A particle is moving with a constant ang...

A particle is moving with a constant angular acceleration of `4 rad//s^(2)` in a circular path. At time `t=0` , particle was at rest. Find the time at which the magnitudes of centripetal acceleration and tangential acceleration are equal.

`(2)/(3) s`

`1/3s`

`(1)/(4) s`

`(1)/(2) s`

Text Solution

Verified by Experts

The correct Answer is:

Centripetal acceleration `=(v^2)/(r ) = a_(R )`
and tangential acceleration `a= r alpha = a_(T )`
It is given that at time t, `|a_(R ) |=|a_(T )|`
` therefore (v^2)/(r ) = r alpha ` But `alpha = 4 rad //s^(2)`
`therefore v^(2) = r^(2)alpha =4r ^(2) " " therefore v=2r`
But ` v= u+ at and u=0 `
`therefore v= at and a = r omega therefore v= r alpha t= 4 rt `
` therefore ` From (1) and (2)
`4 r t = 2r " " therefore t=(1)/(2 ) s`

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