Question

The bob of a pendulum at rest is given a sharp hit to impart a horizontal velocity √(10gl) where l is the length of the pendulum. Find the tension in the string when (a) the string is horizontal, (b) the bob is at its highest point and (c) the string makes an angle of 60° with the upward vertical.

Answer 1

Let tension = T
 Mass ( bob) = m
  velocity = v.
Refer to the attachment

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.
$T = \frac{mv^2}{r}$
$T = m8gl / l$


(b) Assume that the velocity at y = v₃
Similarly like equation (i) , at y and velocity v₃
1/2mv₁² = 1/2mv₃² + mg(2 l )
$1/2 m (log l) = 1/2 mv_{ 3}^2 +2mgl$ 
v₃² = 6mgl

Now thew tension
$T_{y} = mv ^{2} /l$ -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 ,
$T_{z} = \frac{mv ^{2}}{l}$ -mg(cos60°) .
On calculating , we get
 $T_{z}$= 7mg -0.5mg 
$T_{z}$ = 6.5 mg




Hope it Helps :-)

Answer 2

this is answer to your question