Question

Three particles, each of mass 200g, are kept at the corners of an equilateral triangle of side 10cm.Find the moment of inertia of the system about an axis :
a) joining two of the particles and
b) passing through one of the particles and perpendicular to the plane of the particles .​

Answer 1

Answer:

$\large\boxed{\sf{(a)\:1.5\times{10}^{-3}\;\;Kg\:{m}^{2}}}$

$\large\boxed{\sf{(b)\:4\times{10}^{-3}\;\;Kg\:{m}^{2}}}$

Explanation:

It's given that,

Threee particles are kept at the corners of an equilateral triangle .

• Mass of particles = 200 g = 0.2 Kg
• Side of triangle = 10 cm = 0.1 m

Let the equilateral triangle be ∆ABC.

Now, to find the moment of inertia of the system about an axis:

$\bold\red{Note:-}$ Refer to the attachment.

(a) joining two of the particles

Let the particles at B and C are joined to make the axis.

Also, AD perpendicular to axis BC is drawn.

• AD = √3/2 × 10 = 5√3 cm = 5√3/100 m

In this case, B and C lies on the axis,

Therefore, they will not contribute in the moment of inertia.

Only particle at A will contribute in moment of inertia.

Therefore, we will get,

$= > i = m {r}^{2} \\ \\ = > i = 0.2 \times {( \dfrac{5 \sqrt{3} }{100} )}^{2} \\ \\ = > i = 0.2 \times \dfrac{75}{ {10}^{4} } \\ \\ = > i = 15 \times {10}^{ - 4} \\ \\ = > \bold\purple{i = 1.5 \times {10}^{ - 3} \: \: Kg \: {m}^{2} }$

(b) passing through one of the particles and perpendicular to the plane of the particles .

A line perpendicular to plane of ∆ABC is drawn from A.

Thus, particles at B and C will produce a torque on it.

In this case, A lies on the axis,

Therefore, it will not contribute in moment of inertia.

Only, particles at B and C will contribute in moment of inertia.

Therefore, we will get,

$= > i = m {r}^{2} + m {r}^{2} \\ \\ = > i = 2m {r}^{2} \\ \\ = > i = 2 \times 0.2 \times {(0.1)}^{2} \\ \\ = > i = 0.4 \times {10}^{ - 2} \\ \\ = > \bold\purple{i = 4 \times {10}^{ - 3} \: \: Kg \: {m}^{2}}$

Answer 2

Answer:

Therefore, the I distance from the axis (AD) = 13/2 x10 = 5 13 cm.

Therefore moment of inertia about the axis BC will be |= mr? = 200 K (573)2 = 200 x 25 x 3

= 15000 gm - cm2 = 1.5 x 10-3 kg-m2.

b) The axis of rotation let pass through A and I to the plane of triangle Therefore the torque will be produced by mass Band C

Therefore net moment of inertia = 1 = mr+ mr2

= 2 x 200 x102 = 40000 gm-cm2 = 4 *10-3 kg m2.