To understand how to use conservation of angular momentum to solve problems invo
ID: 1478606 • Letter: T
Question
To understand how to use conservation of angular momentum to solve problems involving collisions of rotating bodies. 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_x. 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? 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? Assume that the turntable deccelerated during time Delta t before reaching the final angular velocity ( Delta 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, (T), acting on the bottom disk due to friction with the record?Explanation / Answer
Part A:
By the law of conservation of angular momentum,
initial angular momentum of the two discs = final angular momentum of the two discs
=> It * i + 0 = [It + Ir] * f
=> f = i * [It / (It + Ir)]
Part B:
Final rotational KE
KEf = (1/2) [It + Ir] * (f)2
KEf= (1/2) [It + Ir] * (It*i/(It+Ir))2
KEf= KEi*It/[It + Ir]
Part C:
Average angular deceleration
= [f - i] / (t)
=> average torque on the bottom disc
= (It) * [f - i] / (t).
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