Question

A soccer ball with a radius of $25cm$, is kicked with an initial velocity of $15 m/s$ and rolls without slipping across a level horizontal grass field. If the acceleration of the ball is $-25.0m{s^{-2}}$. Which of the following statements best represents how many rotations the ball makes before coming to rest?
(A) $2.9$ revolutions.
(B) $4.5$ revolutions.
(C) $12.5$ revolutions.
(D) $18$ revolutions.
(E) $24$ revolutions.

Answer 1

The given problem can be solved using one of the four equations of kinetic energy. In this problem we will use the kinematics of the rotational motion formula then, the rotation of the ball before the ball is coming to the rest is determined.

Formula used:
The kinematics of the rotational motion of the ball is given by;
$\omega _f^2 - \omega _i^2 = 2\alpha \theta$
Where, ${\omega _f}$ denotes the final angular velocity of the ball, ${\omega _i}$ denotes the initial angular velocity of the ball, $\alpha$ denotes the acceleration of the ball of the ball.

Complete step by step solution:
The data given in the problem is;
Final velocity, $v = 0\,\,m{s^{ - 1}}$,
Initial velocity, $u = 15\,\,m{s^{ - 1}}$,
Acceleration, $a = - 25.0\,\,m{s^{ - 2}}$,
Radius of the ball, $r = 25cm = 0.25m$
The rotational kinetic energy of the ball is;
$\Rightarrow \omega _f^2 - \omega _i^2 = 2\alpha \theta$
Where, $v = r{\omega _f}$; $u = r{\omega _i}$; $a = r\alpha$;
$\Rightarrow \dfrac{{{v^2}}}{{{r^2}}} - \dfrac{{{u^2}}}{{{r^2}}} = 2\left[ {\dfrac{a}{r}} \right]\theta$
Now substitute the values of $v$,$u$,$a$and $r$ in the above equation;
$\Rightarrow \dfrac{0}{{{{\left( {0.25} \right)}^2}}} - \dfrac{{{{\left( {15} \right)}^2}}}{{{{\left( {0.25} \right)}^2}}} = 2\left[ {\dfrac{{ - 25.0}}{{0.25}}} \right]\theta$
$\Rightarrow 0 - \dfrac{{225}}{{0.0625}} = 2\left[ { - 100} \right]\theta$
$\Rightarrow - 3600 = - 200 \times \theta$
Therefore,
$\Rightarrow \theta = 18$ radian
The number of revolutions that the ball made is;
$\Rightarrow N = \dfrac{\theta }{{2\pi }}$
Where, $N$ is the number of revolutions made by the ball.
Substitute the value of $\theta = 18$ radian;
$\Rightarrow N = \dfrac{{18}}{{2\pi }}$
$\Rightarrow N = 2.9$revolutions.
Therefore, the number of revolutions made by the ball is $N = 2.9$ revolutions before coming to rest.
Hence, the option $N = 2.9$ revolutions is the correct answer.

Thus, the option A is correct.

Note: In proportion to the turning energy or the angular kinetic energy the kinetic energy of the object is obtained by the rotation of that particular object. The number of the revolution is directly proportional to the angle and inversely proportional to the $2\pi$.