Physics System of Particles and Rotational Motion
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New answer posted
a month agoContributor-Level 9
Statement (1) is correct

The component of in the direction of is always perpendicular to , so it does not change.
Statement (2) is correct
The magnitude of does not change with time as .
Statement (3) is correct
The direction of changes with time.
Statement (4) is incorrect.
New answer posted
a month agoContributor-Level 10
L = Iω
I = (mL²/3) + (mL²/3) + [m (√2L)²/12] + m (L/√2)² = mL² [2/3 + 1/6 + 1/2] = (4mL²/3); L = (4/3)mL²ω
New answer posted
a month agoContributor-Level 10
Using parallel axis theorem for the side BC, the moment of inertia of the triangle about an axis passing through A and perpendicular to the plane of the triangle,
I? = 2 [ (1/3)ma²] + [ (1/12)ma² + m (√3a/2)²] = (3/2)ma²
If the angular velocity of the triangle at any instant is ω, the velocity of the vertex B at that instant is aω
Therefore, the velocity of B is maximum at the instant the angular velocity is maximum, i.e. when the side BC becomes horizontal
Let the angular velocity at this instant be ω?
Then, by conservation of energy
Gain in kinetic energy = Loss in potential energy
(1/2)I? ω²? = 3mg (Loss in height of CM of tr
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