

The radius of gyration of the uniform disc about a tangent in its plane = ?

Solution :

Since we know that the moment of inertia disc about central transverse axis is given as :

= \frac{mR^2}{2}}

Here m and R are mass and radius of disc respectively .

Let , K = radius of gyration

∴ {mK^2\:= \frac{mR^2}{2}}

So, K = \frac{R}{\sqrt[2]{2}}

= 2.5 units (radius of gyration is given 2.5 units )

Now by theorem of perpendicular axis :

Ix + Iy = Iz

Where Ix , Iy and Iz are the moment of inertia about x , y and z axis .

∴For a uniform disc , Ix = II

So, Ix = Iy = \frac{mR^2}{2\times 2}

= \frac{mR^2}{4}

Now by theorem of parallel axis :

Moment of inertia about tangent =

{\frac{mR^2}{4} \:+\:{mR^2}}

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

Let K' be the radius of gyration about the tangent , then :

mK'² = \frac{5mR^2}{4}

So, K' = \frac{\sqrt[5]{R}}{2}

Since \frac{R}{\sqrt[2]{2}}

=2.5 units

So, K' = {\frac{\sqrt[2]{5}}{\sqrt[2]{2}}\:×\:2.5}

= √12.5 units

Hence , radius of gyration about a tangent of the disc is

= √12.5 units

Explanation:

I hope it's helpful to you

don't forget to follow me and make my answer brain list

THANK YOU

By

爪尺 卄卂匚Ҝ乇尺