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A disk having initial angular velocity $\omega$ is gently placed on a rough horizontal surface. What is the angular velocity of rotation when pure rolling starts?

I've tried applying conservation of angular momentum about the axis passing through the point of contact of the ring and the surface but it gives me the same angular velocity in the answer. But when I applied $F=ma$ and $\text{torque}=\text{moment of inertia}\cdot\text{angular accelerations}$ and the condition for pure rolling, I got the right answer that is $\omega/3$. So what is the problem with the principle of conservation of angular momentum?

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    $\begingroup$ Some of the angular momentum goes into the ground. $\endgroup$ – ja72 Jan 5 '16 at 14:43
  • $\begingroup$ I believe you must have missed some terms in your alternative approach - but without seeing your work it's impossible to know where you went wrong. Incidentally, if the "correct answer" is $\omega/3$ then perhaps you were using a disk ($I=\frac12 m r^2$) and not, as stated in the problem, a ring ($I = m r^2$)? Is that possible? $\endgroup$ – Floris Jan 5 '16 at 18:23
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The question implies that when the disc is initially put down it is not rolling without slipping yet. In fact, at $t=0$ it's not yet rolling at all.

Simple conservation of momentum doesn't apply here because friction energy is not conserved.

The disc exerts a force $mg$ (its weight) on the surface, which in turn exerts an equal but opposite Normal force $F_N$, which prevents the ring from penetrating the surface:

$$F_N=mg$$

In the simple friction model a friction force $F_f$ is exerted in the opposite sense of motion, acc.:

$$F_f=\mu F_N=\mu mg,$$

where $\mu$ is the friction coefficient.

We can now set up two equations of motion:

1. Translation:

The force $F_f$ causes acceleration:

$$ma=\mu mg$$

So that:

$$v(t)=\mu gt$$

2. Rotation:

The force $F_f$ also causes a torque $\tau$:

$$\tau=I\alpha,$$

where $\tau=-R \mu mg$, $I=\frac{mR^2}{2}$ and $\alpha=\frac{d\omega}{dt}$, so:

$$-R \mu mg=\frac{mR^2}{2} \frac{d\omega}{dt}$$

$$-2\mu g=R\frac{d\omega}{dt}$$

$$\omega(t)=\omega-\frac{2\mu g}{R}t$$


Rolling without slipping occurs when $v(t)=\omega(t) R$, so with the expressions above:

$$\mu gt=\omega R- 2\mu gt$$

$$t=\frac{\omega R}{3\mu g}$$

Inserting this into $\omega(t)$ we get:

$$\large{\omega(t)=\frac{\omega}{3}}$$

The ring will lose half of its initial rotational momentum before rolling without slipping is achieved.

The initial kinetic energy was:

$$K_0=\frac{mR^2\omega^2}{4}$$

The final kinetic energy $K(t)$, including translational energy is:

$$K(t)=\frac{mR^2}{4}\frac{\omega^2}{9}+\frac{m}{2}\frac{\omega^2 R^2}{9}=\frac{mR^2\omega^2}{12}$$

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    $\begingroup$ "the right answer is $\frac{\omega}{3}$ " seems inconsistent with your treatment. $\endgroup$ – Floris Jan 5 '16 at 18:05
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    $\begingroup$ @Floris: I can't find anything wrong with my derivation. Also it seems more 'natural' to conclude that rotational and translational kinetic energies end up equi-partitioned, as I showed. $\endgroup$ – Gert Jan 5 '16 at 18:21
  • $\begingroup$ I agree... I did the math independently and got the same result. I think you need a "disk" ($I=\frac12 m r^2$) to get the factor $\frac13$ of OP. $\endgroup$ – Floris Jan 5 '16 at 18:22
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    $\begingroup$ @Floris: that thought occurred to me too, thinking for a minute I got $I$ for a ring wrong. But it is of course $mR^2$. Thanks for checking. $\endgroup$ – Gert Jan 5 '16 at 18:26
  • $\begingroup$ sorry it is a disk ... but the point is why can't i apply conservation of momentum about the axis passing through the point of contact as there the net torque is zero. $\endgroup$ – Varun Chandra Jan 6 '16 at 15:00
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You are trying to use an axis that goes through the point of contact between the disk and the surface.

But because of friction, the disk is accelerating (decelerating). This means that the axis that goes through this point is also accelerating and is therefore not at rest in any inertial frame.

When you use non-inertial frames for reference, you get fictitious forces that in this case lead to torques.

To avoid this you need to either only use an axis that goes through the center of mass, or you need to use an axis that is fixed in some inertial frame. The axis that goes through the point of contact is neither.

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You consider angular momentum is unchanged if its derivative is nule. But you said so yourself: Torque is not null. Since Torque is the derivative of angular momentum over time, that automatically implies that the angular momentum of the body is changing, so it can't be conserved. Which makes sense, given the fact you have friction.

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because i think since there is no fixed reference axis about which angular momentum can be conserved.. where to choose axis is the problem.. if u choose axis at the point of contact the point itself is moving. if any external frame is asked to come in picture the value of r*v {rcrossv} (where r is the radius vector from the reference axis) in L = Iw+MRcrossV then starts varying...

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  • $\begingroup$ please rectify my answer if required and give valuable comments too... $\endgroup$ – Samar Gaurav Mar 27 '18 at 2:58

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