The base of a record player is a disk that has a mass of 2 kg and a diameter of

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 MR2.

Explanation / Answer

To solve this problem you use conservation of angular momentum

L0=Lf

Idisk*disk=final(Idisk+Irecord)   We are trying to find final for now

final = (Idisk*disk) / (Idisk+Irecord)

Idisk=(Mdisk*Rdisk^2)/2 = (2kg*0.34m^2)/2 = 0.1156kg*m2

Irecord=(Mrecord*Rrecod^2)/2 = (1.3kg * 0.16m^2)/2 = 0.01664kg*m2

disk=50rpm

Now we can plug in the values for the variables

final = (0.1156kg*m2* 50rpm) / (0.1156kg*m2 + 0.01664kg*m2)

final = (5.78kg*m2*rpm) / (0.13224kg*m2)

final = 43.7rpm

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