1. A 3 metre piece of string is threaded through a hole in the table. If equal portions of the string are above and below the table, and the top mass of 2kg rotates at 1rev/sec , how fast does the bottom mass of 1kg rotate? (in rev/sec)

Question

1. A 3 metre piece of string is threaded through a hole in the table. If equal portions of the string are above and below the table, and the top mass of 2kg rotates at 1rev/sec , how fast does the bottom mass of 1kg rotate? (in rev/sec)

2. If, in the above question, the two masses rotate, such that one is directly underneath the other at 1rev/sec , find the lengths of the two parts of the string

Thanks so much in advance :) !!!

Answer 1

1. For equilibrium, there must be equal centripetal force on each mass. The means MRw^2 is the same for each. For equal string lengths from the hole (1.5 m each), the bottom mass must have w^2 twice as high, so that is rotates at sqrt2 = 1.414 rpm

2. Same w in this case. Therefore the product M*R must be the same for each mass.
The lower mass would have twice the length, so it would have to be 2 meters. (The top mass would then be 1 meter)

1. A 3 metre piece of string is threaded through a hole in the table. If equal portions of the string are above and below the table, and the top mass of 2kg rotates at 1rev/sec , how fast does the bottom mass of 1kg rotate? (in rev/sec)

1. For equilibrium, there must be equal centripetal force on each mass. The means M*R*w^2 is the same for each. For equal string lengths from the hole (1.5 m each), the bottom mass must have w^2 twice as high, so that is rotates at sqrt2 = 1.414 rpm

1. For equilibrium, there must be equal centripetal force on each mass. The means MRw^2 is the same for each. For equal string lengths from the hole (1.5 m each), the bottom mass must have w^2 twice as high, so that is rotates at sqrt2 = 1.414 rpm