Question

The mass of thin uniform rod is 200 gm and its
length is 10 cm. Find its moment of inertia and
radius of gyration about axis passing through
its midpoint and perpendicular to its length
[Ans: 2.886 x 10-2 M​

Answer 1

Answer:

the ans of mi is 1.6× 10^-4

and the radius of gyration is 2.8×10^-2

from its eq I= ML^2\12

k=L\2√3

Answer 2

Answer:

The moment of inertia is 1.67 × $10^{-4} kgm^{2}$ and the radius of gyration is 2.88 × $10^{-2}$ m

Explanation:

Given:

• The mass of thin uniform rod is 200 gm
• The length of thin uniform rod is 10 cm

To find: Moment of inertia and Radius of gyration

Solution:

Moment of Inertia:

• The term "moment of inertia" refers to the quantity that describes how a body resists angular acceleration and is calculated by multiplying each particle's mass by its square of distance from the rotational axis.
• Moment of Inertia is also known as the angular mass or rotational inertia.
• The SI unit of moment of inertia is kg /m2.

Moment of inertia of the rod passing through its centre and perpendicular to its length is

$I = \frac{ML^{2} }{12}$

Given that

M = 200 gm

Convert gm into kg

M = 200 gm = 0.2 kg

M = 0.2 kg

L = 10 cm

L  = 0.1 m

Substitute the value of M and L in the Equation

$I = \frac{ML^{2} }{12}$

$I = \frac{0.2 * (0.1)^{2} }{12}$

= $\frac{0.2*0.01}{12}$

= 1.67 × $10^{-4} kgm^{2}$

Moment of inertia =  = 1.67 × $10^{-4} kgm^{2}$

Radius of gyration :

• The imaginary distance from the centroid that the area of cross-section is thought to be focused at in order to get the same moment of inertia is known as the radius of gyration.
• Radius of gyration is denoted by the symbol K.

Formula:

K = $\sqrt{\frac{I}{M} }$

We know that,

I =  = 1.67 × $10^{-4} kgm^{2}$

M = 0.2

Radius of curvature,

K = $\sqrt{\frac{I}{M} }$

K = $\sqrt{\frac{1.67 * 10^{-4} }{0.2} }$

  = 0.0288

  = 2.88 × $10^{-2}$ m

Radius of curvature =  2.88 × $10^{-2}$ m

Final answer:

The moment of inertia about an axis is 1.67 × $10^{-4} kgm^{2}$ and the radius of gyration is 2.88 × $10^{-2}$ m

#SPJ2