(Figure 1) Consider a turntable to be a circular disk of moment of inertia I, rotating at a constant angular velocity w; 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. 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.

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Part B
Ke=
Because of friction, rotational kinetic energy is not conserved while the disks' surfaces slip over each other. What is the final rotational kinetic energy, KF, of the two spinning disks?
Express the final kinetic energy in terms of It, Ir, and the initial kinetic energy K; of the two-disk system. No angular velocities should appear in your answer.
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A
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Part C
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1 of 1
(Figure 1) Consider a turntable to be a circular disk of
moment of inertia I, rotating at a constant angular velocity
w; 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 Ir. 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.
Assume that the turntable deccelerated during time At before reaching the final angular velocity (At 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, (7), acting on the bottom disk due to friction with the record?
Express the torque in terms of I, ₁, wr, and At.
Transcribed Image Text:Figure Part B Ke= Because of friction, rotational kinetic energy is not conserved while the disks' surfaces slip over each other. What is the final rotational kinetic energy, KF, of the two spinning disks? Express the final kinetic energy in terms of It, Ir, and the initial kinetic energy K; of the two-disk system. No angular velocities should appear in your answer. ▸ View Available Hint(s) 9 A Submit Previous Answers Part C * Incorrect; Try Again; 2 attempts remaining 1 of 1 (Figure 1) Consider a turntable to be a circular disk of moment of inertia I, rotating at a constant angular velocity w; 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 Ir. 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. Assume that the turntable deccelerated during time At before reaching the final angular velocity (At 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, (7), acting on the bottom disk due to friction with the record? Express the torque in terms of I, ₁, wr, and At.
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