Answer

Verified

345.3k+ views

**Hint:**Find the equation of motion of the particle and use the conservation of mechanical energy at starting point and at the point of the balance condition. From there find the speed of the particle at the point where the tension in the string is equal to the weight of the particle.

**Formula used:**

Newton's second law is given by,

\[F = ma\]

where \[m\] is the mass of the particle and \[a\] is the acceleration of the particle.

The conservation of mechanical energy is given by,

\[{K_i} + {U_i} = {K_f} + {U_f}\]

where, \[{K_i}\] and \[{U_i}\] are the initial kinetic and potential energy and \[{K_f},{U_f}\] are the final kinetic and potential energy.

**Complete step by step answer:**

We have given here that the particle is given a velocity of \[\sqrt {gl} \] horizontally. Now, the particle will start swinging due to this. So, we have to find the velocity of the particle when the weight of the particle will be equal to the tension in the string. Now, there are two forces acting on the particle vertically one is the centrifugal force and another is the gravitational pull on the particle. Let, the mass of the particle is $m$.

So, the equation of motion of the particle along the wire can be written as, \[T = mg\cos \theta + \dfrac{{m{v^2}}}{l}\]

Now, according to the question this is equal to the weight of the particle hence we can write,

\[mg = mg\cos \theta + \dfrac{{m{v^2}}}{l}\]….(i)

Now, from conservation of mechanical energy we can write,

\[\dfrac{1}{2}m{(\sqrt {gl} )^2} = mgl' + \dfrac{1}{2}m{(v)^2}\]

Now, from the diagram we can see that, \[l' = l - l\cos \theta \].

So, putting this value we have,

\[\dfrac{1}{2}mgl = mgl(1 - \cos \theta ) + \dfrac{1}{2}m{(v)^2}\]

\[\Rightarrow \dfrac{1}{2}mgl - mgl = - mgl\cos \theta + \dfrac{1}{2}m{(v)^2}\]

Upon simplifying we have,

\[gl\cos \theta = \dfrac{1}{2}({v^2} + gl)\]

\[\Rightarrow \cos \theta = \dfrac{1}{{2gl}}({v^2} + gl)\]…….(ii)

Putting these value in equation (i) we have,

\[mg = mg\cos \theta + \dfrac{{m{v^2}}}{l}\]

\[\Rightarrow mg = mg\dfrac{1}{{2gl}}({v^2} + gl) + \dfrac{{m{v^2}}}{l}\]

\[\Rightarrow g = \dfrac{1}{{2l}}({v^2} + gl) + \dfrac{{{v^2}}}{l}\]

Upon simplifying we have,

\[g = \dfrac{{{v^2}}}{{2l}} + \dfrac{g}{2} + \dfrac{{{v^2}}}{l}\]

\[\Rightarrow g - \dfrac{g}{2} = \dfrac{{3{v^2}}}{{2l}}\]

\[\Rightarrow {v^2} = \dfrac{{gl}}{3}\]

\[\therefore v = \sqrt {\dfrac{{gl}}{3}} \]

**Hence, the speed of a particle at the point where the tension in the string equals weight of the particle is \[\sqrt {\dfrac{{gl}}{3}} \].**

**Note:**The point at which the particle starts swinging the weight of the particle may seem equal to the tension in the string. But, at the point of starting the particle has acceleration acting in the horizontal direction and since the tension and the weight of the particle is not equal it has some motion along the vertical along with the horizontal line also.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

How many crores make 10 million class 7 maths CBSE

The 3 + 3 times 3 3 + 3 What is the right answer and class 8 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Give 10 examples for herbs , shrubs , climbers , creepers

Change the following sentences into negative and interrogative class 10 english CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE