Answer
405k+ views
Hint In a vertical circle, the weight of the body affects the tension on the string. At the top the weight provides some of the centripetal force required for circular motion. At the bottom, it increases the centripetal force required. So we need to set the equation for the body and from there we can calculate the tension in the string.
Formula used: In this solution we will be using the following formula;
$ \Rightarrow {F_{NET}} = ma $ , where $ {F_{NET}} $ is the net force acting on a body, $ m $ is the mass of the body, and $ a $ is the acceleration attained by the body.
$ \Rightarrow {a_c} = m{\omega ^2}r $ , where $ {a_c} $ is the centripetal acceleration of a body moving in a circular motion, $ \omega $ is the magnitude of the angular velocity of the body and $ r $ is the radius of the circle.
Complete step by step answer
During a vertical swirl, the weight of the object plays an important role in the determination of the tension on the string. Depending on the immediate location of the body, it can either add to tension or relieve it. To calculate the tension, we analyse the motion using Newton’s second law of motion. Thus,
$ \Rightarrow {F_{NET}} = ma $ where $ {F_{NET}} $ is the net force acting on a body, $ m $ is the mass of the body, and $ a $ is the acceleration attained by the body.
A) At point A on the vertical swirl (top if the circle):
$ \Rightarrow {T_1} + mg = m{a_c} $ where $ T $ is the tension, $ g $ is the acceleration due to gravity, and $ {a_c} $ is the centripetal acceleration directed inward towards the centre of the circle.
Centripetal acceleration in general can be given as
$ \Rightarrow {a_c} = m{\omega ^2}r $ where $ \omega $ is the magnitude of the angular velocity of the body and $ r $ is the radius of the circle.
According to the question, $ \omega = 2rev/s $ , converting to SI unit, we multiply by $ 2\pi $ (since 1 rev subtends $ 2\pi $ radians), hence
$ \Rightarrow \omega = 12.56rad/s $ or $ \omega = 12.56{s^{ - 1}} $ .
Hence, we calculate $ m{a_c} $ from all known variables
$ \Rightarrow m{a_c} = 0.4{\left( {12.56} \right)^2} \times 1.2 = 75.72N $
Hence,
$ \Rightarrow {T_1} + 0.4\left( {10} \right) = 75.72 $
Thus,
$ \Rightarrow {T_1} = 75.72 - 4 $
$ \Rightarrow {T_1} = 71.72N $
B) For point B (bottom of circle)
$ \Rightarrow {T_2} - mg = m{a_c} $ , (since tension and weight point in opposite direction)
Similarly,
$ \Rightarrow {T_2} - 0.4\left( {10} \right) = 75.72 $
Thus,
$ \Rightarrow {T_2} = 75.72 + 4 $
$ \Rightarrow {T_2} = 79.72N $ .
Note
Alternatively we can calculate centripetal force from
$ \Rightarrow m{a_c} = \dfrac{{m{v^2}}}{r} $ where $ v $ is the magnitude of the linear speed.
We calculate $ v $ from $ v = \omega r $ . Thus,
$ \Rightarrow v = 12.56\left( {1.2} \right) = 15.072m/s $
$ \Rightarrow m{a_c} = \dfrac{{0.4{{\left( {15.072} \right)}^2}}}{{1.2}} = 75.72 $ which is identical to the above.
Formula used: In this solution we will be using the following formula;
$ \Rightarrow {F_{NET}} = ma $ , where $ {F_{NET}} $ is the net force acting on a body, $ m $ is the mass of the body, and $ a $ is the acceleration attained by the body.
$ \Rightarrow {a_c} = m{\omega ^2}r $ , where $ {a_c} $ is the centripetal acceleration of a body moving in a circular motion, $ \omega $ is the magnitude of the angular velocity of the body and $ r $ is the radius of the circle.
Complete step by step answer
![seo images](https://www.vedantu.com/question-sets/bee1c586-872a-4f5f-b191-66dcc1bee42c2653420743468915387.png)
During a vertical swirl, the weight of the object plays an important role in the determination of the tension on the string. Depending on the immediate location of the body, it can either add to tension or relieve it. To calculate the tension, we analyse the motion using Newton’s second law of motion. Thus,
$ \Rightarrow {F_{NET}} = ma $ where $ {F_{NET}} $ is the net force acting on a body, $ m $ is the mass of the body, and $ a $ is the acceleration attained by the body.
A) At point A on the vertical swirl (top if the circle):
$ \Rightarrow {T_1} + mg = m{a_c} $ where $ T $ is the tension, $ g $ is the acceleration due to gravity, and $ {a_c} $ is the centripetal acceleration directed inward towards the centre of the circle.
Centripetal acceleration in general can be given as
$ \Rightarrow {a_c} = m{\omega ^2}r $ where $ \omega $ is the magnitude of the angular velocity of the body and $ r $ is the radius of the circle.
According to the question, $ \omega = 2rev/s $ , converting to SI unit, we multiply by $ 2\pi $ (since 1 rev subtends $ 2\pi $ radians), hence
$ \Rightarrow \omega = 12.56rad/s $ or $ \omega = 12.56{s^{ - 1}} $ .
Hence, we calculate $ m{a_c} $ from all known variables
$ \Rightarrow m{a_c} = 0.4{\left( {12.56} \right)^2} \times 1.2 = 75.72N $
Hence,
$ \Rightarrow {T_1} + 0.4\left( {10} \right) = 75.72 $
Thus,
$ \Rightarrow {T_1} = 75.72 - 4 $
$ \Rightarrow {T_1} = 71.72N $
B) For point B (bottom of circle)
$ \Rightarrow {T_2} - mg = m{a_c} $ , (since tension and weight point in opposite direction)
Similarly,
$ \Rightarrow {T_2} - 0.4\left( {10} \right) = 75.72 $
Thus,
$ \Rightarrow {T_2} = 75.72 + 4 $
$ \Rightarrow {T_2} = 79.72N $ .
Note
Alternatively we can calculate centripetal force from
$ \Rightarrow m{a_c} = \dfrac{{m{v^2}}}{r} $ where $ v $ is the magnitude of the linear speed.
We calculate $ v $ from $ v = \omega r $ . Thus,
$ \Rightarrow v = 12.56\left( {1.2} \right) = 15.072m/s $
$ \Rightarrow m{a_c} = \dfrac{{0.4{{\left( {15.072} \right)}^2}}}{{1.2}} = 75.72 $ which is identical to the above.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)