Very interesting case, where energy is not conserved?

Question

This is not a homework problem. I have a bigger, more conceptual doubt behind it.

Applying linear momentum conservation: We get velocity of disc is v (towards right)

Now, friction will also apply a torque which will rotate the disc. I solved the question by two methods:

Method 1: Impulse (Force related)

In the above diagram, $v_t$ is the tangential velocity of the particle and is equal to $\frac{v}{2}$ and the normal velocity $v_n$ is equal to $\frac{v\sqrt{3}}{2}$. By concept of impulse, we can say the following:

$\int fdt$ = $\frac{mv}{2}$ , where f is frictional force.

$\int \tau dt$ = $\int fRdt$ = $\Delta$L , where $\tau$ is torque and L is angular momentum.

Using above equations, we get:

R$\frac{mv}{2}$ = L = Iw, where I is moment of inertia = $\frac{mR^2}{2}$

Solving, we get w = $\frac{v}{R}$

The other method is angular momentum and I got same answer from that method.

The only issue is that the final kinetic energy is greater than initial and I cannot figure out how is this possible.

Initial Kinetic Energy: $\frac{1}{2}$m$v^2$

Final Kinetic Energy: $\frac{1}{2}$m$v^2$ + $\frac{1}{2}$I$w^2$

I think the problem is in friction. It is doing negative work on the particle, which reduces its momentum in tangential direction. However on the disc, it does positive work, which gives in linear momentum, but also it acts as a torque and gives rotational kinetic energy.

The collision is also elastic in normal(radial) direction as velocity of approach = velocity of separation which means total work of normal force on disc and on particle is 0, but there is something weird going on in tangential direction.

How is the final kinetic energy more than initial? I would be very grateful for an detailed explanation.

Answer 1

You are right to be perplexed: the problem is either poorly thought-out or, as the other answer suggests, poorly phrased. The situation described in the problem is basically impossible or, rather, only possible if some sort of chemical or other energy gets converted to mechanical energy during the collision, so that the final kinetic energy ends up being greater than the initial kinetic energy.

Why does the final kinetic energy have to be greater than initial? Because, as you rightly pointed out, conservation of linear momentum requires that the linear velocity of the disc becomes $v$, so its translational kinetic energy is $\frac{mv^2}{2}$.

But conservation of angular momentum requires that the disc also acquires some non-zero rotational energy. Therefore, the final kinetic energy must be greater than $\frac{mv^2}{2}$, i.e. greater than the initial energy.

If the objects are just a normal, "honest" disc and ball, that can't happen.

Answer 2

The problem seems poorly worded; the intended meaning—the only possible self-consistent description—appears to be that the particle comes to rest with respect to the disc after impact.

(Comments on the page discuss at least two other possibilities that can't occur, either by the problem description or Nature's restrictions. The first is that the particle rebounds elastically from the disc and immediately gets stuck on the table to end up at rest. But a frictionless surface provides no such opportunity for sticking. The second is that kinetic energy is gained by the disc and the system cools down. But since cooling a system decreases its entropy and the work done to accelerate something transfers no entropy, entropy must be destroyed, which Nature prohibits.)

The initial momentum (of the particle) is $mv$. Linear momentum is conserved for the objects on a frictionless table, so the final speed of the mass of $2m$ is $\frac{(mv)}{2m}=\frac{v}{2}$.

The initial angular momentum around the disc center (of the particle) is $(mv)\frac{R}{2}$. Angular momentum is conserved as well for the objects on a frictionless table, so the final rotational speed is $(mv)\frac{R}{2}\left(\frac{1}{I}\right)$, where $I$ is the mass moment of inertia for the particle and disc rotating in concert around their common center of mass, which is $R/2$ away from the center. (Thank you to
@Albertus Magnus for identifying this behavior.) From the parallel axis theorem, $I=I_\mathrm{circle}+I_\mathrm{particle}=\left(\frac{mR^2}{2}+\frac{mR^2}{4}\right)+\frac{mR^2}{4}=mR^2$. This angular speed is thus $\frac{v}{2R}$.

The initial kinetic energy is $\frac{1}{2}mv^2$. The final kinetic energy is $\frac{1}{2}(2m)\left(\frac{v}{2}\right)^2+\frac{1}{2}\left(mR^2\right)\left(\frac{v}{2R}\right)^2=\frac{3}{8}mv^2$, which is lower after this inelastic collision.

Answer 3

Your analysis is correct. The initial energy in the system of particle plus disc is the kinetic energy of the particle, $(1/2) m v^2$. The final energy in the system of particle plus disc is the kinetic energy of the disc, which has a contribution $(1/2) m v^2$ from linear motion and also a contribution from rotation, making the total greater than $(1/2) m v^2$. Here I am taking the wording of the question to mean that the particle hits the disc and the disc moves away, while the particle then stays still at the location of the collision (it does not stick to the disc).

The conclusion is that something unusual must happen in order to make the particle end up not moving. If the collision were elastic or totally inelastic the particle would end up moving (either bounced off the disc or stuck to it). We have to conclude that no interaction involving only the disc and particle can give the observed result. There must be a third party involved, such as a nail that pops up and catches the particle, or a lump of glue or something. But if there is a nail or a lump of glue then our momentum analysis is incomplete: we have to account for the momentum of the body to which the nail is attached.

If we insist that there is just the particle and the disc then the conclusion is one of:

either

1. the final state of motion given in the question never occurs, being a physical impossibility for the given system and initial conditions

or

2. the disc or particle is not simply a disc or particle, but has some stored internal energy, which is released when the two collide, having the effect of bringing the particle to rest and providing the requisite energy (imparted to the disc) to make this possible while conserving momentum and angular momentum overall

Answer 4

Let's assign some new variables so that we can determine the more general case.

Let:
u = the initial velocity of the particle.
v = the final velocity of the disk.
m = mass of the particle.
M = mass of the disk.
$\omega_p, \omega_d$ = angular velocity of the particle and disk respectively.
$I_p, I_d$ = moment of inertia of the particle and disk respectively.
r = orthogonal distance from the centre of the disk of the collision.
R = radius of the disk. (In the given example r = R/2)

The initial angular momentum of the particle is $$I_p \ \omega_p = m r^2 \omega_p = m r u$$

Since the particle comes to a stop and the disk is rotating after the collision, we can assume this is intended to be an elastic collision.

The angular momentum of the disk after the collision is $$I_d \ \omega_d = \frac{M R^2}{2} \omega_d$$

Ref: List of moments of inertia:

Since the particle comes to a stop, all the angular momentum is transferred to the disk and we can equate the two:
$$ \frac{M R^2}{2} \omega_d = m r u$$

and solve for the angular velocity of the disk after the collision:

$$ \omega_d = \frac{2 m r u}{M R^2}$$

The linear momentum of the system before and after the collision is also conserved, so: $$mu = Mv$$ $$v = u \frac mM$$

Conservation of energy in an elastic collision requires the total kinetic energy before the collision is equal to the total kinetic energy after the collision, so:

$$ \frac12 m u^2 = \frac12 M v^2 + \frac12 I_d \omega_d^2$$

Substitute the expressions obtained for v, $I_d$ and $\omega_d$ above:

$$ \frac12 m u^2 = m u^2 \left(\frac12 \frac {m}{M} + \frac{m}{M} \frac{r^2}{R^2} \right)$$

This is the general equation for the collision of a particle with a disk, where the particle comes to rest after the collision.

If we set M = m as is implied in the OP, then the only real solution is when r=0, which proves the original question is flawed.

If we set $r = R/2$ and solve we get $$M = \frac32 m$$ which also does not agree with set up in the OP.