Question

Angular speed of a particle increases from 2rad per sec to 4 rad per sec across any two
diametrically opposite positions. Its angular acceleration will be?
A) 6 rad per square second B)6/Π rad per second square C)Π/6. rad per second square. D)Π rad per second square​

Answer 1

Answer:

6 / pi

OR

3.33

Explanation:

hope below explanation help u

Answer 2

Given:

Angular speed of a particle increases from 2 rad per sec to 4 rad per sec across any two

diametrically opposite positions.

To find:

The angular acceleration of the object.

Calculation:

When an object travels from one diametric end to the other in a circular path , it covers an angular displacement of π radians.

Now , applying formula of rotational kinematics:

$\therefore \: {( \omega2)}^{2} = {( \omega1)}^{2} + 2 \alpha \theta$

$= > \: {( 4)}^{2} = {(2)}^{2} + 2 \alpha (\pi)$

$= > \: 16 = 4 + 2 \alpha (\pi)$

$= > \: 12 = 2 \alpha (\pi)$

$= > \: 6 = \alpha (\pi)$

$= > \: \alpha = \dfrac{6}{\pi} \: rad {s}^{ - 2}$

So, the final answer is:

$\boxed{ \bf{ \alpha = \dfrac{6}{\pi} \: rad {s}^{ - 2} }}$