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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.

User Duke Nuke
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1 Answer

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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.

User Marco Ferrari
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