Chapter 9, Problem 104P

(a)

To determine

The speed of the ball just after impact.

Answer to Problem 104P

  $v_0=200$ m/s

Explanation of Solution

Given:

Mass of a uniform solid ball = $20$ g = $0.02$ kg

Radius of the ball = $5$ cm = $0.05$ m

The height above the horizontal surface at which the force is applied on the ball, $h=9$ cm

During the impact, the force (F) increases from $0.0$ N to $40.0$ kN in time (t) = $1.0\times {10}^{-4}$ s and then the force decreases linearly to $0.0$ N in time (t) = $1.0\times {10}^{-4}$ s

Therefore, average force, $F_{av}=20.0$ kN and elapsed time $\Delta t=2.0\times {10}^{-4}$ s.

Formula used:

Applying impulse-momentum theorem to the ball,

  $\begin{array}{c}I=F_{av}\Delta t\\ =\Delta p\\ =m{v}_0\end{array}$

  $\Rightarrow m{v}_0=F_{av}\Delta t$

  $v_0=\frac{F_{av}\Delta t}{m}$  $\left(1\right)$

Where $\Delta p$ is momentum, and $v_0$ is speed of the ball just after the impact.

Calculation:

  

FIGURE: $1$

Substituting numerical values in equation $\left(1\right)$ ,

  $\begin{array}{c}{v}_0=\frac{{\text{20×10}}^{\text{3}}{\text{N×2×10}}^{\text{-4}}\text{s}}{\text{0}\text{.02}\text{kg}}\\ =200\text{m/s}\end{array}$

Conclusion:

The speed of the ball just after impact is $200$ m/s.

(b)

To determine

The angular speed of the ball after impact.

Answer to Problem 104P

  ${\omega }_0=8000$ rad/s

Explanation of Solution

Given:

Mass of a uniform solid ball = $20$ g = $0.02$ kg

Radius of the ball = $5$ cm = $0.05$ m

The height above the horizontal surface at which the force is applied on the ball, $h=9$ cm = $0.09$ m

During the impact, the force (F) increases from $0.0$ N to $40.0$ kN in time (t) = $1.0\times {10}^{-4}$ s and then the force decreases linearly to $0.0$ N in time (t) = $1.0\times {10}^{-4}$ s

Therefore, average force, $F_{av}=20.0$ kN and elapsed time $\Delta t=2.0\times {10}^{-4}$ s.

Formula used:

Applying Newton’s second law in rotational form to ball,

  $F_{av}\left(h-r\right)=I_{cm}\alpha =I_{cm}\frac{{\omega }_0}{\Delta t}$

  $\Rightarrow F_{av}\left(h-r\right)=I_{cm}\frac{{\omega }_0}{\Delta t}$

  ${\omega }_0=\frac{F_{av}\left(h-r\right)\Delta t}{I_{cm}}$  $\left(2\right)$

Where,

  $F_{av}$ is the average force on the ball

  $h$ is the height above the horizontal surface at which the force is applied on the ball

  $r$ is radius of the ball

  $I_{cm}$ is the moment of inertia with respect to an axis through the center of mass of the ball

  $\alpha$ is angular acceleration of the ball

  ${\omega }_0$ is angular speed of the ball after impact

  $\Delta t$ is elapsed time

Moment of inertia with respect to an axis through the center of mass of the ball is

  $I_{cm}=\frac{2}{5}m{r}^2$

Substituting this in equation $\left(2\right)$ ,

  ${\omega }_0=\frac{F_{av}\left(h-r\right)\Delta t}{\frac{2}{5}m{r}^2}$  $\left(3\right)$

From equation $\left(1\right)$ , $v_0=\frac{F_{av}\Delta t}{m}\Rightarrow \Delta t=\frac{m{v}_0}{F_{av}}$

Substituting the expression for $\Delta t$ in equation $\left(3\right)$ ,

  $\begin{array}{c}{\omega }_0=\frac{F_{av}\left(h-r\right)\frac{m{v}_0}{F_{av}}}{\frac{2}{5}m{r}^2}\\ =\frac{5v_0\left(h-r\right)}{2r^2}\end{array}$  $\left(4\right)$

Calculation:

  

FIGURE: 2

Substituting the numerical values in equation $\left(4\right)$ ,

  ${\omega }_0=\frac{\text{5}\left(\text{200 m/s}\right)\left(\text{0}\text{.090 m-0}\text{.050 m}\right)}{\text{2}{\left(\text{0}\text{.050 m}\right)}^{\text{2}}}$

  $=\frac{\text{1000 m/s×0}\text{.04 m}}{{\text{5×10}}^{\text{-3}}{\text{m}}^{\text{2}}}\text{=8000}\text{rad/s}$

Conclusion:

The angular speed of the ball after impact is $8000$ rad/s.

(c)

To determine

The speed of the ball when it begins to roll without slipping.

Answer to Problem 104P

  $\text{257 m/s}$

Explanation of Solution

Given:

Mass of a uniform solid ball = $20$ g = $0.02$ kg

Radius of the ball = $5$ cm = $0.05$ m

The height above the horizontal surface at which the force is applied on the ball, $h=9$ cm = $0.09$ m

During the impact, the force (F) increases from $0.0$ N to $40.0$ kN in time (t) = $1.0\times {10}^{-4}$ s and then the force decreases linearly to $0.0$ N in time (t) = $1.0\times {10}^{-4}$ s

Therefore, average force, $F_{av}=20.0$ kN and elapsed time $\Delta t=2.0\times {10}^{-4}$ s

Coefficient of kinetic friction, ${\mu }_k=0.5$

Formula used:

Constant acceleration equation that relates the speed of the ball to the acceleration and time,

  $v_{cm,x}=v_0+a_{cm,x}t$  $\left(5\right)$

Where,

  $v_{cm,x}$ is speed of the center of mass of the ball in x-direction

  $v_0$ is speed of the ball just after impact

  $a_{cm,x}$ is the acceleration the center of mass of the ball in x-direction

  $t$ is time

Referring to the force diagram shown in figure 3, applying Newton’s second law to the ball,

  $\Sigma F_x=f_k=m{a}_{cm,x}$  $\left(6\right)$

  $\Sigma F_y=F_n-mg=0$  $\left(7\right)$

And

  $\Sigma {\tau }_{cm}=-f_{k}r=I_{cm}\alpha$  $\left(8\right)$

Where,

  $F_x$ is force in the x-direction

  $f_k$ is force of friction in x-direction

  $F_y$ is force in the y-direction

  $f_n$ is force of friction in y-direction

  $m$ is mass of the ball

  $a_{cm,x}$ is the acceleration the center of mass of the ball in x-direction

  $g$ is acceleration due to gravity

  $r$ is radius of the ball

  ${\tau }_{cm}$ is torque about the center of mass of the ball

  $I_{cm}$ is the moment of inertia with respect to an axis through the center of mass of the ball

  $\alpha$ is angular acceleration of the ball

  $g$ is acceleration due to gravity, $g=9.8{\text{m/s}}^{\text{2}}$

But, $f_k={\mu }_k{F}_n$  $\left(9\right)$

Where, ${\mu }_k$ is coefficient of kinetic friction

From equation $\left(7\right)$ , $F_n=mg$

Substituting this in equation $\left(9\right)$ ,

  $f_k={\mu }_{k}mg$

Substituting the expression for $f_k$ in equation $\left(6\right)$ ,

  ${\mu }_{k}mg=m{a}_{cm,x}\Rightarrow a_{cm,x}={\mu }_{k}g$

Substituting $a_{cm,x}$ value in equation $\left(5\right)$ ,

  $v_{cm,x}=v_0+{\mu }_{k}gt$  $\left(10\right)$

From equation $\left(8\right)$ angular acceleration,

  $\alpha =\frac{-f_{k}r}{I_{cm}}$

Substituting for $f_k$ and $I_{cm}$ ,

  $\alpha =\frac{-{\mu }_{k}mgr}{\frac{2}{5}m{r}^2}=\frac{-5{\mu }_{k}g}{2r}$

Now let us write constant-acceleration equation that connects angular speed of the ball to the angular acceleration and time,

  $\omega ={\omega }_0+\alpha t={\omega }_0-\frac{5{\mu }_{k}g}{2r}t$  $\left(11\right)$

When the ball rolls without slipping $v_{cm,x}$ can be written as,

  $v_{cm,x}=r\omega$

From equation $\left(11\right)$ , $\omega ={\omega }_0-\frac{5{\mu }_{k}g}{2r}t$

Hence, $v_{cm,x}=r\left[{\omega }_0-\frac{5{\mu }_{k}g}{2r}t\right]$

  $v_{cm,x}=r{\omega }_0-\frac{5{\mu }_{k}g}{2}t$  $\left(12\right)$

Now equating the expressions $\left(10\right)$ and $\left(12\right)$ ,

  $v_0+{\mu }_{k}gt=r{\omega }_0-\frac{5{\mu }_{k}g}{2}t$

On rearranging,

  ${\mu }_{k}gt+\frac{5{\mu }_{k}g}{2}t=r{\omega }_0-v_0$

  $\frac{7{\mu }_{k}gt}{2}=r{\omega }_0-v_0\Rightarrow 7{\mu }_{k}gt=2\left(r{\omega }_0-v_0\right)$

  $\Rightarrow t=\frac{2\left(r{\omega }_0-v_0\right)}{7{\mu }_{k}g}$  $\left(13\right)$

Calculation:

  

FIGURE:3

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

  $\begin{array}{l}t=\frac{\text{2}\left[\left(\text{0}\text{.05 m}\right)\left(\text{8000 rad/s}\right)\text{-200 m/s}\right]}{\text{7×0}\text{.5×9}{\text{.8 m/s}}^{\text{2}}}\\ t=1\text{1}\text{.66 s}\end{array}$

Substituting the numerical values in equation $\left(10\right)$ and evaluate for $v\left(\text{11}\text{.66 s}\right)$ ,

  $\begin{array}{c}v=\text{200 m/s + 0}\text{.5×9}{\text{.8 m/s}}^{\text{2}}\text{×11}\text{.66 s}\\ \text{=200 m/s + 56}\text{.84 m/s}\\ \text{=257 m/s}\end{array}$

Conclusion:

The speed of the ball when it begins to roll without slipping is $\text{257 m/s}$ .

(d)

To determine

The distance travelled by the ball along the surface before it begins to roll without slipping.

Answer to Problem 104P

  $\text{2}\text{.65}\text{km}$

Explanation of Solution

Given: Coefficient of kinetic friction, ${\mu }_k=0.5$

Formula used:

The distance travelled by the ball in time $t$ is,

  $\Delta x=v_{0}t+\frac{1}{2}{a}_{cm,x}{t}^2$

Since, $a_{cm,x}={\mu }_{k}g$

  $\Delta x=v_{0}t+\frac{1}{2}{\mu }_{k}g{t}^2$  $\left(14\right)$

Where, $v_0$ is speed of the ball just after impact

  $a_{cm,x}$ is the acceleration the center of mass of the ball in x-direction

  $t$ is time

  ${\mu }_k$ is coefficient of kinetic friction

  $g$ is acceleration due to gravity, $g=9.8{\text{m/s}}^{\text{2}}$

Calculation:

  

FIGURE: 4

From the part $\left(a\right)$ and $\left(c\right)$ ,

  $v_0=200$ m/s and $t=11.66$ s

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

  $\begin{array}{c}\Delta x=\left(\text{200 m/s×11}\text{.66 s}\right)\text{+}\frac{\text{1}}{\text{2}}\text{×0}\text{.5×9}{\text{.8 m/s}}^{\text{2}}{\left(\text{11}\text{.66 s}\right)}^{\text{2}}\\ =2649.6\text{m}\\ \text{=2}\text{.65}\text{km}\end{array}$

Conclusion:

The distance travelled by the ball along the surface before it begins to roll without slipping is $\text{2}\text{.65}\text{km}$ .