# Rod rotating, falling and hitting a mass

1. Jul 7, 2015

### Karol

1. The problem statement, all variables and given/known data
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?

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

3. 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 vL 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

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2. Jul 7, 2015

### Nathanael

This isn't valid. The energy of a point particle of mass M rotating at the center of mass is not the same as the energy of a rod rotating.
Use the rotational energy.

This isn't valid either. I think your equation only applies to point particles colliding elastically.
For this scenario, you will need to write a conservation of energy equation and a conservation of angular momentum equation.

I think it should be $\ddot x+\left( \frac{\mu g}{L} \right)x=0$
The acceleration acts in the negative x direction.

3. Jul 7, 2015

### SammyS

Staff Emeritus
Your solution is not valid. It has serious problems.

It is not simply two point masses colliding, so that elastic collision equation cannot be used. (A similar result does hold for angular velocity and moments of inertia.)

The potential energy expression you have is correct.

However, the Kinetic Energy is not simply $\displaystyle \ \frac{1}{2}Mv^2 \$. Rotational Kinetic Energy is also involved.

Solve the collision by conserving angular momentum as well as kinetic energy.

4. Jul 7, 2015

### haruspex

Further to what Nathanael and Sammy wrote, you just have to be a teensy bit careful about the choice of reference axis for conservation of angular momentum through the collision. And you don't need to get into the gory details of the elastic collision, with force and displacement equations. You can treat it as an atomic event, just as you would for the elastic collision of two balls. All that's different is that you use angular momentum here instead of linear momentum.

5. Jul 8, 2015

### Karol

Moment of inertia round the pivot point:
$$I=\frac{1}{3}ML^2+M\frac{L^2}{4}=\frac{7}{12}ML^2$$
Conservation of energy:
$$MgL\left( \frac{1}{2}+\frac{1}{2}\cos\theta \right)=\frac{1}{2}Mv^2+\frac{1}{2}I\omega_0^2$$
$$MgL(1+\cos\theta)=M\omega_0^2\frac{L^2}{4}+\omega_0^2$$
Conservation of momentum before and after the collision round the pivot point and conservation of energy:
$$\left\{ \begin{array} {l} \left( \frac{7}{12}ML^2 \right)\omega_0=\left( \frac{7}{12}ML^2 \right)\omega'+mLv' \\ MgL\left( \frac{1}{2}+\frac{1}{2}\cos\theta \right)=\frac{1}{2} \left( \frac{7}{12}ML^2 \right)\omega'^2+\frac{1}{2}mv' \end{array} \right.$$
The term on the left from the first, momentum, equation is:
$$\left( \frac{7}{12}ML^2 \right)\omega_0=\frac{4MgL(1+\cos\theta)}{ML^2+4I}=\frac{7}{10}Mg(1+\cos\theta)$$
The conservation of momentum equation now:
$$\frac{7}{10}Mg(1+\cos\theta)=\frac{7}{12}ML^2+mLv'\;\rightarrow\omega'=\frac{12}{7ML^2}\left[ Mg(1+\cos\theta)-mLv' \right]$$
I insert this $\omega'$ into the second, the energy, equation:
$$MgL(1+\cos\theta)-\frac{1}{42M}\left[ Mg(1+\cos\theta)-mLv' \right]^2=mv'^2$$
Very complicated and i must be wrong

6. Jul 8, 2015

### Nathanael

The moment of inertia of a rod about it's center is not ML2/3 it is ML2/12

The 0.5Mv2 term should not be there. The 0.5Iω2 term already accounts for all the kinetic energy.

A piece of the rod at distance x will have kinetic energy 0.5(ωx)2dm=0.5(ωx)2(M/L)dx. If we integrate this over the whole length of the rod we get $E_{kinetic}=0.5ω^2\frac{M}{L}\int\limits_0^L x^2dx$ but you see $\frac{M}{L}\int\limits_0^L x^2dx$ is simply the rotational inertia. Thus all the kinetic energy is accounted for in the rotational term.

In general we could say $E_{kinetic}=\int 0.5(ωx)^2 dm = 0.5ω^2\int x^2dm=0.5Iω^2$ because the definition of I is ∫x2dm. So if an object is purely rotating we do not need any 'linear kinetic energy terms' to account for it's kinetic energy. (For point objects, 0.5Iω2 reduces to 0.5Mv2)

7. Jul 8, 2015

### Karol

The moment of inertia round it's end is $I=\frac{1}{3}ML^2$
Conservation of energy before hitting:
$$\frac{1}{2} MgL(1+\cos\theta)=\frac{1}{2} I\omega_0^2\;\rightarrow\omega_0^2=\frac{3g(1+\cos\theta)}{L}$$
At collision:
$$\left\{ \begin{array} {l} (1) \left( \frac{1}{3}ML^2 \right)\omega_0=\left( \frac{1}{3}ML^2 \right)\omega'+mLv' \\ (2)\; MgL\left( \frac{1}{2}+\frac{1}{2}\cos\theta \right)=\frac{1}{2} \left( \frac{1}{3}ML^2 \right)\omega'^2+\frac{1}{2}mv' \end{array} \right.$$
$$(1):\; \omega'=\frac{mLv'-\frac{1}{3}ML^2\omega_0}{\frac{1}{3}ML^2}=\frac{3mLv'-ML^2\omega_0}{ML^2}$$
$$(1)+(2):\; \frac{1}{6}ML^2\frac{(3mLv'-ML^2\omega_0)^2}{M^2L^4}+\frac{1}{2}mv'^2=\frac{1}{2}MgL(1+\cos\theta)$$
Still complicated, should i continue? i want to extract v'

8. Jul 8, 2015

### haruspex

That's dimensionally inconsistent.

9. Jul 9, 2015

### Nathanael

When you solved for ω' in equation (1), your answer is off by a factor of negative one. (It doesn't really matter, though, as ω' is squared in the other equation.)

There are two small things I would do to simplify it:
First, I would write $ω'=ω_0-\frac{3mv'}{ML}$
Second, instead of writing the initial energy as 0.5MgL(1+cosθ) I would write it as 0.5Iω02.
(You have ω0 in the other equation, so you might as well express v' in terms of ω0 instead of mixing ω0 and θ)

But it looks good to me.

I believe it's a typo; if you look in the equation "(1)+(2)" the mistake is fixed.

10. Jul 9, 2015

### Karol

$$\left\{ \begin{array} {l} (1) \left( \frac{1}{3}ML^2 \right)\omega_0=\left( \frac{1}{3}ML^2 \right)\omega'+mLv' \\ (2)\; \frac{1}{2}I\omega_0^2=\frac{1}{2} \left( \frac{1}{3}ML^2 \right)\omega'^2+\frac{1}{2}mv'^2 \end{array} \right.$$
$$(1):\; \omega'=\omega_0-\frac{3mv'}{ML}$$
$$v'=\frac{2M\sqrt{3Lg(1+\cos\theta)}}{3m+M}$$
$$F=-mg\frac{\mu}{L}x=m\ddot x\;\rightarrow\; \ddot x+\left( \frac{\mu g}{L} \right)x=0$$
$$x=A\cos\beta t+B\sin\beta t,\;\beta^2=\frac{\mu g}{L}$$
$$t=0,\; x=0:\; A=0$$
$$\dot x=-A\beta\sin(\beta t)+b\beta\cos(\beta t),\; t=0,\; \dot x=v'\;\rightarrow B=v', \; x=v'\sin(\beta t)$$
When the block stops the velocity zeroes:
$$\dot x=\beta v'\cos(\beta t)=0\;\rightarrow t\triangleq t'=\frac{\pi}{2\omega'}$$
The total distance:
$$s=\int_0^{t'}v'\sin(\beta t)=...=\frac{MLv'}{ML\omega_0-3mv'}$$

11. Jul 9, 2015

### Nathanael

Not the right value for B (the dimensions are not correct).

t' is not quite right

You are integrating x? You only need to evaluate it at t', no need to integrate.

Last edited: Jul 9, 2015
12. Jul 9, 2015

### Karol

$$\dot x=B\beta\cos(\beta t),\; t=0,\; \dot x=v'\;\rightarrow B=\frac{v'}{\beta}=v'\sqrt{\frac{L}{\mu g}}, \; x=v'\sqrt{\frac{\mu g}{L}}\sin(\beta t)$$
$$\dot x=v'\cos\left( \sqrt{\frac{\mu g}{L}} t \right)=0\;\rightarrow t \triangleq t'=\frac{\pi}{2}\sqrt{\frac{L}{\mu g}}$$
$$s=x(t=t')=v'\sqrt{\frac{\mu g}{L}}\sin\left( \sqrt{\frac{\mu g}{L}} \sqrt{\frac{L}{\mu g}} \frac{\pi}{2} \right)=\sqrt{\frac{L}{\mu g}}v'$$

13. Jul 9, 2015

### Nathanael

(I see two mistakes which cancel each other out, such as your equation for x having the wrong dimensions, but I think it is just typos.)

14. Jul 9, 2015

### haruspex

I think it is a little simpler in the post-collision part to consider $\frac 12 m{v'}^2=\int_0^s F.dx=\frac{\mu mgs^2}{2L}$.

15. Jul 9, 2015

### Karol

Which are the mistakes? indeed the dimensions for x aren't correct

16. Jul 10, 2015

### Nathanael

Here you say that $B=\frac{v'}{\beta}$ but then in your solution for x you put in $B=v'\beta$

Here you are basically saying that $v'\beta\cdot1=\frac{v'}{\beta}$

It is just careless mistakes (which cancel out) so I assume it is typos.

17. Jul 10, 2015

### Karol

Thank you all