Chapter 10: Q69P (page 291)

In Fig. $10 - 49$ , a small disk of radius $r = 2.00 cm$ has been glued to the edge of a larger disk of radius $R = 4.00 cm$ so that the disks lie in the same plane. The disks can be rotated around a perpendicular axis through point $O$ at the center of the larger disk. The disks both have a uniform density (mass per unit volume) of $1.40 \times 103 kg / m^{2}$ and a uniform thickness of $5.00 mm$ . What is the rotational inertia of the two-disk assembly about the rotation axis through O?

Short Answer

Expert verified

The rotational inertia of the two-disc assembly about the rotation axis through O is

$6.16 \times 10^{- 5} k g m^{2}$

Step by step solution

Given

Thickness of the disc $d = 5.0 \times 10^{- 3} m$
Radius of big disc $R = 0.04 m$
Radius of small disc $r = 0.02 m$
Density of material $ρ = 1400 kg / m^{3}$
Rotational inertia of wheel $I = 0.050 {kgm}^{2}$
Magnitude of force $P = 3.0 N$

Understanding the concept

Use the parallel axis theorem to obtain the rotational inertia of the two-disc assembly.

Formula:

$I = \frac{1}{2} {mR}^{2} + {md}^{2} m = ρ v = ρ π r^{2} d$

Calculate the rotational inertia of the two-disk assembly about the rotation axis through O

For the rotational inertia of the two-disc assembly about the rotation axis through ,

$I = \frac{1}{2} {MR}^{2} + \frac{1}{2} {mr}^{2} + m {(r + R)}^{2} I = \frac{1}{2} ρ π R^{2} {dR}^{2} + \frac{1}{2} ρ π r^{2} {dr}^{2} + ρ π r^{2} d {(r + R)}^{2} I = (ρ π d) [\frac{1}{2} R^{4} + \frac{1}{2} r^{4} + r^{2} {(r + R)}^{2}] I = (1400 k g / m^{3}) π (5.0 \times 10^{- 3} m) [\frac{1}{2} {(0.04 m)}^{4} + \frac{1}{2} {(0.02 m)}^{4} + {(0.02 m)}^{2} {(0.02 m + 0.04 m)}^{2}] I = 6.16 \times 10^{- 5} {kg.m}^{2}$ Therefore, the rotational inertia of the two-disc assembly about the rotation axis through O is $6.16 \times 10^{- 5} k g m^{2}$