Chapter 9, Problem 101P

(a)

To determine

The distance moved by the ball $s_1$ and time taken $t_1$ to cover the distance $s_1$ before it begin to roll without slipping, and the final speed $v_1$ of the ball.

Answer to Problem 101P

  $s_1=\frac{12v_0^2}{49{\mu }_{k}g}$

  $t_1=\frac{2}{7}\frac{v_0}{{\mu }_{k}g}$

  $v_1=\frac{5}{7}{v}_0$

Explanation of Solution

Given: Mass of the ball is $M$ .

Radius of the ball is $R$ .

The coefficient of kinetic friction between the ball and the floor is ${\mu }_k$ .

At the instant the ball touches the floor it is moving horizontally with a speed $v_0$ , that is the initial velocity.

Formula used:

Velocity $v$ of the ball at any instant of time $t$ is given by

  $v=u+at$  $\left(1\right)$

Equation of kinematics

  $v^2=u^2+2as$   $\left(2\right)$

Where, $v$ is final velocity

  $u$ is initial velocity

  $a$ is acceleration

  $s$ is distance travelled.

Calculation:

  

FIGURE: $1$

In the figure $1$ , $Mg$ is the force due to gravity which is acting downward through the center and

N is the normal reaction force which is acting in the upward direction through the point of contact.

Velocity $v_1$ of the ball at any instant of time $t_1$ can be written by using equation $\left(1\right)$ as,

  $v_1=v_0+a{t}_1$   $\left(3\right)$

In this problem there arise a force of kinetic friction, $f_k={\mu }_{k}N$ acting horizontally through the point of contact and is in the direction opposite to the motion of the ball. This forces of kinetic friction produces horizontal acceleration

  $\begin{array}{c}a=\frac{Force}{Mass}\\ =\frac{{\mu }_{k}N}{M}\\ =\frac{{\mu }_{k}Mg}{M}\\ ={\mu }_{k}g\end{array}$

This acceleration is in the direction opposite to the direction of motion of the ball and hence equation $\left(3\right)$ becomes

  $v_1=v_0-{\mu }_{k}g{t}_1$  $\left(4\right)$

The horizontal force $f_k$ also produces a torque about the center of the ball. This torque has a magnitude of $f_{k}R$ , which is acting in anticlockwise direction and produces anticlockwise angular acceleration $\alpha$ in the ball about its center. This can be written as,

  $f_{k}R=I_{cm}\alpha$  $\left(5\right)$

Where, $I_{cm}$ is the moment of inertia about the center of mass of the ball

  $\begin{array}{c}{f}_k={\mu }_{k}N\\ ={\mu }_{k}Mg\end{array}$

  $I_{cm}=\frac{2}{5}M{R}^2$

Thus, the equation $\left(5\right)$ becomes,

  ${\mu }_{k}MgR=\frac{2}{5}M{R}^2\alpha$

Angular acceleration,

  $\alpha =\frac{5}{2}\frac{{\mu }_{k}g}{R}$  $\left(6\right)$

This angular acceleration sets the ball rotating with angular velocity $\omega$ in anticlockwise direction, at any instant of time $t$ , given by,

  $\omega =\alpha t_1$  $\left(7\right)$

When the ball starts rolling, the velocity

  $v_1=\omega R$  $\left(8\right)$

is satisfied.

Substituting equation $\left(8\right)$ in equation $\left(4\right)$ ,

  $\omega R=v_0-{\mu }_{k}g{t}_1$

Substituting for $\omega$ from equation $\left(7\right)$ , the above equation becomes,

  $\alpha tR=v_0-{\mu }_{k}g{t}_1$

Substituting the expression for $\alpha$ from equation $\left(6\right)$ and simplifying,

  $\frac{5}{2}{\mu }_{k}g{t}_1=v_0-{\mu }_{k}g{t}_1$

  $\frac{5}{2}{\mu }_{k}g{t}_1+{\mu }_{k}g{t}_1=v_0$

  $v_0=\frac{7}{2}{\mu }_{k}g{t}_1$  $\left(9\right)$

  $t_1=\frac{2}{7}\frac{v_0}{{\mu }_{k}g}$

Substituting the value of $v_0$ from equation $\left(9\right)$ in equation $\left(4\right)$ and simplifying,

  $\begin{array}{l}{v}_1=\frac{7}{2}{\mu }_{k}g{t}_1-{\mu }_{k}g{t}_1\\ =\frac{5}{2}{\mu }_{k}g{t}_1\\ =\frac{5}{2}{\mu }_{k}g\left(\frac{2}{7}\frac{v_0}{{\mu }_{k}g}\right)\end{array}$

  $v_1=\frac{5}{7}{v}_0$

The distance moved by the ball $s_1$ before it begin to roll without slipping:

Using the equation $\left(2\right)$ ,

  $v_1^2=v_0^2+2a{s}_1$

Substituting the expressions for $v_1$ and $a$ ,

  ${\left(\frac{5v_0}{7}\right)}^2=v_0^2-2{\mu }_{k}g{s}_1$

  $-2{\mu }_{k}g{s}_1={\left(\frac{5v_0}{7}\right)}^2-v_0^2$

  $-2{\mu }_{k}g{s}_1=\frac{25v_0^2}{49}-v_0^2$

  $-2{\mu }_{k}g{s}_1=\frac{24v_0^2}{49}$

  $s_1=\frac{12v_0^2}{49{\mu }_{k}g}$

Conclusion:

The distance moved by the ball before it begin to roll without slipping is $s_1=\frac{12v_0^2}{49{\mu }_{k}g}$ .

Time taken by the ball to cover the distance $s_1$ before it begins to roll without slipping is $t_1=\frac{2}{7}\frac{v_0}{{\mu }_{k}g}$ .

The final speed of the ball is $v_1=\frac{5}{7}{v}_0$ .

(b)

To determine

The ratio of the final kinetic energy to the initial kinetic energy of the ball.

Answer to Problem 101P

  $\frac{K_f}{K_i}=\frac{5}{7}$

Explanation of Solution

Given: Mass of the ball is $M$ ,

Radius of the ball is $R$

The coefficient of kinetic friction between the ball and the floor is ${\mu }_k$

At the instant the ball touches the floor it is moving horizontally with a speed $v_0$ , that is the initial velocity.

Final velocity of the ball is $v_1$

Formula used:

The final kinetic energy of the ball is given by,

  $K_f=K_{trans}+K_{rot}$

  $K_f=\frac{1}{2}M{v}_1^2+\frac{1}{2}I{\omega }^2$  $\left(10\right)$

Where, $I$ is moment of inertia.

Initial kinetic energy of the ball is given by,

  $K_i=\frac{1}{2}M{v}_0^2$  $\left(11\right)$

Calculation:

Moment of inertia of the ball is given by,

  $I=\frac{2}{5}M{R}^2$

For rotational motion of the ball the velocity,

  $v_1=R\omega \Rightarrow \omega =\frac{v_1}{R}$

Substituting for $I$ and $v_1$ in equation $\left(10\right)$ ,

  $K_f=\frac{1}{2}M{v}_1^2+\frac{1}{2}\left(\frac{2}{5}M{R}^2\right){\left(\frac{v_1}{R}\right)}^2$

  $K_f=\frac{1}{2}M{v}_1^2+\frac{1}{5}M{v}_1^2$

  $K_f=\frac{7}{10}M{v}_1^2$

But, $v_1=\frac{5}{7}{v}_0$

Therefore final kinetic energy of the ball is,

  $K_f=\frac{7}{10}M{\left(\frac{5}{7}{v}_0\right)}^2=\frac{5}{14}M{v}_0^2$  $\left(12\right)$

From equations $\left(11\right)$ and $\left(12\right)$ , the ratio of the final kinetic energy to the initial kinetic energy of the ball is,

  $\begin{array}{l}\frac{K_f}{K_i}=\frac{\frac{5}{14}M{v}_0^2}{\frac{1}{2}M{v}_0^2}\\ \end{array}$

  $\frac{K_f}{K_i}=\frac{5}{7}$

Conclusion:

The ratio of the final kinetic energy to the initial kinetic energy of the ball is $\frac{5}{7}$ .

(c)

To determine

The values of $s_1,t_1$ and $v_1$ for $v_0=8.0$ m/s and ${\mu }_k=0.060$ .

Answer to Problem 101P

  $s_1=27$ m

  $t_1=3.9$ s

  $v_1=5.7$ m/s

Explanation of Solution

Given: $v_0=8.0$ m/s and ${\mu }_k=0.060$

Formula used:

  $s_1=\frac{12v_0^2}{49{\mu }_{k}g}$  $\left(13\right)$

  $t_1=\frac{2}{7}\frac{v_0}{{\mu }_{k}g}$  $\left(14\right)$

  $v_1=\frac{5}{7}{v}_0$  $\left(15\right)$

Calculation:

Substituting the numerical values in equation $\left(13\right)$ ,

  $\begin{array}{l}{s}_1=\frac{\text{12}{\left(\text{8}\text{.0 m/s}\right)}^{\text{2}}}{\text{49×0}\text{.060×9}{\text{.8 m/s}}^{\text{2}}}\\ s_1=27\text{m}\end{array}$

Substituting the numerical values in equation $\left(14\right)$ ,

  $\begin{array}{l}{t}_1=\frac{\text{2}}{\text{7}}\frac{\text{8}\text{.0 m/s}}{\text{0}\text{.060×9}{\text{.8m/s}}^{\text{2}}}\\ t_1=\text{3}\text{.9}\text{s}\end{array}$

Substituting the numerical values in equation $\left(15\right)$ ,

  $v_1=\frac{5}{7}\left(\text{8}\text{.0}\text{m/s}\right)\text{=5}.7\text{m/s}$

Conclusion:

The values of $s_1,t_1$ and $v_1$ are $27$ m, $3.9$ s, and $5.7$ m/s respectively.