A rope of mass M and length L is being rotated about one end in a gravity free space. A pulse is being created at one of the ends. The angle through which rope will be rotated in the time when pulse reaches the opposite end of rope for first time and for the angular velocity of rope 'ω', is

a) π/√2 b) π/2 c) π/(2√2) d) π2
ans is option a
pls explain me how?

Answer :

Option : (A) Pi / root(2)

To solve this problem, we can make use of the conservation of angular momentum. The angular momentum of an object rotating about a fixed point is defined as the product of its moment of inertia (I) and its angular velocity (ω):

L = I * ω

In this case, the rope is being rotated about one end, so the moment of inertia of the rope can be approximated as that of a thin rod rotating about one end, which is given by:

I = (1/3) * M * L^2

where M is the mass of the rope and L is its length.

Now, let's consider the pulse created at one end of the rope. When the pulse reaches the opposite end of the rope for the first time, it has traveled a distance equal to the length of the rope. Let's denote this distance as d.

The linear speed of the pulse can be calculated using the formula:

v = ω * d

The time taken for the pulse to travel this distance can be found using the formula:

t = d / v

Since we are looking for the angle through which the rope rotates during this time, we can use the formula:

θ = ω * t

Substituting the expressions for v and t derived earlier, we get:

θ = (ω * d) * (d / (ω * d)) = 1 radian

Therefore, the angle through which the rope will be rotated when the pulse reaches the opposite end of the rope for the first time is equal to π/√2 radians.

Hence, the correct answer is option a) π/√2.