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A particle begins to move with a tangent...

A particle begins to move with a tangential acceleration of constant magnitude `0.6m//s^(2)` in a circular path . If it slips when total acceleration becomes `1m//s^(2)`, then the angle through which it would have turned before it stars to slip is `:`

A

`1//3 rad`

B

`2//3 rad`

C

`4//3 rad`

D

`2 rad`

Text Solution

Verified by Experts

The correct Answer is:
B
`a_(Net)=sqrt(a_(g)^(2)+a_(c)^(2))`
`omega^(2)=omega_(0)^(2)+2omega theta`
`:' omega_(0)=0`
so `omega^(2)=2 alpha theta`
`omega^(2)R=2(alphaR theta)`
`a_(c)=omega^(2)R=2a_(t)theta`
`1=sqrt(0.36+(1.2xxtheta)^(2))`
`rArr 1-0..36=(1.2theta)^(2)`
`rArr (0.8)/(1.2)=theta`
`rArr theta =(2)/(3) radian`
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