
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
139.8k+ views
Hint: 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 $.
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 $.
Recently Updated Pages
Average fee range for JEE coaching in India- Complete Details

Difference Between Rows and Columns: JEE Main 2024

Difference Between Length and Height: JEE Main 2024

Difference Between Natural and Whole Numbers: JEE Main 2024

Algebraic Formula

Difference Between Constants and Variables: JEE Main 2024

Trending doubts
Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Physics Average Value and RMS Value JEE Main 2025

Charging and Discharging of Capacitor

Electromagnetic Waves Chapter - Physics JEE Main

Collision - Important Concepts and Tips for JEE

A pressure of 100 kPa causes a decrease in volume of class 11 physics JEE_Main

Other Pages
NCERT Solutions for Class 11 Physics Chapter 4 Laws of Motion

NCERT Solutions for Class 11 Physics Chapter 7 Gravitation

NCERT Solutions for Class 11 Physics In Hindi Chapter 1 Physical World

NCERT Solutions for Class 11 Physics Chapter 14 Waves

JEE Advanced 2025 Notes

A water pump of 1 horse power is required to flow water class 11 physics JEE_Main
