-->

Laws Of Motion

Question
CBSEENPH11016964

A wheel mounted on a stationary axle starts from rest and accelerates with angular acceleration tarannm = 2 + 0.4t rad/s2. Find the angular velocity of wheel after 4 seconds of its start.

Solution
The angular acceleration of the wheel depends on time.
Therefore, equation of kinematics of uniform accelerated motion is not applicable. 
We have here, 
straight alpha space equals space 0 space plus space 0.4 space straight t space rad divided by straight s squared space

therefore space dω over dt space equals space 2 space plus space 0.4 space straight t space

rightwards double arrow space dω space equals space left parenthesis 2 space plus space 0.4 space straight t right parenthesis space dt space

On space integrating space both space sides comma space

integral subscript 0 superscript straight omega dω space equals space integral subscript 0 superscript 4 left parenthesis 2 space plus space 0.4 space straight t right parenthesis space dt space

rightwards double arrow space straight omega vertical line subscript 0 superscript straight omega space equals space left parenthesis 2 straight t space plus space 0.2 space straight t squared right parenthesis vertical line subscript 0 superscript 4 space

rightwards double arrow space straight omega space equals space 11.2 space rad divided by sec
This is the required angular velocity of the wheel after t = 4 seconds.