What is the error in the solution for Example 25.3.1(a) on page 11?

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SUMMARY

The discussion centers on the error in the solution for Example 25.3.1(a) from the MIT OpenCourseWare on conservation of angular momentum. The user initially applied conservation of linear momentum incorrectly, leading to the conclusion that the angular speed after the collision, denoted as ##\omega_2##, is ##\frac{mv_0}{m_rl}## instead of the correct ##\frac{3mv_0}{2m_rl}##. The error stems from neglecting the impulse from the pivot during the instantaneous collision, which affects angular momentum. The correct approach involves using conservation of angular momentum about the pivot point to account for the impulse's effect.

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hydroxide0
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Homework Statement


The problem is Example 25.3.1 (a) on page 11 here: http://ocw.mit.edu/courses/physics/8-01sc-physics-i-classical-mechanics-fall-2010/angular-momentum-1/conservation-of-angular-momentum/MIT8_01SC_coursenotes25.pdf.

Homework Equations


3. The Attempt at a Solution [/B]
Here is my (apparently incorrect) solution:

We shall take our system to be the object and rod and use conservation of momentum. (We can do so because the only external forces acting on the system are the force of the ceiling on the rod and the gravitational force on the rod, and immediately before the collision these cancel. So if we assume the collision is nearly instantaneous, then there is no net external force on the system during the collision and momentum is conserved.) Equating the momentum of the system immediately before and after the collision gives
mv_0=m\frac{v_0}{2}+m_rV
where ##V## is the speed of the center of mass of the rod immediately after the collision. Since the rod is uniform, the center of mass is a distance ##l/2## away from the pivot point. Thus if we let ##\omega_2## be the angular speed of the rod immediately after the collision, then we have ##V=\omega_2\frac l2##. Plugging this into the above equation gives ##mv_0=m\frac{v_0}{2}+m_r\omega_2\frac{l}{2}##, which we can solve for ##\omega_2## to get ##\omega_2=\frac{mv_0}{m_rl}##.

But according to the solution in the link ##\omega_2=\frac{3mv_0}{2m_rl}##, so there must be something wrong with this... if anyone can point out the error, it would be much appreciated!
 
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When the mass hits the rod, there is an impulse from the pivot on the top of the rod. This impulse is not known, so you are better off using the conservation of angular momentum about the pivot (as they did in their solution). That way, the impulse force has no torque.
 
But if we assume the collision is instantaneous (as they did in their solution), then there can't be any net impulse on the system, right?
 
Why can't there be an impulse if the collision is instantaneous?

EDIT: I might not have been clear on this: the impulse is in the ##- \hat{i}## direction.
 
Correct me if I'm wrong about this, but since a force acting over zero time causes zero impulse, the horizontal force of the pivot causes zero impulse during the instantaneous collision. Therefore the momentum just before and just after the collision are equal.
 
hydroxide0 said:
Correct me if I'm wrong about this, but since a force acting over zero time causes zero impulse, the horizontal force of the pivot causes zero impulse during the instantaneous collision.
Not if the force is infinite (or very very large), which is usually the case for collisions.
 

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