Discussion Forum

1)

 A uniform circular disc of mass 1.5 kg and radius 0.5m  is initially at rest on a  horizontal frictionless surface. Three forces of equal magnitude  F=0.5N  are applied simultaneously along the three sides of an equilateral triangle XYZ with its vertices on the perimeter of the disc (see figure) . One second after applying the forces, the angular speed of the disc in rad s^-1 is 

Answer:

Option D

Explanation:

  Angular impluse = change in angular momentum

  $\therefore$      $\int_{}^{} \tau dt=I\omega$

    $\Rightarrow$    $ \omega=\frac{\int_{}^{}\tau dt }{I}$

$=\frac{\int_{}^{}3F \sin 30^{0}R dt }{I}$

 substituting the values, we have

$\omega=\frac{3(0.5)(0.5)(0.5)(1)}{\frac{1.5(0.5)^{2}}{2}}=2rad/s$