Question

what is moment of inertia of a 200 kg sphere whose radius is 100 sm?

Answer 1

Given :-

Mass of the sphere = 200 kg

Radius of the sphere = 100 cm

To Find :-

The momentum of inertia of the sphere.

Analysis :-

Here we are given with the mass and radius of the sphere.

In order to find the momentum of inertia substitute the values given in the question accordingly and find the momentum of inertia of the sphere.

Solution :-

We know that,

• m = Mass
• i = Momentum of inertia
• r = Radius

Using the formula,

$\underline{\boxed{\sf Momentum \ of \ inertia \ of \ a \ sphere=\dfrac{2}{5}mr^2 }}$

Given that,

Mass (m) = 200 kg

Radius (r) = 100 cm = 1 m

Substituting their values,

$\sf i=\dfrac{2}{5} \times 200 \times 1^2$

$\sf i=\dfrac{2 }{5} \times 200$

$\sf i=2 \times 40$

$\sf i=80 \ kg \ m^2$

Therefore, the momentum of inertia of the sphere is 80 kg m².

Answer 2

Answer:

Given :-

• A 200 kg sphere whose radius is 100 cm.

To Find :-

• What is the moment of inertia.

Formula Used :-

We know that,

$\sf\boxed{\bold{\large{I\: =\: \dfrac{2}{5}M{R}^{2}}}}$

where,

• I = Moment of Inertia
• M = Mass
• R = Radius

Solution :-

Given :

• Mass = 200 kg
• Radius = 100 cm = 1 m

According to the question by using the formula we get,

↦ I = $\sf\dfrac{2}{5} \times 200 \times {1}^{2}$

↦ I = $\sf\dfrac{2}{5} \times 200 \times 1$

↦ I = $\sf\dfrac{2}{5} \times 200$

↦ I = $\sf\dfrac{2}{\cancel{5}} \times {\cancel{200}}$

↦ I = $\sf2 \times 40$

➦ I = 80 kg m²

$\therefore$ The moment of inertia is 80 kg m² .