Question

A body rotates about a fixed axis with an angular acceleration of $ 3rad/{s^2} $ . The angle rotated by it during the time when its angular velocity increases from 10 rad/s to 20 rad/s (in radians) is
(A) 50
(B) 150
(C) 200
(D) 100

Answer 1

Hint: The rotational equivalent of the equation of motion could be utilized to solve the problem. This can simply be done by replacing all the linear quantities in the equation with their rotational counterpart.

Formula used: In this solution we will be using the following formulae;
 $ {\omega ^2} = \omega _0^2 + 2\alpha \theta $ where $ \omega $ is the final angular speed of a rotating body, $ {\omega _0} $ is the initial angular speed of the body, $ \alpha $ is the angular acceleration of the body, and $ \theta $ is the angular displacement.
 $ {v^2} = {u^2} + 2as $ where $ v $ is the final linear velocity, $ u $ is the initial, $ a $ is the linear acceleration and $ s $ is the distance travelled.

Complete step by step answer:
To solve the above problem, one can use the rotational equivalent of the third equation of motion which can be written as
 $ {\omega ^2} = \omega _0^2 + 2\alpha \theta $ where $ \omega $ is the final angular speed of a rotating body, $ {\omega _0} $ is the initial angular speed of the body, $ \alpha $ is the angular acceleration of the body, and $ \theta $ is the angular displacement.
Hence, from the question, we can insert all known values, as in
 $ {20^2} = {10^2} + 2\left( 3 \right)\theta $
Hence, by calculating, and making $ \theta $ the subject of the formula, we have that
 $ 400 = 100 + 2\left( 3 \right)\theta $
 $ \theta = \dfrac{{400 - 100}}{{2\left( 3 \right)}} = \dfrac{{300}}{6} = 50rad $
Hence, the angular displacement is 50 rad
Thus, the correct option is A.

Note:
For clarity, we can get the rotational equation of motion by simply substituting the rotational counterparts of all the linear quantities in place of the linear quantities. For example, for the third equation of motion, we have
 $ {v^2} = {u^2} + 2as $ where $ v $ is the final linear velocity, $ u $ is the initial, $ a $ is the linear acceleration and $ s $ is the distance travelled.
Hence, we replace $ v $ with $ \omega $ , $ u $ with $ {\omega _0} $ , $ a $ with $ \alpha $ , and $ s $ with $ \theta $ . Hence, we have
 $ {\omega ^2} = \omega _0^2 + 2\alpha \theta $ .