A Bob of simple pendulum is given horizontal velocity of √10gl when it is at lowest position. Find the tension in the string when string rotates by 120°

Now assume that the position of bob is upward horizontal.

(a)  Let the velocity at x = v₂ 
Now K.E at v₁ = 1/2 mv₁² 
K.E at v₂ = 1/2mv₂²
P.E = mgl 
Now kinetic energy at this point is equal to total energy, Equate both
1/2 mv₁² = 1/2 mv₂² + mgl -----→(i)
1/2 m√(10)² = 1/2 mv₂² + mgl
v₂² = 8 gl .
Now the tension at the horizontal position.

 

(b) Assume that the velocity at y = v₃
Similarly like equation (i) , at y and velocity v₃
1/2mv₁² = 1/2mv₃² + mg(2 l )
 
v₃² = 6mgl

Now thew tension
 -mg
 × mg

 

(c)Assume the velocity at z = v₄ .
Again similarly like equation (i) at z = v₄
1/2mv₁² = 1/2mv₄² +mgh    ∵[h = l(1+cosθ)]
Now, 
1/2 mv₁² =1/2 mv₄² +mgl(1+cos60°)
calculating it , we get
v₄² =7gl

Now the tension ,
 -mg(cos60°) .
On calculating , we get
 = 7mg -0.5mg 
 = 6.5 mg

Let tension = T
 Mass ( bob) = m
  velocity = v.

Now assume that the position of bob is upward horizontal.

(a)  Let the velocity at x = v₂ 
Now K.E at v₁ = 1/2 mv₁² 
K.E at v₂ = 1/2mv₂²
P.E = mgl 
Now kinetic energy at this point is equal to total energy, Equate both
1/2 mv₁² = 1/2 mv₂² + mgl -----→(i)
1/2 m√(10)² = 1/2 mv₂² + mgl
v₂² = 8 gl .
Now the tension at the horizontal position.
T = mv^2 /r 
T = m8gl/l 

 

(b) Assume that the velocity at y = v₃
Similarly like equation (i) , at y and velocity v₃
1/2mv₁² = 1/2mv₃² + mg(2 l )
1/2 m(logl) = 1/2 mv3^2 + 2mgl  
v₃² = 6mgl

Now thew tension
 Ty = mv^2 /l -mg
 T = 6mgl/l × mg
T = 5mg 

(c)Assume the velocity at z = v₄ .
Again similarly like equation (i) at z = v₄
1/2mv₁² = 1/2mv₄² +mgh    ∵[h = l(1+cosθ)]
Now,
1/2 mv₁² =1/2 mv₄² +mgl(1+cos60°)
calculating it , we get
v₄² =7gl

Now the tension ,
 T2 = mv^2/l -mg(cos60°) .
On calculating , we get
 T2 = 7mg -0.5mg 
 = 6.5 mg