
A stone tied to the end of a string of 1 m long is whirled in a horizontal circle with constant speed. If the stone makes 22 revolutions in 44 seconds, what would be the magnitude and direction of acceleration of the stone?
(A) \[{\pi ^2}m/{s^2}\] and direction along the tangent to the circle
(B) \[{\pi ^2}m/{s^2}\] and direction along the radius towards the centre.
(C) \[\dfrac{{{\pi ^2}}}{4}m/{s^2}\] and direction along the radius towards the centre.
(D) \[{\pi ^2}m/{s^2}\] and direction along the radius away from the centre.
Answer
232.5k+ views
Hint: The centripetal force is the force always required to make an object to go in a circular motion. The force is directed inwards along the radius of the circular path traced out by the motion.
Formula used: In this solution we will be using the following formulae;
\[f = \dfrac{n}{t}\] where \[f\] is the frequency of revolution of an object in circular motion, \[n\] is the number of revolution, and \[t\] is the time taken to make that revolution.
\[\omega = 2\pi f\] where \[\omega \] is the angular velocity.
\[{a_c} = {\omega ^2}r\] where \[{a_c}\] is the centripetal acceleration of a circulating object, \[r\] is the radius of the circle.
Complete Step-by-Step Solution:
To calculate the acceleration, we must first calculate the angular velocity. It can be given by
\[\omega = 2\pi f\] where \[f\] is the frequency of revolution of an object in circular motion. The frequency can in itself be given as
\[f = \dfrac{n}{t}\] where \[n\] is the number of revolutions, and \[t\] is the time taken to make that revolution.
Hence,
\[\omega = 2\pi \dfrac{n}{t}\]
Hence, by insertion of known values, we get
\[\omega = 2\pi \dfrac{{22}}{{44}}\]
\[ \Rightarrow \omega = \pi \]rad.
The centripetal acceleration can be given as
\[{a_c} = {\omega ^2}r\]
Hence, by inserting all known values, we get
\[{a_c} = {\pi ^2}\left( 1 \right) = {\pi ^2}m/{s^2}\]
The centripetal acceleration is always directed along the radius of the circle towards the centre.
Hence, the correct option is B
Note: For clarity, centripetal acceleration is directed towards the centre because centripetal force is (since acceleration is always directed along the direction of the force.
The tension on the string in this case for example provides the force for the stone to not break away from a circular motion, and the tension is directed along the length of the string. It is directed inwards as it opposes the string going far out. In all cases of a circular motion, some force must be directed inwards to provide the centripetal force to keep the motion circular.
Formula used: In this solution we will be using the following formulae;
\[f = \dfrac{n}{t}\] where \[f\] is the frequency of revolution of an object in circular motion, \[n\] is the number of revolution, and \[t\] is the time taken to make that revolution.
\[\omega = 2\pi f\] where \[\omega \] is the angular velocity.
\[{a_c} = {\omega ^2}r\] where \[{a_c}\] is the centripetal acceleration of a circulating object, \[r\] is the radius of the circle.
Complete Step-by-Step Solution:
To calculate the acceleration, we must first calculate the angular velocity. It can be given by
\[\omega = 2\pi f\] where \[f\] is the frequency of revolution of an object in circular motion. The frequency can in itself be given as
\[f = \dfrac{n}{t}\] where \[n\] is the number of revolutions, and \[t\] is the time taken to make that revolution.
Hence,
\[\omega = 2\pi \dfrac{n}{t}\]
Hence, by insertion of known values, we get
\[\omega = 2\pi \dfrac{{22}}{{44}}\]
\[ \Rightarrow \omega = \pi \]rad.
The centripetal acceleration can be given as
\[{a_c} = {\omega ^2}r\]
Hence, by inserting all known values, we get
\[{a_c} = {\pi ^2}\left( 1 \right) = {\pi ^2}m/{s^2}\]
The centripetal acceleration is always directed along the radius of the circle towards the centre.
Hence, the correct option is B
Note: For clarity, centripetal acceleration is directed towards the centre because centripetal force is (since acceleration is always directed along the direction of the force.
The tension on the string in this case for example provides the force for the stone to not break away from a circular motion, and the tension is directed along the length of the string. It is directed inwards as it opposes the string going far out. In all cases of a circular motion, some force must be directed inwards to provide the centripetal force to keep the motion circular.
Recently Updated Pages
Dimensions of Charge: Dimensional Formula, Derivation, SI Units & Examples

How to Calculate Moment of Inertia: Step-by-Step Guide & Formulas

Circuit Switching vs Packet Switching: Key Differences Explained

Dimensions of Pressure in Physics: Formula, Derivation & SI Unit

JEE General Topics in Chemistry Important Concepts and Tips

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

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

Laws of Motion Class 11 Physics Chapter 4 CBSE Notes - 2025-26

Waves Class 11 Physics Chapter 14 CBSE Notes - 2025-26

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Physics Chapter 11 CBSE Notes - 2025-26

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

