Chapter 9: Q26P (page 378)

Show that moment of inertia of a disk of mass M and radius R is $\frac{1}{2} {MR}^{2}$ . Divide the disk into narrow rings, each of radius r and width dr. The contribution I of by one of these rings is r²dm, where dm is amount of mass contained in that particular ring. The mass of any ring is the total mass times the fraction of the total area occupied by the area of the ring. The area of this ring is approximately $2 πrdr$ . Use integral calculus to add up all the calculations.

Short Answer

Expert verified

It is proved that the moment of inertia of disk is $\frac{1}{2} M R^{2}$ .

Step by step solution

Identification of given data

The mass of disk is $M$ .

The radius of disk is $R$ .

The radius of each ring is $r$ .

The width of each ring is $d r$ .

The mass of each ring is $d m$ .

The moment of inertia of each ring is $I = r^{2} d m$ .

Conceptual Explanation

The moment of inertia of the disk is obtained by calculating mass of each ring then substitute in the formula for mass of each ring.

Determination of moment of inertia of disk

The density of disk is given as:

$ρ = \frac{M}{(π R^{2}) r}$

The volume of each ring is given as:

$V = (2 π r d r) r d V = 2 π r^{2} d r$

The mass of each ring is given as:

$d m = ρ d V d m = (\frac{M}{(π R^{2}) r}) (2 π r^{2} d r) d m = (\frac{2 M}{R^{2}}) (r d r)$

The moment of inertia of disk is calculated as:

$I_{d} = \int_{0}^{R} I I_{d} = \int_{0}^{R} r^{2} d m I_{d} = \int_{0}^{R} r^{2} ((\frac{2 M}{R^{2}}) (r d r)) I_{d} = (\frac{2 M}{R^{2}}) \int_{0}^{R} r^{3} d r$

$I_{d} = (\frac{2 M}{R^{2}}) {(\frac{r^{4}}{4})}_{0}^{R} I_{d} = (\frac{2 M}{R^{2}}) (\frac{R^{4}}{4}) I_{d} = \frac{1}{2} M R^{2}$

Therefore, the moment of inertia of disk is $\frac{1}{2} M R^{2}$ .

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Short Answer

Step by step solution

Identification of given data

Conceptual Explanation

Determination of moment of inertia of disk

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