Radius of gyration of disc of mass 50 g and radius 2.5 cm about an axis passing through its center of gravity and perpendicular to the plane is

Question

Radius of gyration of disc of mass 50 g and radius 2.5 cm about an axis passing through its center of gravity and perpendicular to the plane is (A) 6.54 cm (B) 3.64 cm (C) 1.77 cm (D) 0.88 cm

Tardigrade Admin · Accepted Answer

Here mass M = 50 g and radius R = 2.5 cm Required moment of inertia of the disc is given by I=(MR2/2)=MK2 so,K2=(R2/2)or K=(R/√2)=(2.5/√2)=(2.5√2/2) = 1.767= 1.77 cm

Q. Radius of gyration of disc of mass 50 g and radius 2.5 cm about an axis passing through its center of gravity and perpendicular to the plane is

6.54 cm

3.64 cm

1.77 cm

0.88 cm

Here mass M = 50 g and radius R = 2.5 cm Required moment of inertia of the disc is given by
$I = \frac{M R ^{2}}{2} = M K^{2}$
so, $K^{2} = \frac{R ^{2}}{2} or K = \frac{R}{2} = \frac{2.5}{2} = \frac{2.5 2}{2}$
= 1.767= 1.77 cm

Q. Radius of gyration of disc of mass 50 g and radius 2.5 cm about an axis passing through its center of gravity and perpendicular to the plane is

6.54 cm

3.64 cm

1.77 cm

0.88 cm

Here mass M = 50 g and radius R = 2.5 cm Required moment of inertia of the disc is given by I=2MR2​=MK2 so,K2=2R2​orK=2​R​=2​2.5​=22.52​​ = 1.767= 1.77 cm

Here mass M = 50 g and radius R = 2.5 cm Required moment of inertia of the disc is given by
$I = \frac{M R ^{2}}{2} = M K^{2}$
so, $K^{2} = \frac{R ^{2}}{2} or K = \frac{R}{2} = \frac{2.5}{2} = \frac{2.5 2}{2}$
= 1.767= 1.77 cm