Question

If the radius of gyration of a solid disc of mass 10
kg about an axis is 0.40 m, then the moment of
inertia of the disc about that axis is
(1) 1.6 kg m? (2) 3.2 kg m?
(3) 6.4 kg m (4) 9.5 kg m?​

Answer 1

Hello Mate,

Here is your answer,

Given:

• Radius of gyration(K) of solid disc = 0.4m

• Mass of disc = 10kg

To find:

• Moment of Inertia of the disc

Formulas Used:

• $k = \frac{r}{ \sqrt{2} }$

Where,

k => Radius of gyration

r => Radius of solid disc

• Moment of Inertia of solid disc = $\frac{1}{2} \: m {r}^{2}$

Solution:

First we have to find out the radius of the solid disc.

We Know

$k = \frac{r}{ \sqrt{2} } \\ \\ 0.4 = \frac{r}{ \sqrt{2} } \\ \\ r = \frac{4 \sqrt{2} }{10} \\ \\ r = \frac{2 \sqrt{2} }{5}$

Substitute Radius value in the inertia formula we get,

$\frac{1}{2} \: m {r}^{2} \\ \\ = \frac{1}{2} \times 10 \times \frac{2 \sqrt{2} }{5} \times \frac{2 \sqrt{2} }{5} \\ \\ 2 \sqrt{2} \times \frac{2 \sqrt{2} }{5} \\ \\ = 1.6 \: kgm$

Answer is option (1)

Hope it helps:)

Answer 2

The formula for finding moment of inertia of a body about an axis is different due to their shape or mass distribution, so the moment of inertia of each particle in the body is considered and their sum is taken, thus the moment of inertia of a body is given by,

I = Σ(Mr²)

Suppose the mass is uniformly distributed in the body, then,

I = mΣ(r²)

Here Σ(r²) is replaced by K², then we have,

I = MK²

where K is called radius of gyration.

Given,

M = 10 kg

K = 0.40 m

Then, the moment of inertia is,

I = MK²

I = 10 (0.40)²

I = 10 × 0.16

I = 1.6 kg m²

Hence (1) 1.6 kg m² is the answer.

#answerwithquality

#BAL