Chapter 13, Problem 68PQ

(a)

To determine

The total angular momentum of the initial wheel-clay system using estimated values of masses of clay and wheel, the radius of the wheel, and the density of the clay.

Answer to Problem 68PQ

The total angular momentum of the initial wheel-clay system is $5.05\text{kg}\cdot {\text{m}}^2/\text{s}$ .

Explanation of Solution

Let the radius of pottery wheel is $7\text{in}$ , the  approximate mass of the pottery wheel is $25.0\text{kg}$ , and density of clay is $1500\text{kg}/{\text{m}}^3$ . Estimated mass of a clay vase is $2.50\text{kg}$.

It is given that clay is in the approximate shape of a sphere.

Write the expression for the density of the sphere.

  $\rho =\frac{M_{\text{sphere}}}{V_{\text{sphere}}}$

Here, $\rho$ is the density of the sphere, $M_{\text{sphere}}$ is the mass of the sphere and $V_{\text{sphere}}$ is the volume of the sphere.

Rearrange above equation to get expression of volume of sphere.

  $V_{\text{sphere}}=\frac{M_{\text{sphere}}}{\rho }$                                                                                                          (I)

Write the expression for the volume of sphere.

  $V_{\text{sphere}}=\frac{4}{3}\pi R_{\text{sphere}}^3$

Here, $R_{\text{sphere}}$ is the radius of sphere.

Substitute $\frac{4}{3}\pi R_{\text{sphere}}^3$ for $V_{\text{sphere}}$ in equation (I) to modify equation (I).

  $\begin{array}{c}\frac{4}{3}\pi R_{\text{sphere}}^3=\frac{M_{\text{sphere}}}{\rho }\\ R_{\text{sphere}}=\sqrt[3]{\frac{3M_{\text{sphere}}}{4\rho \pi }}\end{array}$                                                                                              (II)

The wheel is in form of a disk. Thus, consider wheel as disk to find rotational inertia.

Write the expression for the total rotational inertia of wheel-clay system.

  $I_{\text{total,}i}=I_{\text{disk}}+I_{\text{sphere}}$                                                                                                   (III)

Here, $I_{\text{total,}i}$ is the initial rotational inertia of wheel-clay system, $I_{\text{disk}}$ is the rotational inertia of the disk and $I_{\text{sphere}}$ is the rotational inertia of disk.

Write the expression for the rotational inertia of disk.

  $I_{\text{disk}}=\frac{1}{2}{M}_{\text{disk}}{R}_{\text{disk}}^2$                                                                                                  (IV)

Here, $M_{\text{disk}}$ is the mass of the disk and $R_{\text{disk}}$ is the radius of the disk.

Write the expression for the rotational inertia of the sphere.

  $I_{\text{sphere}}=\frac{2}{5}{M}_{\text{sphere}}{R}_{\text{sphere}}^2$                                                                                              (V)

Substitute (IV) and (V) in equation (III) to get $I_{\text{total},i}$ .

  $I_{\text{total},i}=\frac{1}{2}{M}_{\text{disk}}{R}_{\text{disk}}^2+\frac{2}{5}{M}_{\text{sphere}}{R}_{\text{sphere}}^2$                                                                        (VI)

Write the expression for the total angular momentum of the initial wheel-clay system.

  $L_i=I_{\text{total,}i}{\omega }_i$                                                                                                            (VII)

Here, $L_i$ is the total angular momentum of the initial wheel-clay system and ${\omega }_i$ is the initial angular momentum of the system.

Conclusion:

It is given that the angular velocity of the system is $120\text{rpm}$.

Convert radius of disk from $7\text{in}$ to meter.

  $\begin{array}{c}{R}_{\text{disk}}=\left(7\text{in}\text{.}\right)\left(\frac{1\text{m}}{39.37\text{in}\text{.}}\right)\\ =0.178\text{m}\end{array}$

Convert angular momentum $120\text{rpm}$ into $\frac{\text{rad}}{\text{s}}$ .

  $\begin{array}{c}{\omega }_i=\left(120\frac{\text{rev}}{\text{min}}\right)\left(\frac{\text{2}\pi \text{rad}}{1\text{rev}}\right)\left(\frac{1\text{min}}{60\text{s}}\right)\\ =12.56\frac{\text{rad}}{\text{s}}\end{array}$

Substitute $2.50\text{kg}$ for $M_{\text{sphere}}$ and $1500\text{kg}/{\text{m}}^3$ for $\rho$ in equation (II) to get $R_{\text{sphere}}$ .

  $\begin{array}{c}{R}_{\text{sphere}}=\sqrt[3]{\frac{3\left(2.50\text{kg}\right)}{4\left(1500\text{kg}/{\text{m}}^3\right)\left(3.14\right)}}\\ =0.074\text{m}\end{array}$

Substitute $0.178\text{m}$ for $R_{\text{disk}}$, $25.0\text{kg}$ for $M_{\text{disk}}$, $2.50\text{kg}$ for $M_{\text{sphere}}$ and $0.074\text{m}$ for $R_{\text{sphere}}$ in equation (VI) to get $I_{\text{total,}i}$ .

  $\begin{array}{c}{I}_{\text{total,}i}=\frac{1}{2}\left(25.0\text{kg}\right){\left(0.178\text{m}\right)}^2+\frac{2}{5}\left(2.50\text{kg}\right){\left(0.074\text{m}\right)}^2\\ =0.401\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $0.401\text{kg}\cdot {\text{m}}^2$ for $I_{\text{total,}i}$ and $12.56\frac{\text{rad}}{\text{s}}$ for ${\omega }_i$ in equation (VII) to get $L{i}_{}$ .

  $\begin{array}{r}{L}_i=\left(0.401\text{kg}\cdot {\text{m}}^2\right)\left(12.56\frac{\text{rad}}{\text{s}}\right)\\ =5.05\text{kg}\cdot {\text{m}}^2/\text{s}\end{array}$

Therefore, the total angular momentum of the initial wheel-clay system is $5.05\text{kg}\cdot {\text{m}}^2/\text{s}$ .

(b)

To determine

Whether the system need to be sped up or slowed down, or should nothing be done to maintain a constant angular speed, if shape of the clay changed into a tall, cylindrical vase.

Answer to Problem 68PQ

According to conservation of angular momentum, system would speed up due to the change of shape of the clay. The speed increases by $1\%$ .

Explanation of Solution

The shape of clay changed from sphere to cylinder. It is given that radius of the cylindrical vase is one fourth of radius of sphere.

Since shape of clay is cylinder, calculate rotational inertia of cylinder to find rotational inertia of clay.

Write the equation for the total rotational inertia of the wheel-clay system after the shape is changed.

  $I_{\text{total,}f}=I_{\text{disk}}+I_{\text{cylinder}}$                                                                                             (VIII)

Here, $I_{\text{total, }f}$ is the rotational inertia of the system after shape of sphere clay changed to cylindrical shape and $I_{\text{cylidner}}$ is the rotational inertia of cylinder about its own axis.

Write he expression for the rotational inertia of cylinder.

  $I_{\text{cylinder}}=\frac{1}{2}{M}_{\text{cylidner}}{R}_{\text{cylinder}}^2$                                                                                         (IX)

Here, $M_{\text{cylinder}}$ is the mass of clay and $R_{\text{cylinder}}$ is the radius of cylindrical vase.

Substitute (IX) and (IV) in (VIII) and to get $I_{\text{total,}f}$ .

  $I_{\text{total,}f}=\frac{1}{2}{M}_{\text{disk}}{R}_{\text{disk}}^2+\frac{1}{2}{M}_{\text{cylinder}}{R}_{\text{cylinder}}^2$                                                                       (X)

Write the expression for the percentage change in rotational inertia of system.

  $\%\text{change}=\frac{I_{\text{total,}f}-I_{\text{total,}i}}{I_{\text{total,}i}}\times 100$                                                                                (XI)

Write the expression for final angular momentum of the system.

  $L_f=I_{\text{total,}}{}_f{\omega }_f$                                                                                                         (XII)

Here, $L_f$ is the final angular momentum and ${\omega }_f$ is the final angular velocity.

Write the expression for the conservation of angular momentum.

  $L_i=L_f$

Substitute (VII) and (XII) in above equation to get ${\omega }_f$ .

  $I_{\text{total,}}{}_i{\omega }_i=I_{\text{total,}}{}_f{\omega }_f$                                                                                                 (XIII)

Conclusion:

Radius of cylinder is ${\frac{1}{4}}^{\text{th}}$ of sphere.

Calculate radius of cylinder.

  $\begin{array}{c}{R}_{\text{cylinder}}=\frac{1}{4}\left(0.074\text{m}\right)\\ =0.018\text{m}\end{array}$

Substitute $0.178\text{m}$ for $R_{\text{disk}}$, $25.0\text{kg}$ for $M_{\text{disk}}$, $2.50\text{kg}$ for $M_{\text{cylinder}}$ and $0.018\text{m}$ for $R_{\text{cylinder}}$ in equation (X) to get $I_{\text{total,}f}$ .

  $\begin{array}{c}{I}_{\text{total,}f}=\frac{1}{2}\left(25.0\text{kg}\right){\left(0.178\text{m}\right)}^2+\frac{2}{5}\left(2.50\text{kg}\right){\left(0.018\text{m}\right)}^2\\ =0.397\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $0.397\text{kg}\cdot {\text{m}}^2$ for $I_{\text{total,}f}$ and $0.401\text{kg}\cdot {\text{m}}^2$ for $I_{\text{total,}i}$ in (XI) to get $\%\text{change}$.

  $\begin{array}{c}\%\text{change}=\frac{0.397\text{kg}\cdot {\text{m}}^2-0.401\text{kg}\cdot {\text{m}}^2}{0.401\text{kg}\cdot {\text{m}}^2}\times 100\\ =-1.0\%\end{array}$

Thus, rotational inertia of the system is decreases by $1.0\%$ so that from conservation of angular momentum given in equation (XIII), angular momentum of the system increases by $1.0\%$ .

Therefore, according to conservation of angular momentum, system would speed up due to the change of the clay. The speed increases by $1\%$ .