Question

A bar on a hinge starts from rest and rotates with an angular acceleration α = 13 + 7t, where α is in rad/s2 and t is in seconds. Determine the angle in radians through which the bar turns in the first 4.69 s.

Answer 1

Step-by-step explanation:

Angular acceleration of the bar,
`\alpha=(13+7t)\ rad/s^2`..........(1)

We need to find the angle in radians through which the bar turns in the first 4.69 s.

We know that,

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

`\omega` is angular velocity

On integrating equation (1) wrt t

`\omega=\int\limits {\alpha .dt }`

`\omega=13t+(7t^2)/(2)`...............(2)

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

On integrating equation (2) wrt t as :

`\theta=\int\limits{\omega.dt}`

`\theta=\int\limits^(4.69)_0 {({13t+(7t^2)/(2)}).dt}`

`\theta=(13t^2)/(2)+(7t^3)/(6)`

At t = 4.69 s

`\theta=(13(4.69)^2)/(2)+(7(4.69)^3)/(6)`

`\theta=263.32\ radians`

Hence, this is the required solution.