A rolling wheel of $12$ kg is on an inclined plane at position $P$ and connected to a mass of $3$ kg through a string of fixed length and pulley as shown in figure.
Consider $PR$ as friction free surface.
The velocity of centre of mass of the wheel when it reaches at the bottom $Q$ of the inclined plane $PQ$ will be $\dfrac12\sqrt{xgh}\ \text{m s}^{-1}$. The value of $x$ (rounded off to the nearest integer) is ______.
Numerical value type. Enter your answer.
Answer: 3
Both faces make angle $\alpha$ with the horizontal. When the wheel rolls from $P$ down to $Q$ (a drop $h$, distance $h/\sin\alpha$), the 3 kg block is pulled the same distance up $PR$, rising $h$. Both move with the same speed $v$.
Take the wheel as a ring, $I = MR^2$, so its kinetic energy is $Mv^2$:
$$12gh - 3gh = 12v^2 + \frac12(3)v^2 = 13.5v^2$$
$$v^2 = \frac{9gh}{13.5} = \frac{2gh}{3} \Rightarrow v = \frac12\sqrt{\frac83 gh}$$
$x = \dfrac83 \approx 2.67$, which rounds to $3$. (Treating the wheel as a uniform disc gives $x = \dfrac{24}{7} \approx 3.4$, which also rounds to $3$.)
Solution by Sreeraj P, M.Sc Physics