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 increase from 10 to 20rad/s^2 is​

Answer 1

Answer:-

Given:-

$\alpha = 3 \frac{rad}{ {sec}^{2} }$

$w = 10 \frac{ {rad}^{} }{ {sec}^{} }$

and the final angular velocity is

$20 \frac{rad}{sec}$

As you know kinematics equation can be applied in the case of circular motion.

Therefore,

The equation becomes

${w}^{2} (final) - {w}^{2} (initial) = 2 \alpha (theta)$

substituting the values.

${20}^{2} - {10}^{2} = 2 \times 3 \times (theta)$

$400 - 100 = 6 \times (theta)$

$(theta) = \frac{300}{6}$

$(theta) = 50 \: radians$

$\boxed{\boxed{(theta) =50 radians}}$

so, the angular displacement of particle or the angle swept at the centre is 50 radians when it's angular velocity changes from 10 rad/s to 20 rad/s.