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This isn't a homework question, but it might as well be. The problem I have been pondering is:

If a disc (or children's roundabout if you like), of radius r, mass m, is spun around it's center with an initial force F, and thereafter there is the friction force (of either the axle or air resistance or both) of f, then how long will it take to come to a stop?

I have thought about it and have come up with not much. My first way is thus:

$F = ma$, so

$a = F/m$, the initial acceleration ( or should that be $(F-f)/m$ ?).

And then the deceleration is $a = -f/m$.

I'm not sure how to calculate the initial linear velocity, but assuming I have it, $u$, say, then I could say that after time $t$ the velocity is $v = u -at = u -ft/m$, where f is the friction force.

So then the disc would stop spinning when $ t= um/f$. I am aware that this is wrong (well, it might work if we were dealing with linear motion). Straight away it seems wrong because it doesn't take into account the radius of the disc and also the slow down seems linear, when from observation it seems rotating discs slow down and taper off to a standstill. But that is as far as I got. I have tried to use angular motion equations (well $\omega r = v$) but I am stuck at this point, and of course, finding the initial velocity.

Any help is appreciated.

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closed as too localized by dmckee May 31 '13 at 17:02

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

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Try using $\int{d\vec{r} \times \vec{F}} = I \alpha$, were $I$ is the inertia and $\alpha$ the angular acceleration –  Angel Joaniquet Tukiainen May 27 '13 at 15:42
    
Hi JJG, I added the homework tag because it seems you want to be walked through the problem (and because it's mostly a calculation and not conceptual question). See our homework policy for why this applies: meta.physics.stackexchange.com/questions/714/… –  Brandon Enright May 27 '13 at 18:04
    
it actually does help, and for work needs - I need to animate a rotating object. –  alex440 May 27 at 11:31

2 Answers 2

up vote 1 down vote accepted

Assuming 1. your body was given an initial speed, not Force (because forces act over time). I am calling it: $$\dot\theta_0$$ 2. The friction force remains constant. 3. The dis is spinning along its symmetry axis and is full

If a body is spinning then it would remain spinning (1st law of newton).

The constant force is slowing down the rotation of the disc, but to know how long it would take you must know how much the force is needed to change the speed.

You must take in to account the moment of inertia. The moment of inertia is usually a matrix and depends on the directions, but since you have a disk, its easy to find its moment of inertia, assuming the disc is uniform (as in, its not a ring, this matter), its moment of inertia is

$$ I = \frac{m r^2}{2}$$

taken from here, note it matters in which direction it spins.

From this point you can treat this as a free-fall problem where

$$ mg \rightarrow F $$ $$ m \rightarrow I $$ $$ v_0 \rightarrow \dot\theta_0 $$

We solve as we would:

$$ v_0 - at^2 =0 $$ or $$ \dot\theta_0 - \frac{F}{I}t^2 =0 $$

so $$ t =\sqrt{\frac{\dot\theta_0 I}{F}} $$

I am not 100% I didn't miss anything, so please comment so I can fix before down-voting.

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There are a few subtle complications here.

  1. You have the right ideas, but you need the angular forms of all these variables and equations. A spinning object has an angular velocity $\omega$. Its tendency to resist change in $\omega$ is known as its moment of inertia $I$, which Wikipedia tells us is $mr^2/2$ if all the mass is in a uniform disk being spun about its axis of symmetry. Torque $\tau$ is the angular version of force, and it is basically the force times the lever-arm advantage the force has given that is is being applied at some distance off axis. Newton's second law becomes $\tau = I \dot{\omega}$.

  2. Now you speak of an initial force $F$. In order to cause rotational motion, this force must be applied a nonzero distance $d$ from the axis of desired rotation. Only the component of $F$ perpendicular to both the rotation axis and the line from this axis to the point of application will help here, so let's assume all the force is applied in this direction. Then this can be described as a torque of magnitude $Fd$.

  3. However, an initial, instantaneous torque does nothing to move the disk, just as any finite, instantaneous linear force does not move an object. The force must be applied over an extended time. One way to see this is that the product of force and the distance over which it is applied is the imparted energy - if nothing moves during the application of the force, there is no energy given to the object. Instead, we can speak of an impulse. If the angular momentum is $L = I\omega$, then the initial angular momentum obeys $L_0 = Fd\Delta t$ for a force $F$ applied at a distance $d$ from the axis, in the proper direction, and sustained over an interval of time $\Delta t$. Clearly then the initial angular velocity is $$ \omega_o = \frac{2Fd\Delta t}{mr^2}. $$

  4. Now we come to the slowing down part. This is much more complicated. An oversimplified model would have a constant force $f$ applied at a nonzero distance $x$ from the axis of rotation. Then $$ \dot{\omega} = \frac{2fx}{mr^2} $$ (up to a sign). In this case we would have something like $\omega = \omega_0 - \dot{\omega}t$, so the total time spent spinning would be $$ \frac{\omega_0}{\dot{\omega}} = \frac{Fd\Delta t}{fx}. $$

  5. Of course, the friction may be velocity-dependent. Also, it is almost certainly applied over a range of distances from the axis. This latter point simply means the total torque acting to slow the disk must be computed by integrating the torque per unit area (force per unit area, which is presumably constant, times distance from axis) over the contact area. The velocity dependence means $\dot{\omega}$ will depend on $\omega$. If the dependence is simple, the result will be a differential equation that can be solved easily. If it is more complicated, numerical techniques will have to be employed, as there won't be a simple formula.

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