Question

The angular speed of a motor increases from $600$ rpm to $1200$ rpm in $10$s. What is the angular acceleration of the motor:

• A. $600radse{c}^{-2}$
• B. $60\pi radse{c}^{-2}$
• C. $60radse{c}^{-2}$
• D. $2\pi radse{c}^{-2}$

Answer 1

The correct option is D $2\pi radse{c}^{-2}$

Similar to expression used $v=u+at$

We use $\omega ={\omega }_o+\alpha t$ (for rotation)

$\omega =1200rpm$ (rotation per minute)

=$1200\times \frac{2\pi }{60}rad/sec$

= $40\pi rad/sec$

${\omega }_o=600\times \frac{2\pi }{60}$

= $20\pi rad/sec$

Now, $40\pi =20\pi +\alpha \times 10$

$\therefore \alpha =\frac{40\pi -20\pi }{10}$

= $2\pi rad/se{c}^2$