Rod rotating and hitting a mass

[Karol]

Homework Statement

A is the pivot at the upper edge of the rod.
The rod of length L and mass M stays horizontally. it can pivot round it's edge. suddenly it falls, rotates round A and hits a mass m which gains velocity v.
At which angular velocity the rod reaches the mass
What is the rod's angular velocity immediately after it hits?
How much energy is lost during the hit

Homework Equations

moment of inertia of a rod round it's edge: $I_A=\frac{1}{3}ML^2$
Conservation of momentum: $m_1v_1+m_2v_2=m_1v_1'+m_2v_2'$
Kinetic energy of a solid body: $E_k=\frac{1}{2}I\omega^2$

The Attempt at a Solution

The energy of height of the COM in the horizontal position transforms to kinetic:
$$\frac{1}{2}MgL=\frac{1}{2}MV^2~~\rightarrow~~V^2=gL$$
Conservation of momentum, V₁ is the rod's COM velocity after the hit: $MV=MV_1+mv$
Angular velocity, ω₁ is the rod's angular velocity after: $\frac{1}{2}L\omega_1=V_1$
$$\rightarrow~~\omega_1=\frac{2}{ML}(M\sqrt{gL}-mv)$$
The energy loss is:
$$\frac{1}{2}MgL-\left( \frac{1}{2}I_A\omega_1+mv^2 \right)$$

 

[BvU]

Well now, interesting. Is there a question ? Or do you just want a stamp of approval (which really isn't PF business). You wouldn't get it from me anyway, because I miss things like conservation of angular momentum. I also have difficulty with your $V^2 = gL$: it's not as if the COM is dropping over $L/2$ only: there is also rotational kinetic energy to consider !

 

[haruspex]

As BvU observes, you have ignored the rotational energy acquired by the rod.
Also, linear momentum will not be conserved. Can you see why?

The problem as stated does not provide enough information. You need to know the coefficient of restitution in the impact. Probably should assume it is completely inelastic.

 

[Karol]

Conservation of energy between the initial condition and the vertical one:
$$\frac{1}{2}MgL=\frac{1}{2}I_A\omega_0^2~~\rightarrow~~\omega_0^2=\frac{3g}{L}$$
Conservation of angular momentum: $\vec L=I\times \vec{\omega}$
$$I_A\omega_0=I_A\omega+mLv~~\rightarrow~~\omega=\omega_0-\frac{mLv}{I_A}$$

 

[haruspex]

Conservation of energy between the initial condition and the vertical one:
$$\frac{1}{2}MgL=\frac{1}{2}I_A\omega_0^2~~\rightarrow~~\omega_0^2=\frac{3g}{L}$$
Conservation of angular momentum: $\vec L=I\times \vec{\omega}$
$$I_A\omega_0=I_A\omega+mLv~~\rightarrow~~\omega=\omega_0-\frac{mLv}{I_A}$$

That all looks right, except for a possible flaw in the question. The block mass m is shown as having significant height, so its angular momentum about A will be less than mLv. Leaving that aside...

Next, you need to represent the fact that the rod does not penetrate the block. That puts a constraint on the relationship between ω and v. If we assume a fully inelastic collision then you can take that relationship as equality.

 

[Karol]

we don't assume assume inelastic collision since v, the block's velocity, is given.

 

[haruspex]

we don't assume assume inelastic collision since v, the block's velocity, is given.

Good point!

 

[Karol]

The energy that was lost:
Initial energy is $E_i=\frac{1}{2}MgL$
$$\Delta E=\frac{1}{2}MgL-\frac{1}{2}I_A\omega_0^2-\frac{1}{2}mv^2$$

 

[haruspex]

The energy that was lost:
Initial energy is $E_i=\frac{1}{2}MgL$
$$\Delta E=\frac{1}{2}MgL-\frac{1}{2}I_A\omega_0^2-\frac{1}{2}mv^2$$

ω, not ω₀, right?

 

[Karol]

right, Haruspex, thank you very much...