Question

In the given below figure shows an arrangement of $15$ Identical discs that have been glued together in a rod-like shape of length $L = 1.0000 m$ , and (total) mass $M = 100.0 mg$ . The discs are uniform, and the disc arrangement can rotate about a perpendicular axis through its central disc at point $O$ .

$(a)$ What is the rotational inertia of the arrangement about that axis?

$(b)$ If we approximated the arrangement as being a uniform rod of mass $M$ and length $L$ , what percentage error would we make in using the formula used to calculate the rotational inertia?

Accepted Answer

$L = 1.0000 m$ Let us consider three of the discs (starting with the one at the point $O$ ).

$\oplus OO$ .

The first disc (the one at the point $O$ , shown here with the plus sign inside) has rotational inertia,

$I = \frac{1}{2} m R^{2}$ .

The second disc (using the parallel axis theorem) has rotational inertia,

$I = \frac{1}{2} m R^{2} + m h^{2}$ ,

where $h = 2 R$ ,

$I = \frac{1}{2} m R^{2} + m (2 R)^{2}$ .

The third disc has rotational inertia, where $h = 4 R$ ,

$I = \frac{1}{2} m R^{2} + m (4 R)^{2}$ .

If we had considered five of the discs with the one at $O$ in the middle, then the total rotational inertia is
$OO \oplus OO$

$I = 5 (\frac{1}{2} m R^{2}) + 2 (m (2 R)^{2} + m (4 R)^{2})$ .

So, the pattern is now clear, and we can write down the total rotational for the collection of fifteen discs:

$(b)$ .

The generalization to $N$ discs (where $N$ is assumed to be an odd number) is
$I = \frac{1}{6} (2 N^{2} + 1) N m R^{2}$ .

In terms of the total mass $(m = \frac{M}{15})$ , and the total length $(R = \frac{L}{30})$ we obtain,
$I = 0.083519 M L^{2} \approx (0.08352) (0.1000 kg) (1.0000 m)^{2} = 8.352 \times 10^{- 3} kg m^{2} . . . . . . (1)$

The rotational inertia of rod along the axis perpendicular to the length of the rod is,

$I = \frac{1}{12} M L^{2}$

$I = \frac{1}{12} \times (0.1000 kg) \times (1.000 m)^{2}$

$I = 8.333 \times 10^{- 3} kg m^{2} . . . . . . (2)$

Percentage error $= \frac{∆ I}{I_{disc}} = \frac{I_{disc} - I_{rod}}{I_{disc}} = \frac{8.352 \times 10^{- 3} - 8.333 \times 10^{- 3}}{8.352 \times 10^{- 3}} = 0.0022 = 0.22 %$ .

We find our answer to part $(b)$ is $0.22 %$ lower.

Important Questions on Rotation

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