Question

A bar on a hinge starts from rest and rotates with an angular acceleration α=(10+6t) rad/s^2, where t is in seconds. Determine the angle in radians through which the bar turns in the first 4.00s.

Answer 1

144 rad should be the answer your looking for buddy!

Answer 2

`\theta=144\ rad`

Step-by-step explanation:

given,

α=( 10+6 t ) rad/s²

`\alpha =(d\omega)/(dt)`

`d\omega= \alpha dt`

integrating both side

`\omega= \int (10+6 t )dt`

`\omega=10 t+6(t^2)/(2)`

`\omega=10 t+3t^2`

we know

`\omega =(d\theta)/(dt)`

`d\theta= \alpha dt`

integrating both side

`\theta= \int (10 t+3t^2 )dt`

`\theta=10(t^2)/(2)+3(t^3)/(3)`

`\theta=5 t^2+t^3`

now, at t = 4 s θ will be equal to

`\theta=5* 4^2+4^3`

`\theta=144\ rad`