a rotating disc of mass 'm', radius 'r', thickness 'D' can rotate in vacuum with constant angular velocity 'w' . now the disc is placed in water of density 'p', find the time in which its angular velocity becomes half.
Without friction, the angular velocity will not change. In real fife of course it will. Even in the vacuum there is loss of kinetic energy to the bearings. However as far as the information given here, the answer is constant angular velocity forever. or at least for a very long time like the rotation of the earth.
To find the time it takes for the angular velocity of the rotating disc to become half when placed in water, we can use the concepts of rotational dynamics and fluid mechanics.
First, let's consider the forces acting on the disc when it is submerged in water. There are two main forces to consider: the buoyant force and the gravitational force.
1. Buoyant Force: The buoyant force acting on the disc is given by the equation F_b = ρ * V * g, where ρ is the density of water, V is the volume of the displaced water (volume of the disc), and g is the acceleration due to gravity.
The volume of the disc can be calculated as V = π * r^2 * D, where r is the radius of the disc and D is the thickness of the disc.
2. Gravitational Force: The gravitational force acting on the disc is given by F_g = m * g, where m is the mass of the disc and g is the acceleration due to gravity.
Now let's apply Newton's second law of rotational motion, which states that the net torque acting on an object is equal to the moment of inertia times the angular acceleration.
The moment of inertia (I) of the disc can be calculated as I = (1/2) * m * r^2, assuming a uniform disc.
Since the angular velocity (ω) is constant initially, the net torque is zero.
At equilibrium, the buoyant force (F_b) acting on the disc creates an opposing torque which reduces the angular velocity.
The torque due to the buoyant force is given by τ_b = r * F_b, where r is the radius of the disc.
Equating the torque due to the buoyant force to the torque due to the gravitational force, we have:
r * F_b = I * α
Where α is the angular acceleration.
Substituting the expressions for F_b, I, and rearranging the equation, we get:
r * (ρ * V * g) = (1/2) * m * r^2 * α
Simplifying further by substituting V and rearranging the equation, we get:
(ρ * π * r^2 * D * g) = (1/2) * m * r^2 * α
Simplifying further by substituting α = dw/dt (where w is the angular velocity and t is time), we get:
(ρ * π * r^2 * D * g) = (1/2) * m * r^2 * (dw/dt)
Now, we can solve this differential equation to find the time it takes for the angular velocity to become half.
To do this, we can rearrange the equation and integrate both sides:
(2 * ρ * π * r^2 * D * g) / (m * r^2) = ∫[w_0 to w_f] (dw / w)
Where w_0 is the initial angular velocity and w_f is the final angular velocity (half of the initial velocity).
Solving this integral, we get:
(2 * ρ * π * r^2 * D * g) / (m * r^2) = ln(w_f / w_0)
Finally, rearranging the equation to solve for the time (t) it takes to reach the final angular velocity, we get:
t = -(m * r^2) / (2 * ρ * π * r^2 * D * g) * ln(w_f / w_0)
Therefore, the time it takes for the angular velocity of the rotating disc to become half when placed in water is given by this equation, where you need to substitute the values of m (mass of the disc), r (radius of the disc), D (thickness of the disc), ρ (density of water), g (acceleration due to gravity), w_0 (initial angular velocity), and w_f (final angular velocity).