Question

A uniform wheel of mass $10.0kg$ and radius $0.400m$ is mounted rigidly on a massless axle through its center. The radius of the axle is $0.200m$, and the rotational inertia of the wheel-axle combination about its central axis is $0.600kg.m^2$.The wheel is initially at rest at the top of a surface that is inclined at angle $\theta ={30.0}^o$ with the horizontal; the axle rests on the surface while he wheel extends into a groove in the surface without touching the surface. Once released, the axle rolls down along he surface smoothly and without slipping. When the wheel-axle combination has moved down the surface by $2.00m$, what are its rotational kinetic energy?

Answer 1

As the wheel -axle system rolls down the inclined plane by s distance d, the change in potential energy is $△U=-mgd\sin \theta$. BY energy conversion, the total kinetic energy gained is
$-△U=△K=△K_{trans}+△K_{rot}\Rightarrow mgd\sin \theta =\frac{1}{2}m{v}^2+\frac{1}{2}I{w}^2$.
Since the axle rolls without slipping, the angular speed is given by $w=v/r$, where $r$ is the radius of the axle. THe above equation then becomes
$mgd\sin \theta =\frac{1}{2}I{w}^2(\frac{m{r}^2}{I}+1)=△K_{not}(\frac{m{r}^2}{I}+1)$.
(a)With $m=10.0kg,d=2.00m,r=0.200mandI=0.600kg.m^2,$ the rotational
kinetic energy may be obtained as
$△K_{rot}=\frac{mgd\sin \theta }{\frac{m{r}^2}{I}+1}=\frac{\left(10.0kg\right)\left(9.80m/s^2\right)\left(2.00m\right)\sin {30.0}^o}{\frac{\left(10.0kg\right)\left(0.200m\right)}{0.600kg.m^2}}+1$