Discussion Forum

1)

 Two identical uniform discs roll without slipping on two different surfaces AB and CD  (see figure) starting at A and C with linear speeds  v₁ and v₂, respectively, and always remain in contact with the surfaces. If they  reach B and D with the same linear speed  and v₁ =3 m/s, then v₂   in m/s is (g=10 m/s² )

 

Answer:

Option D

Explanation:

 In case of pure rolling, mechanical energy remains constant (as work-done by friction is zero) . Further in case of a disc.

  translational kinetic energy/  rotational kinetic energy =   $\frac{K_{T}}{K_{R}}$

              $=\frac{\frac{1}{2}mv^{2}}{\frac{1}{2}I\omega^{2}}$

         =$\frac{mv^{2}}{\left(\frac{1}{2}mR^{2}\right)\left(\frac{V}{R}\right)^{2}}=\frac{2}{1}$

  or      $K_{r}=\frac{2}{3}$                              ( Total kinetic energy)

 or Total Kinetic energy

                           $K=\frac{3}{2}K_{r}=\frac{3}{2}\left(\frac{1}{2}mv^{2}\right)=\frac{3}{4}mv^{2}$

Decrease in potential energy = increase in kinetic energy

      or        $mgh=\frac{3}{4}m\left(v_{f}^{2}-v_i^2\right)$

  or      $v_{f}=\sqrt{\frac{4}{3}gh+v_{i}^{2}}$

  As final velocity in both cases is same

 So, value of   $\sqrt{\frac{4}{3}gh+v_{i}^{2}}$ should be same in both cases.

$\therefore$   $\sqrt{\frac{4}{3}\times10\times 30+(3)^{2}}=\sqrt{\frac{4}{3}\times10\times 27+(v_{2})^{2}}$

                        $v_{2}=7 m/s$