A steel disc ( S) spins on about a frictionless spindle on a frictionless air table.

The steel disc has a radius of R=0.24m and a mass ms=6.15kg, and may be modelled as solid with a uniform mass distribution.

The steel plate has an initial angular velocity of ω0=4.4rad/s.

A stationary aluminium disc of identical size and shape, and a mass of ma=2.11kg is dropped onto the steel disc from a small height.

After a brief duration the two discs rotate together.
a)Determine the final angular velocity of both disks.
b)Determine the final total kinetic energy of the two discs.

Question

A steel disc ( S) spins on about a frictionless spindle on a frictionless air table.

The steel disc has a radius of R=0.24m and a mass ms=6.15kg, and may be modelled as solid with a uniform mass distribution.

The steel plate has an initial angular velocity of ω0=4.4rad/s.

A stationary aluminium disc of identical size and shape, and a mass of ma=2.11kg is dropped onto the steel disc from a small height.

After a brief duration the two discs rotate together.
a)Determine the final angular velocity of both disks.
b)Determine the final total kinetic energy of the two discs.

anonymous · Answer

I was able to calculate the initial kinetic energy of the rotating steel disc to be equal to 1.715J, but don't know how to move forward. Tried many methods but all turn out wrong.

anonymous · Answer

By stating that the second disk is dropped from a small height, we're being asked to negate it's potential energy. This leaves the total energy in the system as the initial kinetic energy of the spinning disk. KE = KE. The moment of inertia of a thin disk is 0.5*m*r^2. This yields I = 0.177 kg m^2. Rotational kinetic energy is KE = 0.5 I w^2. This yields KE = 1.71J, as you had found before. This is also the final kinetic energy. By substitution we know that the energy with the second disk is KE = 0.5(0.5*2m*r^2)w^2. We know everything except omega. With a bit of algebra, we can find w = 3.11rad/s.