A massless spring of spring constant k is mounted on a turntable of rotation inertia I.the turntable is on a frictionless vertical acle,though initially it is not rotating.the spring is compressed a distance dx from its equilibrium,with a mass m placed against it.when the spring si released,the mass leaves the spring moving at right angles to a line through the center of the turntable,at a distance b from the denter,and slides without friction across the table and off find expression for a)the linear speed of the mass and b) the rotational speed of turntable?

Relevance

The stored energy in the spring is 0.5*k*dx^2. This energy is transferred to the kinetic energy of the mass and rotational energy of the turntable. If v is the mass velocity on release, its kinetic energy is

.5*m*v^2

The rotational energy of the turntable is

0.5*I*w^2, where w = angular velocity of the turntable.

The sum of these must equal the initial energy in the spring.

You then use conservation of angular momentum to get another relation. (Consider angular momentum about the center of the turntable.) The angular momentum of the mass on release is m*v*b; the angular momentum of the turntable is w*I. Since the initial momentum of the system was zero, these must be equal and opposite, so

w*I = -m*v*b; .........this with the energy equivalence

.5*m*v^2 + 0.5*I*w^2 = 0.5*k*dx^2

gives you two equations in two unknowns, w and v.

solve for v in terms of w from the first, plug into the second to get w, then put that back into the first to solve for v.

The rotational speed will come out in radians per second; convert to rpm using one rev = 2π radians, one minute = 60 seconds. rpm = (30/π)*w