Question

Two identical uniform discs roll without slipping on two different surfaces $A B$ and $C D$ (see figures) starting at $A$ and $C$ with linear speeds $v_{1}$ and $v_{2}$, respectively and always remain in contact with the surfaces. If they reach $B$ and $D$ with the same linear speed and $v _{1}=6\, m / s$, then $v _{2}$ in $m / s$ is _____. (Take $g=10\, m / s ^{2}, \sqrt{19}=4.36$ )

Answer 1

Final kinetic energy of both discs is same.
Applying conservation of energy,
$\frac{1}{2} mv _{1}^{2}\left(1+\frac{ K ^{2}}{ R ^{2}}\right)+ mgh _{1}$
$=\frac{1}{2} m v_{2}^{2}\left(1+\frac{K^{2}}{R^{2}}\right)+m g h_{2}$
$\frac{v_{1}^{2}}{2}\left(1+\frac{K^{2}}{R^{2}}\right)+g h_{1}=\frac{v_{2}^{2}}{2}\left(1+\frac{K^{2}}{R^{2}}\right)+g h_{2}$
$\therefore \frac{6^{2}}{2} \times \frac{3}{2}+10 \times 30$
$=\frac{v_{2}^{2}}{2} \times \frac{3}{2}+10 \times 27$
$27+30=\frac{3}{4} v_{2}^{2}$
$\therefore 57 =\frac{3}{4} v_{2}^{2} $
$v_{2}^{2} =\frac{57 \times 4}{3}$
$\therefore v^{2} =76=2 \sqrt{19}$
$\therefore v =8.72\, m / s$