Discussion Forum

A wheel of moment of inertia 2 kg m² is rotating about an axis passing through centre and perpendicular to its plane at a speed 60 rad/s. Due to friction, it comes to rest in 5 minutes. The  angular momentum of the wheel three minutes before  it slops rotating is 

Answer:

Explanation:

Given , l= 2 kg m²

 $\omega_{0}$ = 60 rad/s , $\omega$ =0

 t=5 min=5 x60= 300 s

 From the relation, 

$\omega=\omega_{0}+\alpha t$

$\alpha=\frac{\omega-\omega_{0}}{t}$

 $\Rightarrow $   $\alpha=\frac{0-60}{300}=\frac{-60}{300}=\frac{-1}{5}rad/s^{2}$

 for t=2 min

 = 2 x 60=120 s

 $\omega=\omega_{0}+\alpha t$

= $60-\frac{1}{5}\times 120=60-24$

 $\therefore$    $\omega$ =36 rad/s

 As, angular momentum.  L= l $\omega$

 substituting  the values  in the above relation, 

We get  L= 2 x36= 72 kg m2 /s