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

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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 =...

Unlock
This Answer Now

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Start your 48-Hour Free Trial

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.

Approved by eNotes Editorial Team