Chapter 9, Problem 48P

(a)

To determine

Percentage difference between the moment of inertia in two cases.

Explanation of Solution

Given:

Diameter of each sphere $=d_{sphere}=10\text{ cm}$

Radius of sphere $=r_{sphere}=5\text{ cm}$

Length of the rod $=L=30\text{ cm = 0}\text{.30 m}$

Mass of each sphere $=m_{sphere}=500\text{ g = 0}\text{.500 kg}$

Mass of each rod $=m_{rod}=60\text{ g = 0}\text{.060 kg}$

Moment of inertia of each sphere about its center of mass $=I_{cs}$

Moment of inertia of each sphere about axis of rotation $=I_{sphere}$

Moment of inertia of rod about axis of rotation $=I_{rod}$

Distance of center of each sphere from axis of rotation $=d$

Moment of inertia of the system when spheres are treated as particle $=I_{sys}$

Moment of inertia of the system $=I_{sys}$

Percentage difference $=\Delta D$

Formula Used:

Moment of inertia of each sphere about its center of mass is given as

  $I_{cs}=(0.4)m_{sphere}{r}_{sphere}^2$

Moment of inertia of rod about its center of mass is given as

  $I_{rod}=\left(\frac{1}{12}\right)m_{rod}{L}^2$

According to parallel axis theorem, moment of inertia about the axis of rotation is given as

  $I_{axis}=I_{cm}+m{d}^2$

Where, d is the distance between center of mass and axis of rotation.

Percentage difference in the moment of inertia is given as

  $\Delta D=\frac{(I_{sys}^{\text{'}}-I_{sys})}{(I_{sys}^{\text{'}})}\times 100$

Calculation:

Case 1:

Consider the two spheres as point particles and mass of the rod negligible.

Distance of center of each sphere from axis of rotation is given as

  $\begin{array}{l}d=\left(\frac{L}{2}\right)+r_{sphere}\\ d=\left(\frac{30}{2}\right)+5\\ d=20\text{ cm = 0}\text{.20 m}\end{array}$

Moment of inertia of the system is given as

  $\begin{array}{l}I=2m_{sphere}{d}^2\\ I=2(0.500){(}^{0.20}\\ I=0.04{\text{ kgm}}^2\end{array}$

Case 2:

Moment of inertia of each sphere about its center of mass is given as

  $I_{cs}=(0.4)m_{sphere}{r}_{sphere}^2$

Using parallel axis theorem, moment of inertia of each sphere about axis of rotation is given as

  $\begin{array}{l}{I}_{as}=I_{cs}+m_{sphere}{d}^2\\ I_{as}=(0.4)m_{sphere}{r}_{sphere}^2+m_{sphere}{d}^2\\ I_{as}=(0.4)(0.500){(}^{0.05}+(0.500){(}^{0.20}\\ I_{as}=0.0205{\text{ kgm}}^2\end{array}$

Moment of inertia of rod about its center of mass is given as

  $\begin{array}{l}{I}_{rod}=\left(\frac{1}{12}\right)m_{rod}{L}^2\\ I_{rod}=\left(\frac{1}{12}\right)(0.060){(}^{0.30}\\ I_{rod}=0.00045{\text{ kgm}}^2\end{array}$

Moment of inertia of the system is given as

  $\begin{array}{l}{I}_{sys}^{\text{'}}=2I_{as}+I_{rod}\\ I_{sys}^{\text{'}}=2(0.0205)+(0.00045)\\ I_{sys}^{\text{'}}=0.04145{\text{ kgm}}^2\end{array}$

Percentage difference in the moment of inertia is given as

  $\begin{array}{l}\Delta D=\frac{(0.04145-0.04)}{(0.04145)}\times 100\\ \Delta D=3.5\%\end{array}$

Conclusion:

Hence, Percentage difference in the moment of inertia is $3.5\%$ .

(b)

To determine

The moment of inertia of system will change if solid sphere is replaced with hollow shell.

Explanation of Solution

Given:

Diameter of each sphere $=d_{sphere}=10\text{ cm}$

Radius of sphere $=r_{sphere}=5\text{ cm}$

Length of the rod $=L=30\text{ cm = 0}\text{.30 m}$

Mass of each sphere $=m_{sphere}=500\text{ g = 0}\text{.500 kg}$

Mass of each rod $=m_{rod}=60\text{ g = 0}\text{.060 kg}$

Moment of inertia of each sphere about its center of mass $=I_{cs}$

Moment of inertia of each sphere about axis of rotation $=I_{sphere}$

Moment of inertia of rod about axis of rotation $=I_{rod}$

Distance of center of each sphere from axis of rotation $=d$

Moment of inertia of the system $=I_{sys}^{\text{'}}$

Formula Used:

Moment of inertia of each sphere about its center of mass is given as

  $I_{cs}=(0.4)m_{sphere}{r}_{sphere}^2$

Moment of inertia of rod about its center of mass is given as

  $I_{rod}=\left(\frac{1}{12}\right)m_{rod}{L}^2$

According to parallel axis theorem, Moment of inertia about the axis of rotation is given as

  $I_{axis}=I_{cm}+m{d}^2$

Where, d is the distance between center of mass and axis of rotation.

Calculation:

Moment of inertia of each hollow sphere about its center of mass is given as

  $I_{cs}=(0.67)m_{sphere}{r}_{sphere}^2$

Using parallel axis theorem, moment of inertia of each hollow sphere about axis of rotation is given as

  $\begin{array}{l}{I}_{as}=I_{cs}+m_{sphere}{d}^2\\ I_{as}=(0.67)m_{sphere}{r}_{sphere}^2+m_{sphere}{d}^2\\ I_{as}=(0.67)(0.500){(}^{0.05}+(0.500){(}^{0.20}\\ I_{as}=0.0208{\text{ kgm}}^2\end{array}$

Moment of inertia of rod about its center of mass is given as

  $\begin{array}{l}{I}_{rod}=\left(\frac{1}{12}\right)m_{rod}{L}^2\\ I_{rod}=\left(\frac{1}{12}\right)(0.060){(}^{0.30}\\ I_{rod}=0.00045{\text{ kgm}}^2\end{array}$

Moment of inertia of the system is given as

  $\begin{array}{l}{I}_{sys}^{\text{'}}=2I_{as}+I_{rod}\\ I_{sys}^{\text{'}}=2(0.0208)+(0.00045)\\ I_{sys}^{\text{'}}=0.04205{\text{ kgm}}^2\end{array}$

Conclusion:

Hence, the moment of inertia of the system increases.