Question

The diameter of a thin circular disc of mass 2 kg is 0.2 m.
Its moment of inertia about an axis passing through the
edge and perpendicular to the plane of the disc is
(a) 0.01 kg-m2 (b) 0.02 kg-m2
(c) 0.03 kg-m2 (d) 0.04 kg-m2​

Answer 1

Answer:

Given:

Mass of Circular disc = 2 kg

Diameter = 0.2 m

To find:

Moment of Inertia passing through the edge and perpendicular to plane of disc.

Concept:

Moment of Inertia is actually a physical Quantity that represents the rotational analogue of mass.

Just as mass in Translational Motion , we have Moment of Inertia in Rotational motion.

Calculation:

Moment of Inertia along the centre of the disc and perpendicular to plane is given as :

$\sf{I_{1} = \dfrac{M {R}^{2} }{2}}$

As per Parallel Law of Axes :

Moment of Inertia along the edge parallel to initial axis :

$\sf{I_{2} = \dfrac{M {R}^{2} }{2} + M {R}^{2}}$

$\sf{ = > I_{2} = \dfrac{3M {R}^{2} }{2}}$

Now putting the values as given in the question :

$\sf{ = > I_{2} = \dfrac{3 \times 2 \times { (\dfrac{0.2}{2} )}^{2} }{2}}$

$\sf{ = > I_{2} = 0.03 \: kg {m}^{2} }$

So final answer is

${ \boxed{ \large{ \sf{\blue{ I_{2} = 0.03 \: kg {m}^{2} }}}}}$