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gills
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Homework Statement
Consider a turntable to be a circular disk of moment of inertia I_t rotating at a constant angular velocity w_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.
(a)Because of friction, rotational kinetic energy is not conserved while the disks' surfaces slip over each other. What is the final rotational kinetic energy, K_f, of the two spinning disks?
Express the final kinetic energy in terms of I_t, I_r , and K_i the initial kinetic energy of the two-disk system. No angular velocities should appear in your answer.
(b)Assume that the turntable deccelerated during time [tex]\Delta[/tex]t before reaching the final angular velocity ([tex]\Delta[/tex]t 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,[tex]\tau[/tex] , acting on the bottom disk due to friction with the record?
Express the torque in terms of I_t, w_i, w_f, and [tex]\Delta[/tex]t
Homework Equations
K_i - K_f = [tex]\Delta[/tex]K lost to friction --->
(1/2)I_1*w_1i^2 + (1/2)I_2 * w_2i^2 = (1/2)I_1 * w_1f^2 + (1/2)I_2 * w_2f^2
The Attempt at a Solution
Ok, I've only attempted part (a) at the moment so i need to concentrate on that first. then i'll move onto part (b)
What is confusing is what the question is asking for. delta K is just the initial - final energies, but where does that stand in the solution they are looking for.
This is what i know --->
K_i = (1/2)I_t * w_i^2
K_f = (1/2)w_f^2(I_t + I_r)
How and where do I get rid of the w's and where is delta K in the solution?