Question

a particle suspended by a thread of length 'l' is projected horizontally with velocity √3gl at the lowest point .the height from the bottom at which the tension in the string becomes zero is?

Answer 1

There is no external force act on particle.
so, ∆K.E + ∆P.E = 0
∆K.E is total kinetic energy and ∆P.E is total potential energy.
Let h height from bottom at which tension of thread becomes zero. And that height particle makes Ф angle with vertical .
Then, h = (l + lcosФ)
Now, total potential energy = PEf - P.Ei
= -mg(l + lcosФ) - 0 [ let reference level is bottom ]
= -mgl(1 + cosФ)

Total kinetic energy = K.Ef - K.Ei
= 1/2mv₁² + 1/2mv² ----(1)
v₁ is the velocity of particle at h height and v is the velocity of particle at bottom.

Now, net Force act on at h height in circular motion.
Fnet = mv₁²/l = mgcosФ + T
∵ Tension in thread , T = 0
so, v₁² = glcosФ, put it in equation (1)

Change in kinetic energy = 1/2 mglcosФ + 1/2m{√(3gl)}²
∴1/2mglcosФ + 3mgl/2 - mgl(1 + cosФ) = 0
-1/2mglcosФ + mgl/2 = 0
Hence, cosФ = 1

So, h = l + lcosФ = l + l = 2l

Hence, answer is 2l

Answer 2

Answer:

2l

Explanation:

THE ANSWER IS 2L

H=1+COS THETA

1/2MV1^2+1/2MV^2

H=I+ICOSTHETA=L+L=2L