In a Conical pendulum, a string of length $120cm$ is fixed at rigid support and carries a mass $150g$at the end. If the mass is revolved in a horizontal circle of radius $0.2m$ around a vertical axis, Calculate the tension in the string. $(g = 9.8\dfrac{m}{{{s^2}}})$
Answer
Verified
120k+ views
Hint: In order to solve this equation, we first have to draw a free body diagram of the body. After that, we have to balance the Net force that is acting on the body. All the forces should be in accordance with Newton's third law. We can easily find the solution after balancing all the forces in all the given directions.
Formula Used:
$Centripetal Force$ = $\dfrac{{m{V^2}}}{r}$
$m$ is the mass of body
$V$ is the velocity of body
$r$ is the radius of circle & Secondly, Net force on a body is equal to $0$ i. e. ${F_{net}} = 0$
Complete step by step answer:
Here, a conical pendulum, a string of length $120cm$ & mass of $150g$ & given we have to calculate the Tension $T$ in the String.
In order to understand the Question, Let’s draw a free body diagram of a pendulum.
Free Body Diagram:
Here the mass $m$ is moving in a circular motion of radius $0.2m$ with velocity $V.$ The Force acting on a mass $m$ will be
${F_{net}} = \dfrac{{m{V^2}}}{r}$
Here, the Tension $T$ will be resolved into two components $T\sin \theta $ (in vertical direction) and $ \uparrow \cos \theta $ (in horizontal direction). Now balancing all forces we get
In Vertical Direction:
$T\sin \theta = mg - (i)$
Now,
In Horizontal plane,
$T\cos \theta = $ Horizontal force on a mass $m$ (i.e. Centripetal force)
So,
$T\cos \theta = \dfrac{{m{V^2}}}{r} - (ii)$
So, from eq. we get
$T = \dfrac{{mg}}{{\sin \theta }}$
Now, In $\Delta ABC$, we get $\sin \theta = \dfrac{P}{H}$
So, $\sin \theta = \dfrac{{\sqrt {{l^2} - {R^2}} }}{l}$
Putting the value of $\sin \theta $in eq. $(1)$ we get
$T = \dfrac{{mgl}}{{\sqrt {{l^2} - {R^2}} }}$
Further,
$T = \left( {\dfrac{{150}}{{1000}}} \right)\dfrac{{(9.8) \times 120}}{{\sqrt {{{(120)}^2} - {{(20)}^2}} }}$ $[1kg = 1000g,1m = 100cm]$
$T = 1.52$ Newton.
Hence, the tension on the string will be $1.52$N.
Note: While solving this question, we have to be very careful with directions. Only forces acting in the same direction will balance each other. Also according to Newton's third law, action and reaction occur on different bodies. So we have to be careful while applying it to anybody. the units for every force should be the same.
Formula Used:
$Centripetal Force$ = $\dfrac{{m{V^2}}}{r}$
$m$ is the mass of body
$V$ is the velocity of body
$r$ is the radius of circle & Secondly, Net force on a body is equal to $0$ i. e. ${F_{net}} = 0$
Complete step by step answer:
Here, a conical pendulum, a string of length $120cm$ & mass of $150g$ & given we have to calculate the Tension $T$ in the String.
In order to understand the Question, Let’s draw a free body diagram of a pendulum.
Free Body Diagram:
Here the mass $m$ is moving in a circular motion of radius $0.2m$ with velocity $V.$ The Force acting on a mass $m$ will be
${F_{net}} = \dfrac{{m{V^2}}}{r}$
Here, the Tension $T$ will be resolved into two components $T\sin \theta $ (in vertical direction) and $ \uparrow \cos \theta $ (in horizontal direction). Now balancing all forces we get
In Vertical Direction:
$T\sin \theta = mg - (i)$
Now,
In Horizontal plane,
$T\cos \theta = $ Horizontal force on a mass $m$ (i.e. Centripetal force)
So,
$T\cos \theta = \dfrac{{m{V^2}}}{r} - (ii)$
So, from eq. we get
$T = \dfrac{{mg}}{{\sin \theta }}$
Now, In $\Delta ABC$, we get $\sin \theta = \dfrac{P}{H}$
So, $\sin \theta = \dfrac{{\sqrt {{l^2} - {R^2}} }}{l}$
Putting the value of $\sin \theta $in eq. $(1)$ we get
$T = \dfrac{{mgl}}{{\sqrt {{l^2} - {R^2}} }}$
Further,
$T = \left( {\dfrac{{150}}{{1000}}} \right)\dfrac{{(9.8) \times 120}}{{\sqrt {{{(120)}^2} - {{(20)}^2}} }}$ $[1kg = 1000g,1m = 100cm]$
$T = 1.52$ Newton.
Hence, the tension on the string will be $1.52$N.
Note: While solving this question, we have to be very careful with directions. Only forces acting in the same direction will balance each other. Also according to Newton's third law, action and reaction occur on different bodies. So we have to be careful while applying it to anybody. the units for every force should be the same.
Recently Updated Pages
Difference Between Circuit Switching and Packet Switching
Difference Between Mass and Weight
JEE Main Participating Colleges 2024 - A Complete List of Top Colleges
JEE Main Maths Paper Pattern 2025 – Marking, Sections & Tips
Sign up for JEE Main 2025 Live Classes - Vedantu
JEE Main 2025 Helpline Numbers - Center Contact, Phone Number, Address
Trending doubts
JEE Mains 2025: Check Important Dates, Syllabus, Exam Pattern, Fee and Updates
JEE Main Login 2045: Step-by-Step Instructions and Details
Class 11 JEE Main Physics Mock Test 2025
JEE Main Chemistry Question Paper with Answer Keys and Solutions
JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking
JEE Main Chemistry Exam Pattern 2025
Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs
NCERT Solutions for Class 11 Physics Chapter 7 Gravitation
NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Fluids
Units and Measurements Class 11 Notes - CBSE Physics Chapter 1
JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More