The bob of the pendulum shown in figure describes an arc of a circle in a vertical plane. If the tension in the cord is `2.5` times the weight of the bob for the position shown. Find the velocity and the acceleration of the bob in that position.

Options

A. `16.75ms^(-1) , 5.66ms^-2 `
B. `5.66ms^(-1) , 16.75ms^-2 `
C. `2.88ms^(-1) , 16.75ms^-2 `
D. `5.66ms^(-1) , 8.34ms^-2 `

Answer

A, B

Explanation

`2.5mg-mgcos30^(@)=(mv^(2))/(r)`
:. `11.63g=(v^(2))/(r)=(v^(2))/(2)`
`v=5.66m//s`
`F_(n et)=sqrt((2.5mg)^(2)+(mg)^(2)+(2)(2.5mg)(mg)cos150^(@))`
`=1.7mg`
`a_(n et)=(F_(n et))/(m)=1.7g~~16.75m//s^(2)`