Question

10. A wheel revolving at 300 rpm is slowing down uniformly to 240 rpm while making 20
revolutions. Find () angular acceleration (1) time taken
​

Answer 1

Given:

A wheel revolving at 300 rpm is slowing down uniformly to 240 rpm while making 20

revolutions.

To find:

• Angular acceleration
• Time taken.

Calculation:

$\omega1 = 300 \: rpm = \dfrac{300 \times 2\pi}{60} = 10\pi \: rad {s}^{ - 1}$

$\omega2 = 240 \: rpm = \dfrac{240 \times 2\pi}{60} = 8\pi \: rad {s}^{ - 1}$

Applying equations of kinematics ;

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

$= > {(10\pi)}^{2} = {( 8\pi)}^{2} + 2 \alpha (20 \times 2\pi)$

$= > 100 {\pi}^{2} = 64{\pi}^{2} + 80\pi \alpha$

$= > 36 {\pi}^{2} = 80\pi \alpha$

$= > 36\pi = 80\alpha$

$\boxed{ = > \: \alpha = \dfrac{9\pi}{20} \: rad {s}^{ - 2} }$

Let time taken be t ;

$\omega2 = \omega1 + \alpha t$

$= > 10\pi = 8\pi + \alpha t$

$= > 2\pi = \alpha t$

$= > 2\pi = ( \dfrac{9\pi}{20} ) t$

$= > \: t = \dfrac{40}{9}$

$\boxed{= > \: t = 4.44 \: sec}$

Hope It Helps.