Elastic Collision of a rod which can move and rotate with a particle obliquely

Question

I am facing trouble with the following question

A rod $AC$ of length $l$ and mass $m$ is kept on a horizontal smooth surface. It is free to rotate and move. A particle of same mass $m$ moving on the plane with velocity $v$ strikes rod at $B$ elastically as shown in the figure below.

The question is to find the angular velocity of the rod and its final velocity.

Attempt

I tried basically by applying conservation of energy , linear momentum and angular momentum.

Let the final situation be something like this

Now conservation of energy gives $$1/2mv^2=1/2m(v_5^2+v_6^2)+1/2m(v_3^2+v_4^2)+1/2(1/12ml^2w^2)$$

and conservation of angular momentum $$m(4v/5)(3l/4)=(1/12ml^2w^ 2)w-mv_4(3l/4)$$

and by linear momentum in two directions $$ 3v/4=v_3+v_6$$ and $$4v/5=v_5-v_4$$

But I have 5 unknowns and I could manage only 4 equations. Moreover solving these equations is lengthy and I think there must be some quicker way to solve this being a entrance exam question with time restrictions. Any help would be highly appreciated. Thanks.

Answer 1

Im not solving the whole question but u can also write the missing equation for cofficient of resistution

$e=\frac{\text{Velocity of sepration}}{\text{Velocity of approach}}$

($e=1$ for elastic collision)

And I think this question can also solved by conserving angular momentum about point of collision as Net torque about all the forces are 0 about this point.If you like you can also use Impulse momentum theorem though it will produce same result as yours.

Answer 2

One way to solve this kind of problem is to find the reduced mass of the rod along the contact normal.

Problem Setup

Consider the general situation below. A moving rigid body is impacted by a sphere. The motion of the rigid body is tracked by the location $\vec{r}_1$, velocity $\vec{v}_1$ of the center of mass and the rotational velocity $\vec{\omega}_1$. The impact occurs on some point A with location $\vec{r}_A$, and along the contact normal direction $\hat{n}$. The response depends on the vector $\vec{c} = \vec{r}_1 - \vec{r}_A$ describing the location of the center of mass relative to the contact point.

After the impact, there is a step in motion of the center of mass described by the vectors $\Delta \vec{v}_1$ and $\Delta \vec{\omega}_1$. The impact consists of an exchange of momentum with magnitude $J$ (called an impulse) along the contact normal $\hat{n}$ such that $$\begin{aligned} \Delta \vec{v}_1 & = \tfrac{1}{m_1} \hat{n} J \\ \Delta \vec{\omega}_1 & = \mathrm{I}_1^{-1} ( -\vec{c} \times \hat{n} J) \end{aligned} \tag{1}$$

General 3D solution

The kinematics of the contact point A after the impact defines the impulse magnitude $J$. Before impact, the speed of approach is

$$v_{\rm imp} = \hat{n} \cdot( \vec{v}_1 + \vec{c} \times \vec{\omega}_1 - \vec{v}_2 ) \tag{2}$$

where $\vec{v}_2$ is the velocity of the impacting body. And after the impact the speed of release is equal and opposite of the speed of approach for an elastic contact.

$$ v_{\rm rel} = - v_{\rm imp} $$

or the change in speed must be twice the impact speed

$$ \hat{n} \cdot(\Delta \vec{v}_1 + \vec{c} \times \Delta \vec{\omega}_1 - \Delta \vec{v}_2 ) = - 2 \hat{n} \cdot( \vec{v}_1 + \vec{c} \times \vec{\omega}_1 - \vec{v}_2 ) \tag{3}$$

Use (1) in (3) to find $J$ as

$$ J = \frac{2 \hat{n} \cdot( \vec{v}_1 + \vec{c} \times \vec{\omega}_1 - \vec{v}_2 )}{\frac{1}{m_1}-\hat{n}\cdot(\vec{c}\times \mathrm{I}_1^{-1} \vec{c}\times \vec{n}) + \frac{1}{m_2}} $$

or how I prefer to state it

$$ J = 2\; m_{\rm imp}\; v_{\rm imp} \tag{5} $$

with the reduced mass defined as $$m_{\rm imp} = \left( \tfrac{1}{m_1}-\hat{n}\cdot(\vec{c}\times \mathrm{I}_1^{-1} \vec{c}\times \vec{n}) + \tfrac{1}{m_2} \right)^{-1} \tag{6}$$

Notice that for two spheres the above reduces to $m_{\rm imp} = \left( \tfrac{1}{m_1} + \tfrac{1}{m_2} \right)^{-1}$ which is a well known formula.

Once $J$ is known, then the motion afterward is calculated from (1)

Planar Projection

Now to make the planar projection of the above with $\hat{n} = \pmatrix{n_x \\ n_y}$ and $\vec{c} = \pmatrix{c_x \\ c_y}$

and use it in (6) to get

$$ m_{\rm imp} = \left( \frac{1}{m_1} + \frac{(c_x n_y - c_y n_x)^2}{I_1} + \frac{1}{m_2} \right)^{-1} \tag{7}$$

And the change motion from (1) and (5) in 2D

$$ \begin{aligned} \Delta \vec{v}_1 & = \pmatrix{ \tfrac{n_x}{m_1} J \\ \tfrac{n_y}{m_1} J } \\ \Delta \omega_1 & = \tfrac{n_x c_y - n_y c_x}{I_1} J \end{aligned} \tag{8}$$

Specific Problem

The solution to the question is found by using the following geometry

$$ \begin{aligned} I_1 & = \tfrac{m}{12} \ell^2 \\ \vec{c} & = \pmatrix{-\tfrac{\ell}{4} \\ 0} \\ \vec{n} & = \pmatrix{0 \\ 1} \end{aligned} $$

which results in $m_{\rm imp} = \left( \frac{1}{m_1} + \frac{3}{4 m_1} + \frac{1}{m_2} \right)^{-1}$ and so to get $J$ from (5) and the motion from (1).

The choice of $\hat{n} = \pmatrix{0\\1}$ is because the contact normal has to be normal to both surfaces, and thus perpendicular to the rod. It would be a mistake to assume that $\hat{n}$ must be parallel to the impacting velocity $\vec{v}_2$ which as you can see in my diagram they are at an angle to each other.

Answer 3

If you write the equations shortly after the collision . you obtain:

\begin{align*} \left[ \begin {array}{c} J_{{r}} \left(\dot\varphi _{{{\it rf}}}-\dot \varphi _{{{\it ri}}} \right) =P\sin \left( \alpha \right) L \\ m \left( {\it \dot{x}}_{{{\it rf}}}-{\it \dot{x}}_{{{\it ri }}} \right) =P\,\cos(\alpha)\\ m \left( {\it \dot{y}}_{{{\it rf}}}-{ \it \dot{y}}_{{{\it ri}}} \right) =P\sin \left( \alpha \right) \\ m \left( v_{{{\it bf}}}-v_{{{\it bi}}} \right) =- P\\ \cos(\alpha)\dot{x}_{rf}+\sin(\alpha)\left(L\,\dot{\varphi}_{rf} +\dot{y}_{rf}\right)=v_{bf} \quad\quad (**)\end {array} \right]\tag 1 \end{align*}

with $~L=\frac 14 l~,$ P is the linear momentum

$~,i~$ stay for initial , $~f~$ for final $~r~$ rod and $~b~$ for ball

you have 5 equations (Eq. (1)) for 5 unknowns $~\dot\varphi_{rf}~,v_{bf}~,\dot{x}_{rf}~,\dot{y}_{rf}~,P~$ .

the velocity at the contact point

\begin{align*} \left(\left(\begin{bmatrix} 0 \\ 0 \\ \dot{\varphi}_{rf} \\ \end{bmatrix}\times \begin{bmatrix} L \\ 0 \\ 0 \\ \end{bmatrix}\right)+ \begin{bmatrix} \dot{x}_{re} \\ \dot{y}_{re} \\ 0 \\ \end{bmatrix}\right)^T \begin{bmatrix} \cos(\alpha) \\ \sin(\alpha) \\ 0 \\ \end{bmatrix}=v_{bf}\quad\quad (**) \end{align*}