- #1

Karol

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## Homework Statement

A rod of mass M and length L is in the initial position at angle θ and pivoted at the lower end.

It is released and at arriving to the equilibrium state it hits elastically a mass m that slides on a surface with the friction coefficient ##\frac{\mu}{L}x##

What's the angular velocity when it reached the equilibrium point?

What's the initial velocity of m after being hit?

To what distance will it travel?

## Homework Equations

Differential equation:

$$\ddot x-c^2x=k\;\rightarrow\; x=Ae^{ct}+Be^{-ct}+\frac{k}{c^2}$$

In elastic collision, when one of the masses (V_2=0) is at rest at the beginning, the velocity of that mass after is:

$$V'_2=\frac{2m_1 V_1}{m_1+m_2}$$

## The Attempt at a Solution

Conservation of energy will give the CoM velocity:

$$MgL\left( \frac{1}{2}+\frac{1}{2}\cos\theta \right)=\frac{1}{2}Mv^2$$

$$\omega \frac{L}{2}=v\;\rightarrow \omega=\frac{2v}{L}=\frac{2 \sqrt{ gL(1+\cos\theta) }}{L}$$

The velocity v

_{L}of the lower end of the rod when it hits:

$$v_L=\omega L=2\sqrt{gL(1+\cos\theta)}$$

Elastic collision:

$$v_0=\frac{2Mv_L}{M+m}$$

The acceleration:

$$F=mg\frac{\mu}{L}x=m\ddot x\;\rightarrow\; \ddot x-\left( \frac{\mu g}{L} \right)x=0$$

$$x=Ae^{\sqrt{\frac{\mu g}{L}}}t+Be^{-\sqrt{\frac{\mu g}{L}}}t$$

$$t=0\rightarrow x=0\;\rightarrow\; A+B=0$$

$$\dot x=A\sqrt{\frac{\mu g}{L}}e^{\sqrt{\frac{\mu g}{L}}}t+B\left( -\sqrt{\frac{\mu g}{L}} \right)e^{-\sqrt{\frac{\mu g}{L}}}t$$

$$\dot x(t=0)=v_L\;\rightarrow\; v_L=A\sqrt{ \frac{\mu g}{L} }-B\sqrt{ \frac{\mu g}{L} }\;\rightarrow\;B=\frac{v_L}{-2\sqrt{ \frac{\mu g}{L} }},\; A=-\frac{v_L}{-2\sqrt{\frac{\mu g}{L}}}$$

It stops when the velocity is zero:

$$\dot x=0\;\rightarrow\; 0=A\sqrt{\frac{\mu g}{L}}e^{\sqrt{\frac{\mu g}{L}}}t+B\left( -\sqrt{\frac{\mu g}{L}} \right)e^{-\sqrt{\frac{\mu g}{L}}}t$$

$$e^{2\sqrt{\frac{\mu g}{L}}}=-1$$

It can't be