Homework Help: Acceleration in a non inertial reference frame

1. Jan 28, 2016

bznm

1. The problem statement, all variables and given/known data

A platform rotates with $\omega=10$ rad/s around $z$-axes. A ball is connected, with a yarn to $z$. Its distance to the axes is 15 cm and it rotates with $\omega=10$ rad/s. There isn't friction between platform and ball. Suddenly, the angular velocity of the platform is reduced to $\omega'=2$ rad/s. Find velocity and acceleration of the ball in the system of the platform.

2. Relevant equations

$\vec{a_0}=\vec{a'}+\vec{a_{cc}}+\vec{a_c}$

where $\vec{a'}$=acceleration calculted in a non inertial reference frame, $\vec{a_{cc}}=2\vec{\omega} \times \vec{v'}$, $\vec{a_c}=-\omega ^2r \vec{u_r}$.

So $\vec{a'}=\vec{a_0}-\vec{a_{cc}}-\vec{a_c}$

3. The attempt at a solution
I have written $a'=\omega^2 r-2 w_r v'+w_r^2r$, where $\omega_r=\omega-\omega'$
But the correct formula is $a'=\omega^2r-2\omega'v'-\omega'^2r$

I dont't understand
1) why $a_c$ has to have opposite sign
2) why in $a_c$ and $a_{cc}$ the angular velocity is the angular velocity of the platform insted of relative angular velocity

Thanks for the help

2. Jan 28, 2016

andrewkirk

If $\vec a'$ denotes the acceleration of the ball in an inertial reference frame, and the ball is rotating at rate $\omega$ at radius $r$, then isn't the magnitude of $\vec a'$ simply $\omega^2r$, ie the centripetal acceleration?

3. Jan 28, 2016

Simon Bridge

Forget the equations for a bit and just use physics and observation:

In a set of coordinates that rotates with the platform:
Initially, the ball has constant zero angular velocity, there is no angular acceleration and no centripetal acceleration (the ball just hangs there) but there is a centrifugal force equal and opposite the tension in the yarn which accounts for how it is stationary. This is because the ball is travelling at the same angular velocity as the platform.

Finally: the ball is no longer travelling at the same angular velocity as the platform.
How fast is the ball moving over the platform?
Does the ball have an angular acceleration wrt the platform?
Does it have a centripetal acceleration wrt the platform?
Is there any centrifugal effect present?

Once you've understood what is going on, you can write it out in maths and that will automatically give you the right equations.

4. Jan 29, 2016

bznm

Thanks, Simon. But I have tried to use observation... But I have done a mistake, so I'm trying to use math to solve... but.. :/

1) wrt platform: with omega given by w_r=w-w'.
2) no, w_r= constant
3) it must have centripetal and centrifugal acceleration, bucause the platform is a non inertial frame..

but, I put in the formula w_r instead of w' (in acc and ac) but it is wrong because I don't have to thing about "angular velocity of the ball wrt the platform" but "angular velocity of the platform wrt inertial reference frame"... isn't it?

@andrewkirk : "If ⃗a′ denotes the acceleration of the ball in an inertial reference frame,..." no, it is acceleration of the ball in NON inertial reference frame

Last edited: Jan 29, 2016
5. Jan 29, 2016

andrewkirk

Which leads me to wonder what $\vec a_0$ is.
As Simon says, you'd be better off forgetting those formulas and just thinking about the physics of what is actually happening. What motion does an observer that occupies a fixed point on the rotating platform observe for the ball? I think the problem is a lot simpler than your formula suggests. For instance:
You know the answer to this, since they are both rotating around the same axis and you know the rotational velocity of each. So what is the rotational velocity of the ball relative to the platform?

6. Jan 29, 2016

bznm

a_0 is acceleration in inertial frame

7. Jan 29, 2016

andrewkirk

That's correct bznm.
Now, as to what you wrote after that:
Why do you think that? In the OP you specifically state that you were asked to find the velocity and acceleration 'in the system of the platform', not in an inertial reference frame.
Velocity and acceleration in a reference frame are just time derivatives of the position coordinate functions. You don't have to worry about whether the reference frame is inertial.

8. Jan 29, 2016

bznm

Maybe I have understood what you meant with: "use physics and observation", but I'm not sure.
In this case, in the rotating frame, I can only see centripetal or centrifugal accelerations. Since the angular velocity of the ball is bigger than the platform one, if I'm on the platform, I see the ball run away. So the total acceleration, seen from the platform, has to be centrifugal. Centrifugal acceleration can be written as $v^2/r$ where v is the ball velocity wrt the platform ($v=(w-w') r$), so $a'=(w-w')^2 r$.

I can obtain the same result if I use the formula $\vec{a'}=\vec{a_0}-\vec{a_{cc}}-\vec{a_c}$, but in this formula, I have to put w' in acc and ac because acc and ac are corrective terms only relating to the rotating frame and so the angular velocity that appear in them is the angular velocity that the rotating frame has wrt an inertial frame. Is it correct?

thanks a lot

Last edited: Jan 29, 2016
9. Jan 29, 2016

BvU

Hello bzn,

Seen your last posting and jump in because that's not how things work. In the rotating frame, even when 'standing still', you feel an acceleration away from the axis of rotation with a magnitude of $\omega^2 r$.

In other words, the lab frame and the rotating frame should conclude to e.g. the same tension in the yarn. A kind of relativity principle/postulate .

10. Jan 29, 2016

BvU

Oops have to correct myself. You want the acceleration, not the force on the yarn.

$a'=(w-w')^2 r$ looks reasonable.

11. Jan 29, 2016

bznm

@BvU I'm sorry, I haven't understood where, in my steps, I have made mistakes. Could you show me?

12. Jan 29, 2016

BvU

Messages crossed. I can't show you because you didn't. I was wrong. Too hasty. Happens sometimes.

And $v=(\omega-\omega') r$ looks good too.

13. Jan 29, 2016

bznm

Oh don't worry!! many thanks for your help! :) Could you tell me if this reasoning is correct?
thanks again

14. Jan 29, 2016

andrewkirk

That looks correct. However, the acceleration is centripetal (towards the z axis) not centrifugal (away from the z axis). As seen by an observer stationary on the platform, the ball is always curving towards the z axis away from its instantaneous direction of motion.

15. Jan 30, 2016

bznm

you are right! It is not easy to imagine the situation! Thanks so much!

16. Jan 30, 2016

J Hann

If that were the case why then does an object (in the reference frame of the earth) projected vertically upwards, not land directly on
the spot from which it was launched?

17. Jan 30, 2016

andrewkirk

The Coriolis effect is sufficient to explain that phenomenon. There is no Coriolis effect in the OP situation because the distance of the ball from the axis of rotation is constant.
Interestingly - if you shoot a projectile directly upwards from the North or South Pole, it will land back exactly where it started - because the distance of the object from the axis of rotation is always zero - hence no Coriolis effect.

18. Jan 31, 2016

J Hann

Admitedly, the example I gave is not pertinent to this situation; but
doesn't the ball acquire velocity perpendicular to the radius (and the angular velocity vector)
when the angular velocity of the platform is suddenly reduced?

19. Jan 31, 2016

andrewkirk

Let t1 be the latest time at which the platform's angular velocity is 10 rad/s, and let t2 be the first time at which its angular velocity is 2 rad/s. Then yes, between times t1 and t2 there is an angular acceleration of the ball in the reference frame of the platform.

But, although worded vaguely on this point, my impression is that the question in the OP is asking about the situation after t2. After t2 the ball has no more angular acceleration in the platform's frame, ie its angular velocity is constant in that frame.

20. Jan 31, 2016

J Hann

I agree that the angular velocity (of the ball) is constant after the platform is slowed but you still have
the cross product of the angular velocity of the platform and the velocity of the ball (perpendicular to the radius)
which gives rise to the Coriolis force.

21. Jan 31, 2016

andrewkirk

That Coriolis force is opposed by the tension in the string. The platform observer would expect the tension in the string to be $mr8^2$ based on their observation of the rotational velocity of the ball at 8 rad/s. But when they measure it, they find that it is $mr(10)^2$. The difference between the two is the Coriolis force. Since that force is exactly offset by the 'excess' tension in the string, $\ddot{r}=0$ in the observer's frame as well as in the inertial frame.