10.5 Calculating moments of inertia  (Page 4/9)

Person on a merry-go-round

A 25-kg child stands at a distance $r=1.0\text{m}$ from the axis of a rotating merry-go-round ( [link] ). The merry-go-round can be approximated as a uniform solid disk with a mass of 500 kg and a radius of 2.0 m. Find the moment of inertia of this system.

Strategy

This problem involves the calculation of a moment of inertia. We are given the mass and distance to the axis of rotation of the child as well as the mass and radius of the merry-go-round. Since the mass and size of the child are much smaller than the merry-go-round, we can approximate the child as a point mass. The notation we use is $m_{\text{c}}=25\text{kg},r_{\text{c}}=1.0\text{m},m_{\text{m}}=500\text{kg},r_{\text{m}}=2.0\text{m}$ .

Our goal is to find $I_{\text{total}}={\displaystyle \sum _i{I}_i}$ .

Solution

For the child, $I_{\text{c}}=m_{\text{c}}{r}^2$ , and for the merry-go-round, $I_{\text{m}}=\frac{1}{2}{m}_{\text{m}}{r}^2$ . Therefore

Significance

The value should be close to the moment of inertia of the merry-go-round by itself because it has much more mass distributed away from the axis than the child does.

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Rod and solid sphere

Find the moment of inertia of the rod and solid sphere combination about the two axes as shown below. The rod has length 0.5 m and mass 2.0 kg. The radius of the sphere is 20.0 cm and has mass 1.0 kg.

Strategy

Since we have a compound object in both cases, we can use the parallel-axis theorem to find the moment of inertia about each axis. In (a), the center of mass of the sphere is located at a distance $L+R$ from the axis of rotation. In (b), the center of mass of the sphere is located a distance R from the axis of rotation. In both cases, the moment of inertia of the rod is about an axis at one end. Refer to [link] for the moments of inertia for the individual objects.

1. $I_{\text{total}}={\displaystyle \sum _i{I}_i}=I_{\text{Rod}}+I_{\text{Sphere}}$ ;
   $I_{\text{Sphere}}=I_{\text{center of mass}}+m_{\text{Sphere}}{(L+R)}^2=\frac{2}{5}{m}_{\text{Sphere}}{R}^2+m_{\text{Sphere}}{(L+R)}^2$ ;
   $I_{\text{total}}=I_{\text{Rod}}+I_{\text{Sphere}}=\frac{1}{3}{m}_{\text{Rod}}{L}^2+\frac{2}{5}{m}_{\text{Sphere}}{R}^2+m_{\text{Sphere}}{(L+R)}^2;$
   $I_{\text{total}}=\frac{1}{3}(2.0\text{kg}){(0.5\text{m})}^2+\frac{2}{5}(1.0\text{kg})(0.2{\text{m})}^2+(1.0\text{kg}){(0.5\text{m}+0.2\text{m})}^2;$
   $I_{\text{total}}=(0.167+0.016+0.490)\text{kg}·{\text{m}}^2=0.673\text{kg}·{\text{m}}^2.$
   
2. $I_{\text{Sphere}}=\frac{2}{5}{m}_{\text{Sphere}}{R}^2+m_{\text{Sphere}}{R}^2$ ;
   $I_{\text{total}}=I_{\text{Rod}}+I_{\text{Sphere}}=\frac{1}{3}{m}_{\text{Rod}}{L}^2+\frac{2}{5}{m}_{\text{Sphere}}{R}^2+m_{\text{Sphere}}{R}^2$ ;
   $I_{\text{total}}=\frac{1}{3}(2.0\text{kg}){(0.5\text{m})}^2+\frac{2}{5}(1.0\text{kg})(0.2{\text{m})}^2+(1.0\text{kg}){(0.2\text{m})}^2$ ;
   $I_{\text{total}}=(0.167+0.016+0.04)\text{kg}·{\text{m}}^2=0.223\text{kg}·{\text{m}}^2.$

Significance

Using the parallel-axis theorem eases the computation of the moment of inertia of compound objects. We see that the moment of inertia is greater in (a) than (b). This is because the axis of rotation is closer to the center of mass of the system in (b). The simple analogy is that of a rod. The moment of inertia about one end is $\frac{1}{3}m{L}^2$ , but the moment of inertia through the center of mass along its length is $\frac{1}{12}m{L}^2$ .

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