A mass m on a friction less table is attached to a hanging mass M by a cord through a hole in the table . Then the angular speed with which m must spin for M stay at rest will be
A
`(1)/(2pi)sqrt((ML)/(mg))`
B
`(1)/(pi)sqrt((Mg)/(mL))`
C
`(1)/(pi) sqrt((ML)/(Mg))`
D
`(1)/(2pi)sqrt((Mg)/(mL))`
Text Solution
Verified by Experts
The correct Answer is:
D
The weight Mg is acting downwards .M will remain stationary if the C.F force `(mv ^2)/(L ) ` acting on m balance the weight mg =`(mv^(2))/(L) = m L omega ^(2) = mL . 4 pi^(2) n^(2) ` ` therefore n^(2)=(mg )/(4pi ^(2) mL ) " " therefore n=(1)/(2pi) sqrt((Mg)/(mL ))`
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