Question

A bowling ball encounters a 0.760 m vertical rise on the way back to the ball rack, as the drawing illustrates. Ignore frictional losses and assume that the mass of the ball is distributed uniformly. The translational speed of the ball is 5.10 m/s at the bottom of the rise. Find the translational speed at the top.
1 m/s

Answer 1

Concepts and reasons

The main concept of this question is conservation of energy.

Initially, find the expression for initial and final kinetic energy of the ball. After that apply the law of conservation of the energy and solve for the transitional speed at the top.

Fundamentals

The kinetic energy ${K_t}$ of an object moving with velocity $v$ is given as follows:

${K_t} = \frac{1}{2}m{v^2}$

Here, $m$ is the mass of the object.

The rotational kinetic energy ${K_r}$ of an object rotating with angular velocity $\omega$ is given by following expression:

${K_r} = \frac{1}{2}I{\omega ^2}$

Here, $I$is the moment of inertia of the object.

The relation between rotational speed and linear speed is given as follows:

$v = r\omega$

The potential energy $U$ at height $h$ from the reference level is,

$U = mgh$

Here, $g$ is the acceleration due to gravity.

The moment of inertia $I$ of solid sphere is given by following expression:

$I = \frac{2}{5}m{r^2}$

Here, $r$is the radius of the sphere.

The initial kinetic energy of the ball is given as follows:

${E_i} = \frac{1}{2}{I_i}\omega _i^2 + \frac{1}{2}m{v^2} + mg{h_i}$

The relation between rotational speed $\omega$ and linear speed is given as follows:

$\omega = \frac{v}{r}$

The moment of inertia $I$ of solid sphere is given by following expression:

$I = \frac{2}{5}m{r^2}$

Substitute the expression for $I$ and $\omega$in the expression of kinetic energy as follows:

${E_i} = \frac{1}{2}\left( {\frac{2}{5}m{r^2}} \right){\left( {\frac{{{v_i}}}{r}} \right)^2} + \frac{1}{2}mv_i^2 + mg{h_i}$

On the same line, the expression for final kinetic energy is given as follows:

${E_f} = \frac{1}{2}\left( {\frac{2}{5}m{r^2}} \right){\left( {\frac{{{v_f}}}{r}} \right)^2} + \frac{1}{2}mv_f^2 + mg{h_f}$

Equate the expression for initial and final kinetic energy and solve for the final transitional speed as follows:

$\begin{array}{c}\\{E_i} = {E_f}\\\\\frac{1}{2}\left( {\frac{2}{5}m{r^2}} \right){\left( {\frac{{{v_i}}}{r}} \right)^2} + \frac{1}{2}mv_i^2 + mg{h_i} = \frac{1}{2}\left( {\frac{2}{5}m{r^2}} \right){\left( {\frac{{{v_f}}}{r}} \right)^2} + \frac{1}{2}mv_f^2 + mg{h_f}\\\\\frac{7}{{10}}v_i^2 + g{h_i} = \frac{7}{{10}}v_f^2 + g{h_f}\\\\v_f^2 = v_i^2 - \frac{{10}}{7}g\left( {{h_f} - {h_i}} \right)\\\end{array}$

The expression for the final transitional speed of the ball is,

${v_f} = \sqrt {v_i^2 + \frac{{10}}{7}g\left( {{h_i} - {h_f}} \right)}$

Substitute $5.10{\rm{ m/s}}$ for ${v_i}$ , $9.8{\rm{ m/}}{{\rm{s}}^2}$ for $g$ and $0.760{\rm{ m}}$ for ${h_f} - {h_i}$ as follows:

$\begin{array}{c}\\{v_f} = \sqrt {{{\left( {5.10{\rm{ m/s}}} \right)}^2} - \frac{{10}}{7}\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {0.760{\rm{ m}}} \right)} \\\\ = 3.92{\rm{ m/s}}\\\end{array}$

Ans:

The final transitional speed of the ball is $3.92{\rm{ m/s}}$.