The radius of gyration of a uniform disc about a line perpendicular to the disc equals its radius. Find the distance of the line from the centre.
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♻♻Here Is Your Answer♻♻
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The moment of inertia about the center and ⊥ to the plane of the disc of radius (r) and mass (m) is = mr^2.
➡According to the question,
The radius of gyration of the disc about a point = radius of the disc.
Therefore mk^2 = ½ mr^2 + md^2
(K = radius of gyration about acceleration point, d = distance of that point from the centre)
➡K^2 = r^2/2 + d2
➡r^2 = r^2/2 + d2 (K = r)
➡r^2/2 = d^2
➡d = r / √2...Ans..✔
♻♻Here Is Your Answer♻♻
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
The moment of inertia about the center and ⊥ to the plane of the disc of radius (r) and mass (m) is = mr^2.
➡According to the question,
The radius of gyration of the disc about a point = radius of the disc.
Therefore mk^2 = ½ mr^2 + md^2
(K = radius of gyration about acceleration point, d = distance of that point from the centre)
➡K^2 = r^2/2 + d2
➡r^2 = r^2/2 + d2 (K = r)
➡r^2/2 = d^2
➡d = r / √2...Ans..✔
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