Question

Find the moment of inertia of semicircular wire of mass 'M'and radius 'R' about an axis through the centre and perpendicular to the plane.

Answer 1

Answer on Question #51294, Physics, Mechanics | Kinematics | Dynamics

Find the moment of inertia of semicircular wire of mass 'M' and radius 'R' about an axis through the centre and perpendicular to the plane.

Solution:

The definition of moment of inertia is

$$I = \int R ^ {2} d m$$

where $R$ is the distance of the mass element $dm$ from the axis about which the moment is to be computed.

$$\pi R = l$$

Thus,

$$R = \frac {l}{\pi}$$

The mass element $dm$ is

$$dm = \rho * dl$$

where $\rho$ is the linear density of the wire, and $dl$ the element of length along the wire:

$$dl = R d \theta$$ $$dm = \rho (R d \theta)$$ $$I = \int_ {0} ^ {\pi} R ^ {2} \rho (R d \theta) = R ^ {3} \rho \int_ {0} ^ {\pi} d \theta = R ^ {3} \rho \pi$$

To compare with the table value, the mass of the wire $M = \pi R\rho$, so

$$I = m R ^ {2}$$

Answer: $I = m R^2$

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