## Cylinder Rolling Down a Slope

As an object rolls down a slope gravitational potential energy is changed into kinetic energy, If the body is not of negligible size the body will also have energy because it is rotating about it;s own centre. This is because the body has inertia
$I$
and such a body rotating with angular velocity
$\omega$
has rotational energy
$\frac{1}{2} I \omega^2$
.
A body starting from rest at the top of a slop, in moving a vertical distance
$h$
, will gain kinetic energy
$\frac{1}{2}mv^2$
and rotational energy
$\frac{1}{2}I \omega^2$
and we can use conservation of energy to write

$mgh= \frac{1}{2}mv^2+ \frac{1}{2}I \omega^2$

For a solid cylinder about an axis about the line of symmetry through its centre
$I=mr^2$
and we can use this and the relationship
$v= \omega r$
to write the above conservation of energy equation as
$mgh= \frac{1}{2}mv^2+ \frac{1}{2}\frac{1}{2}mr^2 )\frac{v}{r}_ \omega^2= \frac{3}{4} mv^2$

For a slope inclined at an angle
$\$
to the horizontal, as it travels a distance
$x$
it falls a distance
$h=x sin \$
. The equations becomes
$mg x sin \ = \frac{3}{4}mv^2 \rightarrow xg sin \ = \frac{3}{4}v^2$
.
Differentiating with respect to
$t$
gives
$vg sin \theta = \frac{3}{2}va \rightarrow g sin \theta =\frac{3}{2}a$