Question

A playful child spins one of his mother's plastic plates on his air hockey table. The plate is circular with a mass of 3.41 kg and it has a radius of 34.1 cm. It is rotating with an angular speed of 21.0 revolutions per second. As he continues to play, he then carefully drops a second, smaller plate with mass 1.80 kg and radius 18.0 cm directly down on top of the rotating plate such that they line up perfectly. Assuming that we can model the plates as disks (I = (1/2) MR²) and that the friction between the lower plate and the air hockey table is negigible, then what is the final angular speed of the two plates?

Answer 1

Answer:

Given, mass of the first circular plate is M1 = 3.41 kg and its radius R1 = 34.1 cm = 0.341 m and its initial angular spped is 1,i = 21 revolutions/s = 21 (2) rad/s = 131.88 rad/s.

Mass of the second circular plate is M2 = 1.80 kg, radius R2 = 18.0 cm = 0.18 m. The initail angular speed of the second plate is zero.i.e.,2,i = 0 rad/s.

We need to calculate the final or combined angular speed of the two plates. Assuming the plates are disks, so moment of inertia of the disk is I = 1/2 MR2

Using conservation of angular momentum, Linitail = Lfinal

I11,i + I22.i = (I1 + I2) f

I11,i = (I1 + I2) f    Since 2,i = 0 rad/s

(1/2 M1R12) 1,i = ( 1/2 M1R12 + 1/2 M2R22 ) f

or f = (1/2 M1R12) 1,i / [( 1/2 M1R12 + 1/2 M2R22 )]

or f = M1R12 1,i / (M1R12 + M2R22)

         = (3.41 kg) (0.341 m)2 (131.88 rad/s) / [(3.41 kg) (0.341 m)2 + (1.80 kg) (0.18 m)2]

         = (52.29 / 0.454) rad/s

Therefore, f = 115.17 rad/s. The rotation will continue in the same direction.