Question

Innocent and you need the radius of gyration of a uniform disc about its central and transfer axis is under root 2.5 its radius of gyration about the tangent in its plane {in the same unit (complete must be

Answer 1

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☆Given :

The radius of gyration of a uniform disc about its central and transverse axis =2.5 units

☆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, $K=\frac{R}{\sqrt{2} } =2.5 units$   (radius of gyration is given 2.5 units )

Now by theorem of perpendicular 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 $\frac{R}{\sqrt{2} }=2.5 units$

So, $K^'=\frac{\sqrt{5} }{\sqrt{2} }$ × $2.5$

          $=\sqrt{12.5} units$

☆Hence , The radius of gyration about a tangent of the disc is $=\sqrt{12.5} units$

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Hope it Helps :)

         

Answer 2

Explanation:

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Given :

The radius of gyration of a uniform disc about its central and transverse axis =2.5 units

☆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}

2

mR

2

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

Let , K = radius of gyration

∴ mK^2=\frac{mR^2}{2}mK

2

=

2

mR

2

So, K=\frac{R}{\sqrt{2} } =2.5 unitsK=

2

R

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

Now by theorem of perpendicular axis :

Moment of inertia about tangent = \frac{mR^2}{4} +mR^2=\frac{5mR^2}{4}

4

mR

2

+mR

2

=

4

5mR

2

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} }=2.5 units

2

R

=2.5units

So, K^'=\frac{\sqrt{5} }{\sqrt{2} } × 2.52.5

=\sqrt{12.5} units=

12.5

units

☆Hence , The radius of gyration about a tangent of the disc is =\sqrt{12.5} units=

12.5

units

hope this will help u