Elastic collision of particle and rotating disc

In summary: Since no external forces act on the system, the angular momentum will be conserved. For elastic collision, the kinetic energy of the system stays constant....then the angular momentum of the disk must also be conserved. However, the angular momentum is not conserved in the textbook answer.
  • #1
PiEpsilon
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Homework Statement
A uniform thin disc of radius ##R##, mass ##M## is mounted on a universal bearing permitting rotation about any axis. Initially it spins about a vertical axis (z-direction) with angular velocity ##\omega_0##, as shown in Fig 16-20. A small mass ##m## with velocity ##v_o## in the positive z-direction collides elastically with the rim of the wheel and rebounds in the negative z-direction.

a) What is the direction of the angular momentum ##\vec L## of the disc after the collision?

b) Describe the motion of the figure axis. Indicate by sketch the trajectory of the point of intersection of the figure axis with a unit sphere as seen from the top.
Relevant Equations
none
IMG_89D4AF30EB17-1.jpeg

Consider the system of the mass and uniform disc.
Since no external forces act on the system, the angular momentum will be conserved. For elastic collision, the kinetic energy of the system stays constant.Measuring angular momentum from the hinge:
##\vec L_i = Rmv_0 \space\hat i + I \omega_0 \space\hat k##
##\vec L_f = -Rmv_f \space\hat i + \vec L##

where ##I=\frac12MR^2## and ##\hat i, \hat j, \hat k## are unit vectors in ##x,y,z## direction respectively.

By conservation of energy:
##\frac12mv_0^2+\frac12I\omega_0^2 = \frac12 mv_f^2 +\frac12 I\omega_f^2 \implies m(v_0^2−v_f^2)=I(\omega_f^2−\omega_0^2)##

The angular impulse acting on a disc during the mass collision produces the torque perpendicular to the angular momentum. Therefore only the direction of disc's angular momentum changes, while magnitude stays the same.
Thus:
##I\omega_0 = I\omega_f \implies \omega_0^2 = \omega_f^2##
From conservation equation we then conclude, that ##v_0=v_f## and since the problem states that the mass bounces in exactly the opposite direction, ##\vec v_f = -\vec v_0##

By conservation of angular momentum:
##Rmv_0\space\hat i+ I\omega_0 \space\hat k= −Rmv_0\space \hat i+ \vec L\implies \vec L = 2Rmv_0\space\hat i + \frac12MR^2\omega_0 \space\hat k##

The answer in the textbook is given as:
##\frac{L_x}{L_z} = 4mv_0/MR\omega_0## which agrees with above.

The problem arises when one inspects the angular impulse argument (which, considering the fact that the (internal) force from the collision, acting on a disk is normal to the surface, so angular momentum magnitude shouldn't change at all times, even if the mass does not rebound in the same direction, is really compelling).
However clearly ##\lvert\vec L_0\lvert \not = \lvert\vec L\rvert##

Where is the flaw?
 
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  • #2
I think the flaw is assuming that the impulsive torque is perpendicular to the angular momentum of the disk during the collision. The impulsive torque is perpendicular to the initial angular momentum of the disk, but not perpendicular to the final angular momentum of the disk (immediately after the impulse). The transition from the initial angular momentum to the final angular momentum takes place while the torque is acting. So, I don't think it is correct to say that the torque acts perpendicular to the instantaneous angular momentum while the angular momentum is changing from the initial to the final value.

It can be confusing when we think of the angular impulse as acting at only one instant of time, rather than acting over a very short interval of time.

It might help to consider the case where the initial angular speed of the disk is extremely small so that the initial angular momentum is extremely small. (Imagine ##\omega_0## to be so small that we can't notice any rotation even after a few minutes.) In this case, we would certainly expect the collision to alter the magnitude of the angular momentum of the disk.
 
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  • #3
TSny said:
The impulsive torque is perpendicular to the initial angular momentum of the disk, but not perpendicular to the final angular momentum of the disk (immediately after the impulse).
Wouldn't it be a bit of both? If we consider the collision taking place over some interval, the change in axis of the disc takes place over that interval. It should be a reasonable approximation to split it 50-50.
 
  • #4
TSny said:
I think the flaw is assuming that the impulsive torque is perpendicular to the angular momentum of the disk during the collision. The impulsive torque is perpendicular to the initial angular momentum of the disk, but not perpendicular to the final angular momentum of the disk (immediately after the impulse). The transition from the initial angular momentum to the final angular momentum takes place while the torque is acting. So, I don't think it is correct to say that the torque acts perpendicular to the instantaneous angular momentum while the angular momentum is changing from the initial to the final value.

It can be confusing when we think of the angular impulse as acting at only one instant of time, rather than acting over a very short interval of time.

It might help to consider the case where the initial angular speed of the disk is extremely small so that the initial angular momentum is extremely small. (Imagine ##\omega_0## to be so small that we can't notice any rotation even after a few minutes.) In this case, we would certainly expect the collision to alter the magnitude of the angular momentum of the disk.

I am still unsure I follow.

Consider the rotating disk system. Then the only force producing torque is from the 'bouncing' particle.
Since it is a contact force, it is normal to the disc surface at all times during the collision. I have ignored any torques arising from friction forces between particle and disk, since we know there are none (because particle bounces exactly in opposite direction).
In short I don't see any torque components acting in line with angular momentum, only perpendicular to it.

There is also an issue with conservation of energy. If the answer provided by the textbook is right and the speed of the particle does not change during the collision, it implies that angular speed of the disk must remain the same.
Since moment of inertia of the disk remains constant, the angular momentum magnitude should stay the same too. (##\lvert\vec L_0\rvert = I\omega_0 = I\omega_f = \lvert\vec L\rvert \iff \omega_0 = \omega_f##)

@ I see the issue with considering impulse as force acting in an infinitesimal time interval:
Let a particle of mass ##m## move with speed ##v_0## horizontally.
After the vertical impulse ##J##, which is perpendicular to initial momentum, the magnitude of momentum of the particle changes, despite force acting is perpendicular to the initial momentum.

However the energy consideration (in the original problem) still implies that angular momentum magnitude of the disc should not change.
 
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  • #5
Let ##v_0## and ##v## be the initial and final speeds of the particle (both positive numbers). The force that acts on the disk during the collision is normal to the disk. So, the torque on the disk is parallel to the disk. Orient the x-axis to coincide with the direction of the torque. Let ##L_x## be the angular momentum picked up by the disk due to the collision.

Conservation of angular momentum of the system: $$m(v+v_0)R = L_x$$

Conservation of energy: $$\frac 1 2 m(v_0^2-v^2) = \frac{L_x^2}{2I_x}$$

Solve these for ##v##: $$v =v_0 \frac{1-4m/M}{1+4m/M} \approx v_0(1-8m/M)$$If ##m## is small enough compared to ##M## that ##8m/M## can be neglected compared to 1, then ##v \approx v_0##. In the following we will retain terms of first order in ##m/M##.

$$L_x = m(v+v_0)R \approx 2mRv_0(1-4m/M)$$

$$\frac{L_x}{L_z} \approx \frac{4mv_0}{MR\omega_0}(1-4m/M)$$

$$\omega_x = \frac{L_x}{I_x} \approx \frac {8mv_0}{MR}(1-4m/M)$$

$$\Delta K_{\rm disk} = \frac 1 2 I_x \omega_x^2 = \frac{L_x^2}{2I_x} \approx \frac{8m^2v_0^2}{M}(1-8m/M)$$

The change in the magnitude of the angular momentum of the disk is $$\Delta L_{\rm disk} = \sqrt{L_z^2+L_x^2} - L_z \approx \frac{L_x^2}{2L_z} \approx \frac{4m^2v_0^2}{M\omega_0}(1-8m/M)$$

From the last two equations, we see there will be a change in the kinetic energy and the magnitude of angular momentum of the disk (even to "zeroth order" where ##8m/M## is neglected compared to 1).
 
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  • #6
$$J(\boldsymbol\omega^+-\boldsymbol\omega^-)=\boldsymbol R\times \boldsymbol F;\quad \boldsymbol\omega^-=\boldsymbol\omega_0$$
$$m(\boldsymbol v^+-\boldsymbol v^-)=-\boldsymbol F,\quad \boldsymbol v^-=\boldsymbol v_0$$
$$\frac{1}{2}(m|\boldsymbol v^-|^2+(J\boldsymbol\omega^-,\boldsymbol\omega^-))=
\frac{1}{2}(m|\boldsymbol v^+|^2+(J\boldsymbol\omega^+,\boldsymbol\omega^+))$$
 
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  • #7
The problem becomes nontrivial if we assume that there is no slipping between the particle and the disk during the collision
 
  • #8
TSny said:
Let ##v_0## and ##v## be the initial and final speeds of the particle (both positive numbers). The force that acts on the disk during the collision is normal to the disk. So, the torque on the disk is parallel to the disk. Orient the x-axis to coincide with the direction of the torque. Let ##L_x## be the angular momentum picked up by the disk due to the collision.

Conservation of angular momentum of the system: $$m(v+v_0)R = L_x$$

Conservation of energy: $$\frac 1 2 m(v_0^2-v^2) = \frac{L_x^2}{2I_x}$$

Solve these for ##v##: $$v =v_0 \frac{1-4m/M}{1+4m/M} \approx v_0(1-8m/M)$$If ##m## is small enough compared to ##M## that ##8m/M## can be neglected compared to 1, then ##v \approx 2v_0##. In the following we will retain terms of first order in ##m/M##.

$$L_x = m(v+v_0)R \approx 2mRv_0(1-4m/M)$$

$$\frac{L_x}{L_z} \approx \frac{4mv_0}{MR\omega_0}(1-4m/M)$$

$$\omega_x = \frac{L_x}{I_x} \approx \frac {8mv_0}{MR}(1-4m/M)$$

$$\Delta K_{\rm disk} = \frac 1 2 I_x \omega_x^2 = \frac{L_x^2}{2I_x} \approx \frac{8m^2v_0^2}{M}(1-8m/M)$$

The change in the magnitude of the angular momentum of the disk is $$\Delta L_{\rm disk} = \sqrt{L_z^2+L_x^2} - L_z \approx \frac{L_x^2}{2L_z} \approx \frac{4m^2v_0^2}{M\omega_0}(1-8m/M)$$

From the last two equations, we see there will be a change in the kinetic energy and the magnitude of angular momentum of the disk (even to "zeroth order" where ##8m/M## is neglected compared to 1).
Thank you for an elaborate answer. The result agrees completely if one assumes ##m << M##.

I have suspected that the final speed of the particle must be different but the textbook answer put me off the track.
I guess it is a lesson to check it only after fully understanding the problem.
 
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Related to Elastic collision of particle and rotating disc

1. What is an elastic collision?

An elastic collision is a type of collision in which the total kinetic energy of the system is conserved. This means that the total energy before and after the collision remains the same.

2. How does a particle collide with a rotating disc?

In a collision between a particle and a rotating disc, the particle will hit the disc at a certain angle and with a certain velocity. The disc will then exert a force on the particle, causing it to change direction and possibly transfer some of its energy to the disc.

3. What factors affect the outcome of an elastic collision between a particle and a rotating disc?

The outcome of an elastic collision between a particle and a rotating disc is affected by factors such as the mass and velocity of the particle, the radius and speed of the rotating disc, and the angle at which the particle collides with the disc.

4. How is the conservation of momentum applied in an elastic collision between a particle and a rotating disc?

In an elastic collision between a particle and a rotating disc, the total momentum of the system before and after the collision remains the same. This means that the momentum of the particle and the disc may change, but their combined momentum will remain constant.

5. What is the difference between an elastic collision and an inelastic collision?

In an elastic collision, the total kinetic energy of the system is conserved, while in an inelastic collision, some of the kinetic energy is lost. In an inelastic collision, the objects may also stick together after the collision, while in an elastic collision, they will bounce off each other.

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