A long cylinderical rod is welded to a thin circular disc of diameter 0.5 m at a point on its circumference. The rod is in the same plane as that of the disc and forms a tangent to the disc. The radius of gyration of the disc about the rod (in m ) is

Question

A long cylinderical rod is welded to a thin circular disc of diameter 0.5 m at a point on its circumference. The rod is in the same plane as that of the disc and forms a tangent to the disc. The radius of gyration of the disc about the rod (in m ) is (A) (1/4) (B) √(5/8) (C) (1/2) (D) 2 √2

Tardigrade Admin · Accepted Answer

As we know, moment of inertia of disc tangential to the rim and parallel to diameter,

I = \frac{5}{4} M R^{2}

Now, radius of gyration,

k = \frac{I}{M} = \frac{\frac{5}{4} M R ^{2}}{M} = \frac{5}{4} R

As, given diameter of disc,

R = 0.5 m

so,

k = \frac{5}{4} \times 0.5

\Rightarrow k = \frac{5}{8}

Q. A long cylinderical rod is welded to a thin circular disc of diameter $0.5 m$ at a point on its circumference. The rod is in the same plane as that of the disc and forms a tangent to the disc. The radius of gyration of the disc about the rod (in $m$ ) is

$\frac{1}{4}$

$\frac{5}{8}$

$\frac{1}{2}$

$22$

Q. A long cylinderical rod is welded to a thin circular disc of diameter 0.5m at a point on its circumference. The rod is in the same plane as that of the disc and forms a tangent to the disc. The radius of gyration of the disc about the rod (in m ) is

41​

85​​

21​

22​

As we know, moment of inertia of disc tangential to the rim and parallel to diameter, I=45​MR2 Now, radius of gyration, k=MI​​=M45​MR2​​=45​​R As, given diameter of disc, R=0.5m so, k=45​​×0.5 ⇒k=85​​

Q. A long cylinderical rod is welded to a thin circular disc of diameter $0.5 m$ at a point on its circumference. The rod is in the same plane as that of the disc and forms a tangent to the disc. The radius of gyration of the disc about the rod (in $m$ ) is

$\frac{1}{4}$

$\frac{5}{8}$

$\frac{1}{2}$

$22$