Two particles A and B move anticlockwise with the same speed v in a circle of radius R and are diametrically opposite to each other. At `t=0`, A is given a constant accelerartion (tangential) `a_(t)=(72v^(2))/(25 pi R)`. Calculate the time in which a collides with B, the angle traced by A, its angular velocity and radial acceleration at the time of collision.

Answer

`(5piR)/(6v), (11pi)/6, (17v)/(5R), (289 v^(2))/(25 R)`

Explanation

According to
`theta=1/2xx(72v^(2))/(25piR)xxt^(2)=piR rArr t=(5pi R)/(6v)`
Using `Rtheta=vt+(1/2)(72v^(2))/(25piR)xx(25pi^(2)R^(2))/(36v^(2))`
`a_(T)=(72v^(2))/(25piR)`
`Rtheta=(v5piR)/(6v)+piR=11/6 pi`
Angular velocity :`omega=omega_(v)+alphat`
`=v/R+(72v^(2))/(25piR^(2))xx(5piR)/(6v)=v/R+(12v)/(5R)=(17v)/(5R)`
Angular acceletration `alpha=omega^(2)R=(289v^(2))/(25R)`