Ball hitting a hanging rod: Velocity and Energy Loss Analysis

[Karol]

Homework Statement

A rod of length L hangs on a nail. a ball of mass m hits and sticks to it. the rod rotates to angle θ.
What is the velocity of the ball and what's the loss of energy.

Homework Equations

Kinetic energy of a rigid body: $E=\frac{1}{2}I\omega^2$
Angular momentum: $I\omega$

The Attempt at a Solution

Location of C.O.M:
$$x_{c.m.}=\frac{mL+M\frac{L}{2}}{m+M}=\frac{L}{m+M}\left( m+\frac{M}{2} \right)$$
The potential energy at the inclined position, the final position:
$$E_f=(m+M)g\cdot y_{c.m.}=(m+M)\frac{L}{m+M}\left( m+\frac{M}{2} \right)g\cos\alpha=Lg\left( m+\frac{M}{2} \right)\cos\alpha$$
E_f equals the kinetic energy after the hit:
$$E_f=\frac{1}{2}I\omega^2\rightarrow Lg\left( m+\frac{M}{2} \right)\cos\alpha=(mL^2+\frac{1}{3}ML^2)\omega^2\Rightarrow \omega^2=\frac{g\left( m+\frac{M}{2} \right)\cos\alpha}{L\left( m+\frac{1}{3}M \right)}$$
Conservation of angular momentum:
$$mvL=I\omega \Rightarrow mvL=\left( mL^2+\frac{1}{3} ML^2 \right) \omega$$
$$\Rightarrow v=\frac{ L\left( m+\frac{1}{3}M \right) \omega }{m}=\sqrt{Lg\left( m+\frac{1}{3}M \right)\left(m+\frac{M}{2} \right) }$$
$$\Delta E=Lg\left( m+\frac{M}{2} \right)\cos\alpha-\frac{1}{2}mv^2=Lg\left( m+\frac{M}{2} \right) \left[ \cos\alpha-\frac{1}{2m}\left( m+\frac{1}{3}M\right) \right]$$
Is it true?

 

[haruspex]

What baseline are you taking for the PE? As theta (alpha) increases, does the PE increase or decrease?

 

[Karol]

$$E_f=(m+M)g\cdot y_{c.m.}=(m+M)\frac{L}{m+M}\left( m+\frac{M}{2} \right)g(1-\cos\alpha)=Lg\left( m+\frac{M}{2} \right)(1-\cos\alpha)$$
$$E_f=\frac{1}{2}I\omega^2\rightarrow Lg\left( m+\frac{M}{2} \right)(1-\cos\alpha)=(mL^2+\frac{1}{3}ML^2)\omega^2\Rightarrow \omega^2=\frac{g\left( m+\frac{M}{2} \right)(1-\cos\alpha)}{L\left( m+\frac{1}{3}M \right)}$$
$$v=\frac{ L\left( m+\frac{1}{3}M \right) \omega }{m}=\sqrt{Lg\left( m+\frac{1}{3}M \right)\left(m+\frac{M}{2} \right)(1-\cos\alpha) }$$
$$\Delta E=Lg\left( m+\frac{M}{2} \right)(1-\cos\alpha)-\frac{1}{2}mv^2=Lg\left( m+\frac{M}{2} \right)(1-\cos\alpha) \left[1-\frac{m}{2}\left( m+\frac{M}{3}\right) \right]$$

 

[haruspex]

The first two lines look fine.
In the third you seem to have lost a 1/m. (Check the dimensions.)
In the middle expression of the last line you seem to have lost an entire factor, but maybe that was just a transcription error.
The final expression is dimensionally inconsistent (inside the square brackets you have a constant term minus a mass² term).

 

[Karol]

$$v=\frac{ L\left( m+\frac{1}{3}M \right) \omega }{m}=\frac{1}{m}\sqrt{Lg\left( m+\frac{1}{3}M \right)\left(m+\frac{M}{2} \right)(1-\cos\alpha) }$$
I don't think i missed a factor:
$$\Delta E=E_f-\frac{1}{2}mv^2$$
$$\Delta E=Lg\left( m+\frac{M}{2} \right)(1-\cos\alpha)-\frac{1}{2}mv^2=Lg\left( m+\frac{M}{2} \right)(1-\cos\alpha) \left[1-\frac{1}{2m}\left( m+\frac{M}{3}\right) \right]$$
And the units are correct, at least

 

[ehild]

$$E_f=\frac{1}{2}I\omega^2\rightarrow Lg\left( m+\frac{M}{2} \right)(1-\cos\alpha)=(mL^2+\frac{1}{3}ML^2)\omega^2$$

Haven't you lost an 1/2?

 

[Karol]

$$E_f=\frac{1}{2}I\omega^2\rightarrow Lg\left( m+\frac{M}{2} \right)(1-\cos\alpha)=\frac{1}{2}(mL^2+\frac{1}{3}ML^2)\omega^2\Rightarrow \omega^2=\frac{2g\left( m+\frac{M}{2} \right)(1-\cos\alpha)}{L\left( m+\frac{1}{3}M \right)}$$
And i understand that the rest is good

 

[ehild]

$$E_f=\frac{1}{2}I\omega^2\rightarrow Lg\left( m+\frac{M}{2} \right)(1-\cos\alpha)=\frac{1}{2}(mL^2+\frac{1}{3}ML^2)\omega^2\Rightarrow \omega^2=\frac{2g\left( m+\frac{M}{2} \right)(1-\cos\alpha)}{L\left( m+\frac{1}{3}M \right)}$$
And i understand that the rest is good

With the factor 2, it is correct.

 

[Karol]

Thank you ehild and haruspex (factor of 2...)