The moment of inertia of a uniform ring of mass m and radius r about an axis, AA touching the ring tangentially and lying in the plane of the ring only, as shown in figure , is

Question

The moment of inertia of a uniform ring of mass m and radius r about an axis, AA touching the ring tangentially and lying in the plane of the ring only, as shown in figure , is (A)  (5/4)mr2  (B)  (5/2)mr2  (C)  (1/4)mr2  (D)  (1/2)mr2

Tardigrade Admin · Accepted Answer

By parallel axis theorem, I=Icm+mh2 =(mr2/4)+mr2 =(5/4)mr2

Q. The moment of inertia of a uniform ring of mass $m$ and radius r about an axis, $AA$ touching the ring tangentially and lying in the plane of the ring only, as shown in figure , is

$\frac{5}{4} m r^{2}$

$\frac{5}{2} m r^{2}$

$\frac{1}{4} m r^{2}$

$\frac{1}{2} m r^{2}$

By parallel axis theorem, $I = I_{c m} + m h^{2}$ $= \frac{m r ^{2}}{4} + m r^{2}$ $= \frac{5}{4} m r^{2}$

Q. The moment of inertia of a uniform ring of mass m and radius r about an axis, AA touching the ring tangentially and lying in the plane of the ring only, as shown in figure , is

45​mr2

25​mr2

41​mr2

21​mr2