Calculating angular velocity of rolling object with just gravity? From what I have learned, you can calculate what the angular velocity of an object will be based on its potential energy.
Say there is a situation where:


*

*acceleration due to gravity = 10 m/s²

*friction = infinite (object is in pure rolling motion)

*we know objects current state (velocity, angular velocity, etc..)

*slope angle = 13° above horizontal


And I need to find:


*

*objects angular velocity one second from current time


As you can see, I don't think I can use the potential energy conversion formula (maybe I can, but I am not seeing it).
Is there a way that I could find the 'future' angular velocity (in this case 1 second ahead) by just knowing the current 'state' of the object and that gravity will be affecting it?
Edit:
In my physics simulation, I find that given these values:


*

*gravity = g = 10 m/s²

*friction = μ = infinite

*radius of sphere = r = 0.5m

*angle of incline = θ = 13°


The angular acceleration is roughly 3 radians/s².
I just realized that if I could find the angular acceleration, I could then predict the angular velocity. Could someone help me find a formula to get the angular acceleration, with just knowing these values? 
 A: The moment of a solid sphere:
$I_{sphere}=\frac{2}{5}M{r_{sphere}^2}=0,4M{r_{sphere}}^2$
The torque applied:
$\tau=F_gsin(13)r_{sphere}=10M\times0,2r_{sphere}=0,2M r_{sphere}$ 
Just like $F=ma$, so $a=\frac{F}{m}$ in the linear case, we can write in the circular case:
$\alpha=\frac{\tau}{I}$, where $\alpha$ is the angular acceleration. In this particular case, this becomes:  
$\alpha=\frac{0,2M r_{sphere}}{0,4M{r_{sphere}}^2}=\frac{1}{2r_{sphere}}$ (unit $\frac{rad}{{sec}2}$).
Because $r_{sphere}=0,5$ we get  
$\alpha=1\frac{rad}{{sec}^2}$
Now suppose the sphere rolls (initial velocities zero) for $10sec$ before arriving down the inclined plane. Obviously the sphere has an angular velocity (speed) of $\omega=10\frac{rad}{sec}$. The rotational energy of a body, in this case, a sphere, is
$E_{rot}=\frac{1}{2}I_{sphere}{\omega}^2$
We saw before the moment of inertia, in this case, is $0,4M{r_{sphere}}^2$ which for $r_{sphere}=0,5$ becomes $M$ (unit $\frac{kgm^2}{{sec}^2}$ or $Nm$). So the rotational energy is $5M$.
For evaluating the length of the plane ($l$) we have to compute $l=\frac{1}{2}\alpha t^2$ (like $s=\frac{1}{2}at^2$ in the linear motion case expressed im meters instead radians) so $l=50rad$. One radial corresponds to 57,3 degrees and because $r_{sphere}=0,5m$, $l=50\times57,3\times0,5=1432,5m$.
Now we can calculate the potential energy of the sphere on top of the plane. The height of the plane is $h=lsin(13)=1432,5\times0,2=286,5m$ The mass of the sphere is $M$. $E_{potsphere}=Mgh=2865M(J)$.
That's quite a difference with $5M(J)$! So almost ALL potential energy which the sphere has on the top is transformed in linear kinetic energy ($2860M(J)$) and only $5M(J)$ in rotational energy.
A: I found the angular acceleration using this formula:
-5g/7r * sin θ
where


*

*g = acceleration due to gravity

*r = radius of sphere

*sin θ = angle of incline


Then used that to get velocity by multiplying it by time.
A: Yes, of course you can apply energy conservation or by simple work done by rotating force is given as
The mechanical work applied during rotation is the torque ($\tau$) times the rotation angle ($\theta$): 
$$W = \tau \theta$$
so that the work done is equal to $\frac{1}{2}I\omega^2$. 
