Chapter 8, Problem 72GP

(a)

To determine

To find: The linear speed of the yo-yo when it reaches the end of its $1.0\text{m}$ long string by using the law of conservation of energy.

Answer to Problem 72GP

The linear speed is $3.15\text{m/s}$ .

Explanation of Solution

Given data:

Mass of each disc, $m_1+m_2=m=0.05\text{kg}+0.05\text{kg}=0.1\text{kg}$

Diameter of disc, $d=0.075\text{m}$

Radius of the disc, $r=\frac{0.075}{2}=0.035\text{m}$

Mass of cylindrical hub, $m_3=0.0050\text{kg}$

The total mass of yo-yo $M=m_1+m_2+m_3=0.1+0.005=0.105\text{kg}$

Length of the string, $l=1\text{m}$

Formula used:

According to law of conservation of energy,

Gravitational Potential energy = Rotational K.E + Linear K.E

  $\begin{array}{l}2Mgh=M{v}^2+m{v}^2+\left(\frac{m_3{v}^2}{2}\right)\\ 2Mgh=v^2(M+m+\frac{m_3}{2})\end{array}$

Here,

  $v$ is the linear velocity,

  $M$ is the total mass,

  $h$ is the height,

  $m_1,m_2$ is the mass of disc,

  $m_3$ is the mass of cylindrical hub.

Calculation:

According to law of conservation of energy,

Gravitational Potential energy = Rotational K.E + Linear K.E

  $Mgh=\frac{1}{2}M{v}^2+\left(\frac{1}{2}{I}_1{\omega }^2+\frac{1}{2}{I}_2{\omega }^2\right)\left(\frac{1}{2}{I}_3{\omega }^2\right)$ ,

  $m_1=m_{{}_2};\therefore I_1=I_2=I$ ,

  $I=m{r}^{{}^2}$ ,

  $2Mgh=M{v}^2+m{r}^2{\left(\frac{v}{r}\right)}^2+\left(\frac{m_3{r}^2}{2}\right){\left(\frac{v}{r}\right)}^2$

  $\begin{array}{l}2Mgh=M{v}^2+m{v}^2+\left(\frac{m_3{v}^2}{2}\right)\\ 2Mgh=v^2(M+m+\frac{m_3}{2})\end{array}$

Now substitute the values in the equation,

  $2\times 0.105\text{ kg}\times \text{9}{\text{.81 m/s}}^2\times 1\text{ m = }{v}^2\left(0.105\text{ kg}+0.1\text{ kg+}\frac{0.005}{2}\right)$ ,

  $\begin{array}{l}2.0601=v^2(0.2075)\\ v^2=\frac{2.0601}{0.2075}\\ v=3.15\text{ m/s}\end{array}$

The linear speed of yo-yo is $3.15\text{m/s}$ .

Conclusion: 

From law of conservation of energy, the linear speed is $3.15\text{m/s}$ .

(b)

To determine

To Find: The fraction of rotational kinetic energy.

Answer to Problem 72GP

The fraction is $0.658$ .

Explanation of Solution

Given data:

Mass of each disc $m_1+m_2=m=0.05\text{kg}+0.05\text{kg}=0.1\text{kg}$

Diameter of disc $d=0.075\text{m}$

Radius of the disc $r=\frac{0.075}{2}=0.035\text{m}$

Mass of cylindrical hub $m_3=0.0050\text{kg}$

The total mass of yo-yo $M=m_1+m_2+m_3=0.1+0.005=0.105\text{kg}$

Length of the string $l=1\text{m}$

Formula used:

  $\left(\frac{K{E}_{rotational}}{K{E}_{total}}\right)=\frac{m{r}^2{\left(\frac{v}{r}\right)}^2+\left(\frac{m_3{r}_3{}^2}{4}\right){\left(\frac{v}{r_3}\right)}^2}{\frac{1}{2}M{v}^2+m{v}^2+\left(\frac{m_3{v}^2}{4}\right)}$

  $\begin{array}{l}\\ \left(\frac{K{E}_{rotational}}{K{E}_{total}}\right)=\frac{\left(m+\frac{m_3}{4}\right)}{\left(\frac{1}{2}M+m+\frac{m_3}{4}\right)}\end{array}$

Here,

  $K{E}_{rotational}$ is the kinetic energy due to rotation.

  $K{E}_{total}$ is the total kinetic energy.

  $v$ is the linear velocity.

  $M$ is the total mass.

  $m$ is the mass of disc.

Calculation:

For the ratio of kinetic energy is rotational to the total kinetic energy.

  $\left(\frac{K{E}_{rotational}}{K{E}_{total}}\right)=\frac{m{r}^2{\left(\frac{v}{r}\right)}^2+\left(\frac{m_3{r}_3{}^2}{4}\right){\left(\frac{v}{r_3}\right)}^2}{\frac{1}{2}M{v}^2+m{v}^2+\left(\frac{m_3{v}^2}{4}\right)}$

  $\begin{array}{l}\left(\frac{K{E}_{rotational}}{K{E}_{total}}\right)=\frac{m{v}^2+\left(\frac{m_3{v}^2}{4}\right)}{\frac{1}{2}M{v}^2+m{v}^2+\left(\frac{m_3{v}^2}{4}\right)}\\ \left(\frac{K{E}_{rotational}}{K{E}_{total}}\right)=\frac{v^2+\left(m+\frac{m_3}{4}\right)}{v^2\left(\frac{1}{2}M+m+\frac{m_3}{4}\right)}\end{array}$

  $\begin{array}{l}\\ \left(\frac{K{E}_{rotational}}{K{E}_{total}}\right)=\frac{\left(m+\frac{m_3}{4}\right)}{\left(\frac{1}{2}M+m+\frac{m_3}{4}\right)}\end{array}$

Substituting the values in the equation,

  $\begin{array}{l}\\ \left(\frac{K{E}_{rotational}}{K{E}_{total}}\right)=\frac{\left(0.1+\frac{0.005}{4}\right)}{\left(\frac{1}{2}\times 0.105+0.1+\frac{0.005}{4}\right)}\end{array}$

  $\begin{array}{l}\\ \left(\frac{K{E}_{rotational}}{K{E}_{total}}\right)=\frac{0.10125}{0.15375}\end{array}$

  $\begin{array}{l}\\ \left(\frac{K{E}_{rotational}}{K{E}_{total}}\right)=0.658\end{array}$

Conclusion: 

The fractional kinetic energy is $0.658$ .