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A mass attached to one end of a string crosses top - most point on a vertical circle with critical speed. Its centripetal acceleration when string becomes horizontal will be (where, g=gravitational acceleration)A. glancing angleB. 3gC. 4gD. 6g

Answer» Correct Answer - B
We know that,
velocity of particle at top - most point on vertical circle,
`(v_(top))=sqrt(3rg)` . . . (i)
But centripetal acceleration `(a_(c))=(v^(2))/(r)`
`:." "a_(c)(sqrt(3rg)^(2))/(r)" "` [from Eq. (i)]
`a_(c)=(3rg)/(r)`
`a_(c)=3g`


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