Consider a turntable to be a circular disk of moment of inertia I t rotating at
ID: 1476495 • Letter: C
Question
Consider a turntable to be a circular disk of moment of inertia It rotating at a constant angular velocity 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 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.
Part A
What is the final angular velocity, f, of the two disks?
Express f in terms of It, Ir, and i.
Part B
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 Ki of the two-disk system. No angular velocities should appear in your answer.
Part C
Assume that the turntable deccelerated during time t before reaching the final angular velocity ( 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, , acting on the bottom disk due to friction with the record?
Express the torque in terms of It, i, f, and t.
Explanation / Answer
Part A:
Angular momentum, given by L = I, is a conserved quantity. Therefore, the angular momentum of the turntable-record system before and after collision will be the same:
L = I(t) (i) = I(t) (f) + I(r) (f)
Solve for (f) to get:
I(t) (i) = [I(t) + I(r)] (f)
(f) = I(t) (i) / [I(t) + I(r)]
Part B:
Rotational kinetic energy is given by K = ½I². Therefore:
K(f) = ½ [I(t) + I(r)] (f)²
Substitute for (f) from part one and simplify:
K(f) = ½ [I(t) + I(r)] { I(t) (i) / [I(t) + I(r)] }²
K(f) = ½ [I(t) + I(r)] I(t)² (i)² / [I(t) + I(r)]²
K(f) = ½ I(t)² (i)² / [I(t) + I(r)]
Now, find (i)² in terms of K(i):
K(i) = ½ I(t) (i)²
(i)² = 2K(i) / I(t)
Substitute this back into the K(f) expression and simplify:
K(f) = ½ I(t)² [2K(i) / I(t)] / [I(t) + I(r)]
K(f) = K(i) I(t) / [I(t) + I(r)].