- #1

PiEpsilon

- 22

- 2

- 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

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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