# Question #738fe

Dec 5, 2016

Given

$m \to \text{mass of the hoop} = 1 k g$

$r \to \text{radius of the hoop} = 2 m$

$h \to \text{height of the inclined plane} = 100 m$

$H \to \text{initial height of the center of mass}$
$= h + r = \left(100 + 2\right) m = 102 m$

Let

$\omega \to \text{angular velocity of the hoop at bottom}$

$v \to \text{linear velocity of the hoop at bottom}$

$\text{the moment of inertia of the hoop } I = m {r}^{2}$

$g \to \text{acceleration due to gravity} = 9.8 m {s}^{-} 2$

The initial PE of the hoop $m g H$

The KE of the hoop at the bottom will be equal to the sum of its traslational and rotational KE .This total KE should be equal to its initial PE by the law of conservation energy. So

$\frac{1}{2} m {v}^{2} + \frac{1}{2} I {\omega}^{2} = m g H$

$\implies \frac{1}{2} m {v}^{2} + \frac{1}{2} m {r}^{2} {\omega}^{2} = m g H$

$\implies \frac{1}{2} m {v}^{2} + \frac{1}{2} m {v}^{2} = m g H$

$\implies {v}^{2} = g H$

$\implies v = \sqrt{g H} = \sqrt{9.8 \times 102} \approx 31.6 m {s}^{-} 1$
But I think, this is not correct the height covered by the hoop should be $h = 100 m$ because when the hoop reaches at the bottom of the inclined plane its center of mass remains at 2m height.
$v = \sqrt{g h} = \sqrt{9.8 \times 100} = 31.3 m {s}^{-} 1$