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For the arrangement in the Figure, the p...

For the arrangement in the Figure, the particle `M_(1)` attached to one end of string which moves on a horizantal table in a circle of radius `l/2` (where `l` is the length of the string) with constant angular speed `omega`. The other end of the string attached to to mass `M_(2)` which rest on a vertical rod. When the rod collapse, the acceleration of mass `M_(2)` at that instant

A

`g`

B

`(omega^(2)l)/2`

C

`(2M_(2)g-M_(1)lomega^2)/(2(M_(1)+M_(2))`

D

`(M_(2)g+M_(1)lomega^(2))/(M_(1)+M_(2))`

Text Solution

Verified by Experts

The correct Answer is:
C
From Newton's second law
`T-M_(1)omega^(2)1/2=M`__(1)alpha and `M_(2)g-T=M_(2)a`
After solving above equations, we get
`a=(2M_(2)g-M_(1)lomega^(2))/(2(M_(1)+M_(2))`
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