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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?

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  • $\begingroup$ 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. $\endgroup$ – drvrm Aug 25 '18 at 5:32
  • $\begingroup$ 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. $\endgroup$ – drvrm Aug 25 '18 at 5:36
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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.

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

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  • $\begingroup$ 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. $\endgroup$ – Thomas Fritsch Aug 5 '19 at 10:54
  • $\begingroup$ I simply apply work energy theorm work done by all forces.change in kinetic energy $\endgroup$ – Yuvraj Aug 5 '19 at 11:34
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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.

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