Question

the bob of a pendulum at rest is given a sharp hit to impart a horizontal velocity of (10gl)1/2 where l is length of pendulum. Find the speed of pendulum when its tangential acceleration is g

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.




(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




Hope it Helps :-