Question

the angular velocity of a particle changes from 69 to 71 rpm in 30 sec.its angular acceleration in rev/min us equal to​

Answer 1

$\huge\underline{\underline{\bf \green{Question-}}}$

The angular velocity of a particle changes from 69 to 71 rpm in 30 sec.its angular acceleration in rev/min us equal to

$\huge\underline{\underline{\bf \green{Solution-}}}$

$\large\underline{\underline{\sf Given:}}$

• Angular velocity ${\sf \omega_1=69\:rpm}$
• ${\sf \omega_2=71\:rpm}$
• Time ( t ) = 30 s

$\large\underline{\underline{\sf To\:Find:}}$

• Angular Acceleration $(\alpha)$

❏ $\omega_1$ in rps ➝

$\implies{\sf \omega_1=69\:rpm }$

$\implies{\sf \omega_1=69×\dfrac{2π}{60} }$

$\implies{\bf \omega_1=2.3π\:rps }$

❏ $\omega_2$ in rps ➝

$\implies{\sf \omega_2=71\:rpm }$

$\implies{\sf \omega_2=71×\dfrac{2π}{60}}$

$\implies{\bf \omega_2=2.36π\:rps }$

✪$\large{\boxed{\bf \blue{\alpha=\dfrac{\triangle \omega}{\triangle t}} }}$

$\implies{\sf \dfrac{(2.36-2.3)π}{30}}$

$\implies{\sf \dfrac{0.06π}{30} }$

$\implies{\sf 0.002π}$

$\implies{\bf \red{\alpha = 2×10^{-3}π\:rad/s^2}}$

$\huge\underline{\underline{\bf \green{Answer-}}}$

Angular Acceleration is ${\bf \red{2×10^{-3}πrad/s^2}}$