Question

A body rotates about a fixed axis with an angular acceleration of 3 rad/s square. The angular rotated by it during the time when its angular velocity increases from 10 rad / s to 20 rad /s (in angular)is

Answer 1

$\huge{\bold{\underline{\underline{Correct question:- }}}}$

A body rotates about a fixed axis with an angular acceleration of 3 rad/s square. The angle rotated by it during the time when its angular velocity increases from 10 rad / s to 20 rad /s (in angleis)

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

Let the angle rotated by the body be θ.

$\large{ \alpha = 3 \: rad/sec^2}$

$\large{ \omega_o = 10 \: rad/sec}$

$\large{ \omega = 20 \: rad/sec}$

$\huge{\bold{\underline{Explanation:-}}}$

As we know third Kinematic equation I.e.

$\large{\bold{v^2 - u^2 = 2as}}$

Converting it to angular form,

$\large{\bold{ \omega^2 - \omega_o^2 = 2 \alpha \theta}}$

Now, Substituting the values,

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

$\large{400 - 100 = 6 \times \theta}$

$\large{300 = 6 \times \theta}$

$\large{ \theta = \frac{300}{6}}$

$\large{ \theta = \frac{ \cancel{300}}{ \cancel{6}}}$

$\huge{\boxed{\boxed{ \theta = 50 \degree}}}$

Note:-

$\large{ \star \: \omega_o}$ = Initial angular velocity.

$\large{ \star \: \omega}$ = Final angular velocity.

$\large{ \star \: \alpha}$ = Angular acceleration.

$\large{ \star \: \theta}$ = Angle rotated by the body.

Answer 2

$\huge{\underline{\underline{\mathbb{\pink{ANSWER:-}}}}}$

$\bold\purple{According\:to\:question:-}$

Body rotates with angular accelaration, α = 1 rad/s²

Angular velocity ω = 5 rad/s

Change in angular velocity ω' = 15 rad/s

Now we have to find θ = ?

$\bold\purple{From\:the\:above\:condition\:we\:have:-}$

$w {}^{'2} = w {}^{2} + 2a \: θ$

15² = 5²+ 2 × 1 × θ

2θ =225-25

2θ=200

θ = 100 rad .