Two identical uniform discs roll without slipping on two different surfaces as shown in figure. If they reach points $B$ and $D$ with the same linear speed, then $v_{2}$ (in $m/s$ ) is
[Take $g = 10 {m/s}^{2}$ ]

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Solution

As they roll without slipping, this means work done by friction i.e $W_{f} = 0$
Condition for pure rolling,
$v = r \times ω \Rightarrow ω = \frac{v}{r}$
Hence, we can apply conservation of mechanical energy.
For surface $A B$ , taking reference of $P E = 0$ at $B$
$M . E_{i} = M . E_{f}$
$(K E_{Rot})_{i} + (K E_{Trans})_{i} + P E_{i} = (K E_{Rot})_{f} + (K E_{Trans})_{f} + P E_{f}$
$\Rightarrow \frac{1}{2} I ω^{2} + \frac{1}{2} m v_{1}^{2} + m g h = \frac{1}{2} I ω^{' 2} + \frac{1}{2} m v^{' 2}$
$\Rightarrow (\frac{1}{2} \times \frac{m r^{2}}{2} \times \frac{v_{1}^{2}}{r^{2}}) + \frac{1}{2} m v_{1}^{2} + m g h = \frac{1}{2} \times \frac{m r^{2}}{2} \times \frac{v^{' 2}}{r^{2}} + \frac{1}{2} m v^{' 2} + 0$
[ $I = \frac{m r^{2}}{2}$ for disc.]
$\Rightarrow 3 v_{1}^{2} + 4 g h = 3 v^{' 2} . . . (1)$
where $v_{1} = 3 m/s, h = 30 m$
$\Rightarrow 3 \times (3)^{2} + 4 \times 10 \times 30 = 3 \times (v^{'})^{2}$
$\Rightarrow v^{' 2} = 409 . . . (2)$

Similarly for surface $C D$ , [Taking refrence of $P E = 0$ at $D$ ], since discs are identical, from Eq. $(1)$ we can write,
$3 v_{2}^{2} + 4 g h^{'} = 3 v^{' 2}$
Final velocity is same for both discs i.e $v^{'}$ . Substituting from Eq. $(2)$ ,
$3 (v_{2})^{2} + 4 \times 10 \times 27 = 3 \times 409$
$∵ h^{'} = 27 m$
$\Rightarrow v_{2}^{2} = 409 - 360$
$\Rightarrow v_{2}^{2} = 49$
$∴ v_{2} = 7 m/s$
The speed $v_{2}$ is $7 m/s$

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