(a) A yo-yo is made of two solid cylindrical disks, each of mass 0.050 kg and diameter 0.075 m, joined by a (concentric) thin solid cylindrical hub of mass 0.0050 kg and diameter 0.013 m. Use conservation of energy to calculate the linear speed of the yo-yo just before it reaches the end of its 1.0-m-long string, if it is released from rest. (b) What fraction of its kinetic energy is rotational?

The rotational inertia of a particle is called its moment of inertia. A rotating particle consists of many particles located at different distances from the axis of rotation. The sum of the mass of each particle (m) multiplied by the square of their distances (r) from the axis of rotation gives the moment of inertia (I), i.e.,

\(I = \sum {m{r^2}} \)

In this problem,there are two solid cylindrical disks and one solid cylindrical hub. The moment of inertia of a solid cylinder of mass m and radius r rotating about its axis is \(I = \frac{1}{2}m{r^2}\).

The mass of each solid cylindrical disk is\(M = 0.050\;{\rm{kg}}\).

The diameter of each solid cylindrical disk is\(D = 0.075\;{\rm{m}}\). So, the radius of the disk is:

\(R = \frac{{0.075}}{2} = 0.0375\)

The mass of the thin solid cylindrical hub is\(m = 0.0050\;{\rm{kg}}\).

The diameter of the thin solid cylindrical hub is 0.013 m. So, the radius of the hub is:

\(r = \frac{{0.013\;{\rm{m}}}}{2} = 0.0065\;{\rm{m}}\)

The height covered by the yo-yo is equal to the length of the string, i.e., \(h = 1.0\;{\rm{m}}\).

Since the yo-yo consists of two disks and one hub,the total mass of the yo-yo is:\(\begin{aligned}{c}m' = m + 2M\\ = 0.0050\;{\rm{kg}} + 2\left( {0.050\;{\rm{kg}}} \right)\\ = 0.105\;{\rm{kg}}\end{aligned}\)

The final linear speed of the yo-yo just before it reaches the end of the string is equal to the speed of the center mass of the yo-yo. Let it be \({v_{{\rm{cm}}}}\).

The moment of inertia of the center of mass of the yo-yo is:

\(\begin{aligned}{c}I = {I_{hub}} + 2{I_{{\rm{disk}}}}\\ = \frac{1}{2}m{r^2} + 2\left( {\frac{1}{2}M{R^2}} \right)\\ = \frac{1}{2}\left( {0.0050\;{\rm{kg}}} \right){\left( {0.0065\;{\rm{m}}} \right)^2} + 2\left[ {\frac{1}{2}\left( {0.050\;{\rm{kg}}} \right){{\left( {0.0375\;{\rm{m}}} \right)}^2}} \right]\\ = 7.04 \times {10^{ - 5}}\;{\rm{kg}} \cdot {{\rm{m}}^2}\end{aligned}\)

Theangular velocity \(\omega \) of the yo-yo is related to the linear speed of the yo-yo by the following relation:

\({v_{{\rm{cm}}}} = r\omega \)

Since the yo-yo covers height hduring its motion, the change in its potential energy is\(\Delta PE = m'gh\).

When the yo-yo rolls about the string, it possesses both rotational and translational motions. Thus, the final kinetic energy of the yo-yo is the sum of rotational kinetic energy and translational kinetic energy.

\(\begin{aligned}{c}KE = K{E_{{\rm{rotational}}}} + K{E_{{\rm{translational}}}}\\ = \frac{1}{2}I{\omega ^2} + \frac{1}{2}m'v_{{\rm{cm}}}^2\\ = \frac{1}{2}I\frac{{v_{{\rm{cm}}}^2}}{{{r^2}}} + \frac{1}{2}m'v_{{\rm{cm}}}^2\\ = \frac{{v_{{\rm{cm}}}^2}}{2}\left( {\frac{I}{{{r^2}}} + m'} \right)\end{aligned}\)

Since the yo-yo is released from the rest, its initial kinetic energy will be zero. So, the change in kinetic energy will be equal to the final kinetic energy of the yo-yo, i.e.,

\(\Delta KE = \frac{{v_{{\rm{cm}}}^2}}{2}\left( {\frac{I}{{{r^2}}} + m'} \right)\)

According to the law of conservation of energy, the change in the potential energy of the yo-yo will be equal to the sum of the change in total kinetic energy of the yo-yo, i.e.,

\(\begin{aligned}{c}\Delta PE = \Delta KE\\m'gh = \frac{{v_{{\rm{cm}}}^2}}{2}\left( {\frac{I}{{{r^2}}} + m'} \right)\end{aligned}\) ... (i)

Thus, the velocity of the center of mass of the yo-yo can be written as:

\(\begin{aligned}{c}v_{{\rm{cm}}}^2 = \frac{{2m'gh}}{{\left( {\frac{I}{{{r^2}}} + m'} \right)}}\\ = \frac{{2\left( {0.105\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {1.0\;{\rm{m}}} \right)}}{{\left( {\frac{{7.04 \times {{10}^{ - 5}}\;{\rm{kg}} \cdot {{\rm{m}}^2}}}{{{{\left( {0.0065\;{\rm{m}}} \right)}^2}}} + 0.105\;{\rm{kg}}} \right)}}\\ = 1.16\;{{\rm{m}}^2}{\rm{/}}{{\rm{s}}^2}\\{v_{{\rm{cm}}}} = 1.1\;{\rm{m/s}}\end{aligned}\)

Thus, the speed of the yo-yo just before it reaches the end of the string is 1.1 m/s.

The fraction of kinetic energy, which is rotational, can be calculated by dividing the rotational kinetic energy by the total kinetic energy.

\(\begin{aligned}{c}\frac{{K{E_{{\rm{rotational}}}}}}{{KE}} = \frac{{\frac{1}{2}I{\omega ^2}}}{{\frac{{v_{{\rm{cm}}}^2}}{2}\left( {\frac{I}{{{r^2}}} + m'} \right)}}\\ = \frac{{\frac{1}{2}I\frac{{v_{{\rm{cm}}}^2}}{{{r^2}}}}}{{\frac{{v_{{\rm{cm}}}^2}}{2}\left( {\frac{I}{{{r^2}}} + m'} \right)}}\end{aligned}\)

Using equation (i) in the above expression, you will get:

\(\begin{aligned}{c}\frac{{K{E_{{\rm{rotational}}}}}}{{KE}} = \frac{{\frac{1}{2}I\frac{{v_{{\rm{cm}}}^2}}{{{r^2}}}}}{{m'gh}}\\ = \frac{{\frac{1}{2}\frac{{\left( {7.04 \times {{10}^{ - 5}}\;{\rm{kg}} \cdot {{\rm{m}}^2}} \right) \times \left( {1.1\;{\rm{m/s}}} \right)}}{{{{\left( {0.0065\;{\rm{m}}} \right)}^2}}}}}{{\left( {0.105\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {1.0\;{\rm{m}}} \right)}}\\ = 0.94\end{aligned}\)

Thus, the fraction of kinetic energy which is rotational is 0.94.