- #1

~angel~

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Consider a turntable to be a circular disk of moment of inertia I_t rotating at a constant angular velocity omega_i around an axis through the center and perpendicular to the plane of the disk (the disk's "primary axis of symmetry"). The axis of the disk is vertical and the disk is supported by frictionless bearings. The motor of the turntable is off, so there is no external torque being applied to the axis. Another disk (a record) is dropped onto the first such that it lands coaxially (the axes coincide). The moment of inertia of the record is I_r. The initial angular velocity of the second disk is zero.

There is friction between the two disks.

After this "rotational collision," the disks will eventually rotate with the same angular velocity.

What is the final angular velocity, omega_f, of the two disks?

Express omega_f in terms of I_t, I_r, and omega_i.

Because of friction, kinetic energy is not conserved while the disks' surfaces slip over each other. What is the final kinetic energy, K_f, of the two spinning disks?

Express the final kinetic energy in terms of I_t, I_r, and the initial kinetic energy K_i of the two-disk system. No angular velocities should appear in your answer.

Assume that the turntable deccelerated during time deltat before reaching the final angular velocity ( deltat is the time interval between the moment when the top disk is dropped and the time that the disks begin to spin at the same angular velocity). What was the average torque, \avg{\tau}, acting on the bottom disk due to friction with the record?

Express the torque in terms of I_t, omega_i, omega_f, and deltat.

Ok, I know the answer to the first question, which is (I_t*omega_i)/(I_t+I_r). But I am lost on the whole KE thing. My answer was ((I_t+I_r)*(I_t*K_i))/(2*(I_t^2+2*I_t*I_r+I_r^2)), which was incorrect.

Any help would be appreciated for the last question as well.

Thanks in advance.