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.

Sol:

Given:

Mass = 200g = 0.2 Kg

Sides = 10cm = 0.1 m

The perpendicular distance, AD = \(\frac{\sqrt{3}}{2} * 10\) \( \Rightarrow 5\sqrt{3 cm} \)

(A) Joining Two Of The Particles

Moment of inertia along BC

I = mr2

= \(0.2 (5\sqrt{3})^{2} * 10^{-4}\)

I = 1.5 – 10-3 Kg m2

(B) Passing Through One Of The Particles And Perpendicular To The Plane Of The Particles.

I = mr2 + mr2

I = 4 * 10-3 Kg m2

Explore more such questions and answers at BYJU’S.

Was this answer helpful?

 
   

0 (0)

(0)
(0)

Choose An Option That Best Describes Your Problem

Thank you. Your Feedback will Help us Serve you better.

Leave a Comment

Your Mobile number and Email id will not be published. Required fields are marked *

*

*

BOOK

Free Class

Ask
Question