System of Particles and Rotational Motion
- Three particles, each of mass m gram, are situated at the vertices of an equilateral triangle ABC of side l 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
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IAX = m(AB)² + m(OC)²
= ml² + m (l cos 60º)²
= ml² + ml²/4 = 5/4 ml²
Correct Option: D
IAX = m(AB)² + m(OC)²
= ml² + m (l cos 60º)²
= ml² + ml²/4 = 5/4 ml²
- The ratio of the radii of gyration of a circular disc about a tangential axis in the plane of the disc and of a circular ring of the same radius about a tangential axis in the plane of the ring is
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Circular disc [1]Iy2 = MR² 4 ∴ I'y1 = MR² + MR² = 5 MR² 4 4 
Circular ring [2]∴ I'y2 = MR² + MR² = 3 MR² 2 2
I'y1 = MK1² , I'y2 = MK2²∴ K1² = I'y1 ⇒ K1 : K2 = √5 : √6 K2² I'y2
Correct Option: D
Circular disc [1]Iy2 = MR² 4 ∴ I'y1 = MR² + MR² = 5 MR² 4 4
Circular ring [2]∴ I'y2 = MR² + MR² = 3 MR² 2 2
I'y1 = MK1² , I'y2 = MK2²∴ K1² = I'y1 ⇒ K1 : K2 = √5 : √6 K2² I'y2
- The moment of inertia of a uniform circular disc of radius R and mass M about an axis touching the disc at its diameter and normal to the disc is
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M.I. of a uniform circular disc of radius ‘R’ and mass ‘M’ about an axis passing though C.M. and normal to the disc is
LC.M = 1 MR² 2
From parallel axis theorem,IT = IC.M + MR² = 1 MR² + MR² = 3 MR² 2 2
Correct Option: C
M.I. of a uniform circular disc of radius ‘R’ and mass ‘M’ about an axis passing though C.M. and normal to the disc is
LC.M = 1 MR² 2
From parallel axis theorem,IT = IC.M + MR² = 1 MR² + MR² = 3 MR² 2 2
- A thin rod of length L and mass M is bent at its midpoint into two halves so that the angle between them is 90°. The moment of inertia of the bent rod about an axis passing through the bending point and perpendicular to the plane defined by the two halves of the rod is:
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Mass of each part = M/2
Length of each part = L/2
Total M.I. = Sum of M.I.s of both parts= 
M 
L ² × 1 + M L ² × 1 2 2 3 2 2 3 = 2 × M × L² × 1 = ML² 2 4 3 12
Correct Option: B
Mass of each part = M/2
Length of each part = L/2
Total M.I. = Sum of M.I.s of both parts= M L ² × 1 + M L ² × 1 2 2 3 2 2 3 = 2 × M × L² × 1 = ML² 2 4 3 12
- Four identical thin rods each of mass M and length l, form a square frame. Moment of inertia of this frame about an axis through the centre of the square and perpendicular to its plane is:
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Moment of inertia of a thin rod of length l about an axis passing through centre and perpendicular to the rod = (1/12)Ml²
Thus moment of inertia of the frame.ml² + ml² = 4ml² = ml² 12 4 12 3
Total M.I. = 4 × (ml²/3)Correct Option: D
Moment of inertia of a thin rod of length l about an axis passing through centre and perpendicular to the rod = (1/12)Ml²
Thus moment of inertia of the frame.ml² + ml² = 4ml² = ml² 12 4 12 3
Total M.I. = 4 × (ml²/3)
