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Q.
A solid cylinder rolls up an inclined plane of inclination $\theta$ with an initial velocity $v$. How far does the cylinder go up the plane?
System of Particles and Rotational Motion
Solution:
Let the cylinder go up the plane upto a height $h$. Let $M$ and $R$ be the mass and radius of the cylinder respectively. According to law of conservation of mechanical energy, we get
$\frac{1}{2}Mv^2 + \frac{1}{2} I\omega^2 = Mgh$
$\frac{1}{2}Mv^{2} +\frac{1}{2}\frac{MR^{2}}{2} \omega^{2} = Mgh $
$\left(\because {\text{For a solid cylinder}}, I= \frac{1}{2}MR^{2}\right) $
$\frac{1}{2}Mv^{2} +\frac{1}{4}MR^{2}\omega^{2} = Mgh $
$ \frac{1}{2}Mv^{2}+\frac{1}{4}Mv^{2} = Mgh \quad\left(\because v= R\omega\right) $
$ \frac{3}{4}Mv^{2} = Mgh$
$ h= \frac{3v^{2}}{4 g}\quad...\left(i\right)$
Let $s$ be distance travelled by the cylinder up the plane.Then
$sin \,\theta = \frac{h}{s}$ or $s = \frac{h}{sin\, \theta} = \frac{3\, v^{2}}{4\, g\, sin\, \theta} $ (Using$(i)$)
