The base of a record player is a disk that has a mass of 2 kg and a diameter of 34 cm. A child spins it so it rotates at a constant rate of 50 times a minute. Neglect friction and all that stuff that complicates problems! Then, the child drops on a record (also a disk) with mass 1.3 kg and a diameter of 32 cm. How many times a minute will they spin together if the record player base and the record stick together (assume there is no slipping when the record lands on the base)? Remember, for a solid disk or cylinder, I = ½ MR^2.

Question

The base of a record player is a disk that has a mass of 2 kg and a diameter of 34 cm.  A child spins it so it rotates at a constant rate of 50 times a minute.  Neglect friction and all that stuff that complicates problems!  Then, the child drops on a record (also a disk) with mass 1.3 kg and a diameter of 32 cm.  How many times a minute will they spin together if the record player base and the record stick together (assume there is no slipping when the record lands on the base)?  Remember, for a solid disk or cylinder, I = ½ MR^2.

I did it as follows: 1/2(2)(.17^2) * 5.236 rad = 1/2(3.3)(.17^2) * w And got 30 rpm.

Do I have the right values though for the mass and diameter of the final system, and do I need to account for anything else in the conservation of angular momentum expression?

R_scott · Answer

the record has a smaller radius 1/2 * 2 * .17^2 * 50 rpm = 1/2 * ω * [(2 * .17^2) + (1.3 * .16^2)]