Question

Consider a flywheel whose mass M is distributed almost equally between a heavy, ring-like rim of radius R and a concentric disk-like feature of radius R/2. Other parts of the flywheel, such as spokes, etc, have negligible mass. The best approximation for α, if the moment of inertia of the flywheel about its axis of rotation is expressed as αMR², is _________

Answer 1

Concept:

Moment of inertia:

• Moment of inertia of a body about a given axis is the sum of the product of masses of all the particles of the body and squares of their respective perpendicular distances from the axis of rotation.
• Moment of inertia plays the same role in rotational motion as mass plays in linear motion.
• Moment of inertia of some of the bodies are mentioned below:

Body	Axis	Moment of inertia
Ring	Passing through the centre and perpendicular to the plane of the ring	MR2
Disc	Passing through centre and perpendicular to the plane of the disc	\(\frac{{M{R^2}}}{2}\)
Hollow cylinder	Axis of cylinder	MR2
Solid Cylinder	Axis of cylinder	\(\frac{{M{R^2}}}{2}\)
Hollow sphere	Diameter	\(\frac{2}{3}M{R^2}\)
Solid sphere	Diameter	\(\frac{2}{5}M{R^2}\)

 

Calculation:

Given:

Mass of rim = \(\frac{M}{2}\), Mass of disk = \(\frac{M}{2}\)

Radius of rim = R, Radius of disk = \(\frac{R}{2}\)

∴ Moment of inertia (I)

I = (Mass moment of inertia due to mass at rim) + (Mass moment of inertia due to mass at disk)

\(\begin{array}{l} I = \frac{M}{2}{R^2} + \frac{1}{2} \times \left( {\frac{M}{2}} \right){\left( {\frac{R}{2}} \right)^2}\\ \Rightarrow I = \frac{{M{R^2}}}{2} + \frac{{M{R^2}}}{16} = 0.5625M{R^2} \end{array}\)

∴ α = 0.5625