# 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?

• As you are considering both rotational and translational motion... I think one can use the energy conservation, off course in pure rolling. so change in P.E. can be equated to translational +rotational energy. – drvrm Aug 25 '18 at 5:32
• As regards finding angular velocity - one can use I.ang. accn. = torque , pl. look up<farside.ph.utexas.edu/teaching/301/lectures/node108.html> for an intro. – drvrm Aug 25 '18 at 5:36

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.

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 $$\frac{1}{2}I\omega^2$$.

• I tried to improve grammar and formatting, but still don't fully understand what you mean to say. Please correct further to make it clear. – Thomas Fritsch Aug 5 '19 at 10:54
• I simply apply work energy theorm work done by all forces.change in kinetic energy – Yuvraj Aug 5 '19 at 11:34

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.