Question

Question: (II) A potter’s wheel is rotating around a vertical axis through its center at a frequency of \({\bf{1}}{\bf{.5}}\;{{{\bf{rev}}} \mathord{\left/{\vphantom {{{\bf{rev}}} {\bf{s}}}} \right.} {\bf{s}}}\). The wheel can be considered a uniform disk of mass 5.0 kg and diameter 0.40 m. The potter then throws a 2.6-kg chunk of clay, approximately shaped as a flat disk of radius 7.0 cm, onto the center of the rotating wheel. What is the frequency of the wheel after the clay sticks to it? Ignore friction.

Short answer

The frequency of the wheel after the clay sticks to it is \(1.41\;{{{\rm{rev}}} \mathord{\left/{\vphantom {{{\rm{rev}}} {\rm{s}}}} \right.} {\rm{s}}}\).

Step-by-step solution

Determination of angular momentum

The angular momentum of a rotating object is described as the product of its angular frequency and moment of inertia. The expression for the angular momentum is as follows:

\(L = I\omega \)

Here, \(I\) is the moment of inertia and \(\omega \) is the angular frequency.

Given information

Given data:

The initial angular frequency of the system is\({\omega _{\rm{i}}} = 1.5\;{{{\rm{rev}}} \mathord{\left/{\vphantom {{{\rm{rev}}} {\rm{s}}}} \right.} {\rm{s}}}\).

The mass of the wheel is\({M_{\rm{w}}} = 5.0\;{\rm{kg}}\).

The diameter of the wheel is\({d_{\rm{w}}} = 0.40\;{\rm{m}}\).

The mass of the clay is\({M_{\rm{c}}} = 2.6\;{\rm{kg}}\).

The radius of the clay is \({r_{\rm{c}}} = 7.0\;{\rm{cm}}\).

Calculate the radius of the potter’s wheel

The radius of the potter’s wheel can be calculated as:

\(\begin{aligned}{l}{r_{\rm{w}}} &= \frac{{{d_{\rm{w}}}}}{2}\\{r_{\rm{w}}} &= \frac{{\left( {0.40\;{\rm{m}}} \right)}}{2}\\{r_{\rm{w}}} &= 0.20\;{\rm{m}}\end{aligned}\)

Calculate the final angular frequency

The expression for the moment of inertia of the wheel before the clay is thrown can be written as:

\({I_{\rm{i}}} = \frac{1}{2}{M_{\rm{w}}}r_{\rm{w}}^2\)

The expression for the moment of inertia of the wheel after the clay is thrown can be written as:

\({I_{\rm{f}}} = \frac{1}{2}{M_{\rm{w}}}r_{\rm{w}}^2 + \frac{1}{2}{M_{\rm{c}}}r_{\rm{c}}^2\)

Now, apply the conservation of angular momentum to calculate the final angular frequency.

\(\begin{aligned}{c}{I_{\rm{i}}}{\omega _{\rm{i}}} &= {I_{\rm{f}}}{\omega _{\rm{f}}}\\{\omega _{\rm{f}}} &= \frac{{{I_{\rm{i}}}{\omega _{\rm{i}}}}}{{{I_{\rm{f}}}}}\\{\omega _{\rm{f}}} &= \frac{{\left( {\frac{1}{2}{M_{\rm{w}}}r_{\rm{w}}^2} \right){\omega _i}}}{{\left( {\frac{1}{2}{M_{\rm{w}}}r_{\rm{w}}^2 + \frac{1}{2}{M_{\rm{c}}}r_{\rm{c}}^2} \right)}}\\{\omega _{\rm{f}}} &= \frac{{\left( {{M_{\rm{w}}}r_{\rm{w}}^2} \right){\omega _i}}}{{\left( {\left( {{M_{\rm{w}}}r_{\rm{w}}^2} \right) + \left( {{M_{\rm{c}}}r_{\rm{c}}^2} \right)} \right)}}\end{aligned}\)

Substitute the values in the above expression.

\(\begin{aligned}{c}{\omega _{\rm{f}}} &= \frac{{\left( {\left( {5.0\;{\rm{kg}}} \right){{\left( {0.20\;{\rm{m}}} \right)}^2}} \right)\left( {1.5\;{{{\rm{rev}}} \mathord{\left/{\vphantom {{{\rm{rev}}} {\rm{s}}}} \right.} {\rm{s}}}} \right)}}{{\left( {\left\{ {\left( {5.0\;{\rm{kg}}} \right){{\left( {0.20\;{\rm{m}}} \right)}^2}} \right\} + \left\{ {\left( {2.6\;{\rm{kg}}} \right){{\left\{ {\left( {7.0\;{\rm{cm}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}\;{\rm{m}}}}{{{\rm{1}}\;{\rm{cm}}}}} \right)} \right\}}^2}} \right\}} \right)}}\\{\omega _{\rm{f}}} &= 1.41\;{{{\rm{rev}}} \mathord{\left/{\vphantom {{{\rm{rev}}} {\rm{s}}}} \right.} {\rm{s}}}\end{aligned}\)

Thus, the frequency of the wheel after the clay sticks to it is \(1.41\;{{{\rm{rev}}} \mathord{\left/{\vphantom {{{\rm{rev}}} {\rm{s}}}} \right.} {\rm{s}}}\).