Question

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 $\text{m/s}$) is
[Take $g=10{\text{m/s}}^2$]

Answer 1

As they roll without slipping, this means work done by friction i.e $W_f=0$
Condition for pure rolling,
$v=r\times \omega \Rightarrow \omega =\frac{v}{r}$
Hence, we can apply conservation of mechanical energy.
For surface $AB$, taking reference of $PE=0$ at $B$
$M.E_i=M.E_f$
$(K{E}_{\text{Rot}}{)}_i+(K{E}_{\text{Trans}}{)}_i+P{E}_i=(K{E}_{\text{Rot}}{)}_f+(K{E}_{\text{Trans}}{)}_f+P{E}_f$
$\Rightarrow \frac{1}{2}I{\omega }^2+\frac{1}{2}m{v}_1^2+mgh=\frac{1}{2}I{\omega }^{\prime 2}+\frac{1}{2}m{v}^{\prime 2}$
$\Rightarrow (\frac{1}{2}\times \frac{m{r}^2}{2}\times \frac{v_1^2}{r^2})+\frac{1}{2}m{v}_1^2+mgh=\frac{1}{2}\times \frac{m{r}^2}{2}\times \frac{v^{\prime 2}}{r^2}+\frac{1}{2}m{v}^{\prime 2}+0$
[$I=\frac{m{r}^2}{2}$ for disc.]
$\Rightarrow 3v_1^2+4gh=3v^{\prime 2}...\left(1\right)$
where $v_1=3\text{m/s},h=30\text{m}$
$\Rightarrow 3\times (3{)}^2+4\times 10\times 30=3\times (v^{\prime }{)}^2$
$\Rightarrow v^{\prime 2}=409...\left(2\right)$

Similarly for surface $CD$, [Taking refrence of $PE=0$ at $D$], since discs are identical, from Eq.$\left(1\right)$ we can write,
$3v_2^2+4g{h}^{\prime }=3v^{\prime 2}$
Final velocity is same for both discs i.e $v^{\prime }$. Substituting from Eq.$\left(2\right)$,
$3(v_2{)}^2+4\times 10\times 27=3\times 409$
$\because h^{\prime }=27\text{m}$
$\Rightarrow v_2^2=409-360$
$\Rightarrow v_2^2=49$
$\therefore v_2=7\text{m/s}$
The speed $v_2$ is $7\text{m/s}$