# Rod rotating, falling and hitting a mass

• Karol
In summary: Now you see that the equation 0.5Mv2 is not needed. It just adds in a term that is already accounted for in the 0.5Iω2 term. If you include the 0.5Mv2 term you are saying that the rod is both translating and rotating at the same time. But it is not. It is only rotating. The fact that it is rotating means that all the kinetic energy is accounted for in the 0.5Iω2 term.That is why it is wrong to write 0.5Mv2 in this problem. You are saying the rod
Karol

## 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 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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Karol said:
Conservation of energy will give the CoM velocity:
$$MgL\left( \frac{1}{2}+\frac{1}{2}\cos\theta \right)=\frac{1}{2}Mv^2$$
$$\omega \frac{L}{2}=v\;\rightarrow \omega=\frac{2v}{L}=\frac{2 \sqrt{ gL(1+\cos\theta) }}{L}$$
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.

Karol said:
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:
$$v_0=\frac{2Mv_L}{M+m}$$
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.
Karol said:
The acceleration:
$$F=mg\frac{\mu}{L}x=m\ddot x\;\rightarrow\; \ddot x-\left( \frac{\mu g}{L} \right)x=0$$
I think it should be ##\ddot x+\left( \frac{\mu g}{L} \right)x=0##
The acceleration acts in the negative x direction.

Karol said:

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

It can't be
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.

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.

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

Karol said:
Moment of inertia round the pivot point:
$$I=\frac{1}{3}ML^2+M\frac{L^2}{4}=\frac{7}{12}ML^2$$
The moment of inertia of a rod about it's center is not ML2/3 it is ML2/12

Karol said:
Conservation of energy:
$$MgL\left( \frac{1}{2}+\frac{1}{2}\cos\theta \right)=\frac{1}{2}Mv^2+\frac{1}{2}I\omega_0^2$$
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)

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'

Karol said:
At collision:
##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'##
That's dimensionally inconsistent.

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.

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

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

Karol said:
$$\dot x=-A\beta\sin(\beta t)+b\beta\cos(\beta t),\; t=0,\; \dot x=v'\;\rightarrow B=v', \; x=v'\sin(\beta t)$$
Not the right value for B (the dimensions are not correct).

Karol said:
When the block stops the velocity zeroes:
$$\dot x=\beta v'\cos(\beta t)=0\;\rightarrow t\triangleq t'=\frac{\pi}{2\omega'}$$
t' is not quite right

Karol said:
The total distance:
$$s=\int_0^{t'}v'\sin(\beta t)=...=\frac{MLv'}{ML\omega_0-3mv'}$$
You are integrating x? You only need to evaluate it at t', no need to integrate.

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

(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.)

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}##.

Karol and Nathanael
Nathanael said:
(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.)
Which are the mistakes? indeed the dimensions for x aren't correct

Karol said:
$$\dot x=B\beta\cos(\beta t),\; t=0,\; \dot x=v'\;\rightarrow B=\frac{v'}{\beta}=v'\sqrt{\frac{L}{\mu g}}, \; x=v'\sqrt{\frac{\mu g}{L}}\sin(\beta t)$$
Here you say that ##B=\frac{v'}{\beta}## but then in your solution for x you put in ##B=v'\beta##

Karol said:
$$s=x(t=t')=v'\sqrt{\frac{\mu g}{L}}\sin\left( \sqrt{\frac{\mu g}{L}} \sqrt{\frac{L}{\mu g}} \frac{\pi}{2} \right)=\sqrt{\frac{L}{\mu g}}v'$$
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.

Karol
Thank you all

## 1. How does the rotation of a rod affect its falling and impact on a mass?

The rotation of a rod can significantly impact its falling and impact on a mass. When a rod rotates, it creates angular momentum, which is a measure of its rotational motion. This angular momentum can affect the trajectory of the rod as it falls and can also change the amount of force it exerts on the mass upon impact.

## 2. What factors influence the impact force of a rod falling and hitting a mass?

Several factors can influence the impact force of a rod falling and hitting a mass. These include the mass and velocity of the rod, the angle at which it falls, the density and composition of the mass, and the surface it is falling onto. Additionally, the length and distribution of the rod's weight can also affect the impact force.

## 3. How does the shape of a rod affect its falling and impact on a mass?

The shape of a rod can play a significant role in its falling and impact on a mass. A longer and thinner rod will experience more air resistance as it falls, causing it to reach a lower velocity and exert less force upon impact. A shorter and thicker rod, on the other hand, will experience less air resistance and can reach higher velocities and exert more force upon impact.

## 4. Can a rod's rotation be controlled to minimize its impact force on a mass?

Yes, a rod's rotation can be controlled to minimize its impact force on a mass. By adjusting the rod's angle, velocity, and shape, its rotational motion can be manipulated to reduce the impact force on the mass. Additionally, adding a weight to one end of the rod can also help control its rotation and minimize the impact force.

## 5. How is the kinetic energy of a rod affected by its rotation when falling and hitting a mass?

The kinetic energy of a rod is affected by its rotation when falling and hitting a mass. The rotational kinetic energy of the rod adds to its linear kinetic energy, resulting in a greater total kinetic energy. This can cause the rod to exert a higher impact force on the mass upon collision. However, if the rotation is controlled and the rod's shape is optimized, the rotational kinetic energy can be reduced, resulting in a lower impact force on the mass.

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