
A ball of mass 0.25kg attached to the ends of a string of length 1.96m is rotating in a horizontal circle. The string will break, if tension is more than $25N$. What is the maximum velocity with which the ball can be rotated?
Answer
144.3k+ views
Hint: Centrifugal Force acts on every object moving in a circular path when viewed from a rotating frame of reference. The centrifugal force also depends on the mass of the object, the distance from the centre of the circle and also the speed of rotation. If the object has more mass, the force of the movement and the speed of the object will be greater. If the distance is far from the centre of the circle the force of the movement will be more.
Formula Used:
The mathematical formula for Centrifugal force is given as the negative product of mass (in kg) and tangential velocity (in meters per second) squared, divided by the radius (in meters). This implies that on doubling the tangential velocity, the centripetal force will be quadrupled. Mathematically it is written as:
\[{F_c} = \dfrac{{ - m{v^2}}}{r}\]
In this mathematical formula, $m$ is the mass of the object, $v$ is the velocity of the object and $r$ is the radius.
Complete step by step answer:
In this above numerical problem, the mass of the ball is given equal to $0.25kg$ and the force of tension is given equal to $25N$. Now, the radius of the string is given equal to $1.96m$.
Now, we can use the velocity as the subject of the formula and thus, the expression becomes:
${v^2} = \dfrac{{F \times r}}{m}$
Now, putting the values of force, radius and mass of the ball in the above expression, we get:
${v^2} = \dfrac{{25 \times 1.96}}{{0.25}} = 196$
Now to obtain the maximum velocity with which the ball rotates, we have to find the square root of the above value obtained. Thus, we get:
$v = 14m{s^{ - 1}}$
This is the maximum value with which the ball can rotate in the string before the string breaks.
Note: If an object is moving in a circle and experiences an outward force then this force is called the centrifugal force. The object has the direction along the centre of the circle from the centre approaching the object.
Formula Used:
The mathematical formula for Centrifugal force is given as the negative product of mass (in kg) and tangential velocity (in meters per second) squared, divided by the radius (in meters). This implies that on doubling the tangential velocity, the centripetal force will be quadrupled. Mathematically it is written as:
\[{F_c} = \dfrac{{ - m{v^2}}}{r}\]
In this mathematical formula, $m$ is the mass of the object, $v$ is the velocity of the object and $r$ is the radius.
Complete step by step answer:
In this above numerical problem, the mass of the ball is given equal to $0.25kg$ and the force of tension is given equal to $25N$. Now, the radius of the string is given equal to $1.96m$.
Now, we can use the velocity as the subject of the formula and thus, the expression becomes:
${v^2} = \dfrac{{F \times r}}{m}$
Now, putting the values of force, radius and mass of the ball in the above expression, we get:
${v^2} = \dfrac{{25 \times 1.96}}{{0.25}} = 196$
Now to obtain the maximum velocity with which the ball rotates, we have to find the square root of the above value obtained. Thus, we get:
$v = 14m{s^{ - 1}}$
This is the maximum value with which the ball can rotate in the string before the string breaks.
Note: If an object is moving in a circle and experiences an outward force then this force is called the centrifugal force. The object has the direction along the centre of the circle from the centre approaching the object.
Recently Updated Pages
Difference Between Rows and Columns: JEE Main 2024

Difference Between Natural and Whole Numbers: JEE Main 2024

How to find Oxidation Number - Important Concepts for JEE

How Electromagnetic Waves are Formed - Important Concepts for JEE

Electrical Resistance - Important Concepts and Tips for JEE

Average Atomic Mass - Important Concepts and Tips for JEE

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

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

Degree of Dissociation and Its Formula With Solved Example for JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry
