The radius of gyration of a uniform rod of length l , about an axis passing through a point l 4 away from the centre of the rod, and perpendicular to it, is:

M l 2 12 + M l 4 2 = M K 2 l 2 12 + l 2 16 = K 2 K = 7 48 l

Physics - Rotation Question with Solution | TestHub

PhysicsRotationCalculation of M.I.Easy2 minPYQ_2020

PhysicsEasysingle choice

The radius of gyration of a uniform rod of length $l,$ about an axis passing through a point $\frac{l}{4}$ away from the centre of the rod, and perpendicular to it, is:

Options:

Answer:

Solution:

$\frac{M l^{2}}{12} + M {(\frac{l}{4})}^{2} = M K^{2}$
$\frac{l^{2}}{12} + \frac{l^{2}}{16} = K^{2}$
$K = \sqrt{\frac{7}{48}} l$

Stream:JEESubject:PhysicsTopic:RotationSubtopic:Calculation of M.I.

⏱ 2mℹ️ Source: PYQ_2020

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