ring and string

2 Answers

we have same tension in both parts ( as it is the same string)

cancelling forces in horizontal drirection

T+TcosÎ¸=mw²(3L)

TsinÎ¸=mg

dividing

(1+cosÎ¸)/sinÎ¸ =w²(3L/g)

further cosÎ¸=3/5 sinÎ¸=4/5

2=w²(3L/10)

w²=20/3L

we get w and hence all the value of speed tension etc.

ROHAN

PLZ SEE THREAD TENSION!!!!!!!!!!!!!!..............THERE WE HAVE DISCUSSED......TENSION IN THE STRING!...........IT SHOULD BE DIFFERENT!

AFTER READING IT POST HERE!

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