Two identical uniform discs roll without slipping on two different surfaces $A B$ and $C D$ starting at $A$ and $C$ with linear speed $v_{1}$ and $v_{2}$ , respectively, and always remain in contact with the surfaces. If they reach $B$ and $D$ with the same linear speed and $v_{1} = 3 m / s$ then $v_{2}$ in $m / s$ is $(g = 10 m / s^{2})$

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Solution

Both the discs having final speed same then,
the energy at point $A$ ,
$K_{t o t a l} = K_{t r} + K_{r o t} + m g l = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2} + m g l = \frac{1}{2} m v^{2} + \frac{1}{2} (\frac{m R^{2}}{2}) (\frac{v}{R})^{2} + m g l$
Since velocity of both the discs are same at B , using conservation of energy,
$\frac{3}{4} m v_{1}^{2} + m g (30) = \frac{3}{4} m v_{2}^{2} + m g (27)$
Putting $v_{1} = 3$ and solving
$v_{2}^{2} = 49$
$\Rightarrow v_{2} = 7 m / s$

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