Three particles, each of mass m gram, are situated at the
vertices of an equilateral triangle ABC of side/cm (as shown
in the figure). The moment of inertia of the system about a
line AX perpendicular to AB and in the plane of ABC, in
gram-cm² units will be
(a) 3/2 m????²
(b) 3/4 m????²
(c) 2 m????²
(d) 5/4 m????²
Answers
Answered by
4
Answer:
5ml²/4
Explanation:
Mass of three articles = m (Given)
Following the moment of inertia -
System = mara²+mbrb²+mcrc²
= ma(0)²+m(l)²+m(lsin30°)²
Sin 30° = d/l
or 1/2 = d/l
or d = l/2
Thus, ml²+ml²/4
= 5ml²/4
Therefore, the moment of inertia of the system about a line AX perpendicular to AB and in the plane of ABC, in gram-cm² units will be 5ml²/4
Answered by
1
Answer:
5ml²/4
Explanation:
Mass of three articles = m (Given)
Following the moment of inertia -
System = mara²+mbrb²+mcrc²
= ma(0)²+m(l)²+m(lsin30°)²
Sin 30° = d/l
or 1/2 = d/l
or d = l/2
Thus, ml²+ml²/4
= 5ml²/4
Therefore, the moment of inertia of the system about a line AX perpendicular to AB and in the plane of ABC, in gram-cm² units will be 5ml²/4
Similar questions