# A drill starts from rest. After 3.54 s of constant angular acceleration, the drill turns at a rate of 2668 rad/s. Find the drill's angular acceleration and the angle through which the drill rotates during this period The relations between angular velocity, angular acceleration and angular displacement are the same as those for their linear equivalents.

If an object starts with an angular velocity u and there is a constant rate of angular acceleration of A, the angular velocity after t seconds is given as v =...

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The relations between angular velocity, angular acceleration and angular displacement are the same as those for their linear equivalents.

If an object starts with an angular velocity u and there is a constant rate of angular acceleration of A, the angular velocity after t seconds is given as v = u + A*t

Here, the initial angular velocity is 0. After 3.54 s, the angular velocity is 2668 rad/s.

This gives: 2668 = A*3.54

The angular acceleration is 2668/3.54 = 753.67 rad/s^2

The angular displacement in this duration is u*t + (1/2)*A*t^2

=> (1/2)*(2668/3.54)*(3.54)^2 = 1334*3.54 = 4722.36 radians.

The angular displacement of the drill is 4772.36 radians.

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