Advanced Classical Mechanics/Compound ballistic pendulum with spinning ball

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The easiest way to create motion with minimum friction is to roll balls down a ramp. For that reason it is worthwhile to develop theories for such systems. Here we consider the ballistic pendulum. This calculation models the stable configuration with a completely inelastic collision. However, it is not difficult to generalized these results to include partially elastic collisions, or even the knockdown of a block of wood by a ball rolled down a ramp.

The intent of this essay is for advanced physics majors to understand the derivation, and for beginners to utilize the results of these calculations. It should not be difficult to develop Matlab or Python codes that beginners could use in order to correlate classroom experiments to theory.

Warning: It must emphasized that the assumptions made regarding an "inelastic" collision made here might not always hold. See below for further discussion.

Unit vectors and multiple coordinate systems

Vector notation becomes more powerful when two or more coordinate systems are used, especially when the orthogonality of the unit vectors permit simple transformations between the coordinate systems. We illustrate this with the impulse delivered to the block by the ball. The impulse to the block is proportional to the final minus initial velocity of the ball:

$\overrightarrow{J}=m{\overrightarrow{v}}_i-m{\overrightarrow{v}}_f=J_x\hat{i}+J_y\hat{j}=J_1\widehat{e_1}+J_2\widehat{e_1}$

Using relations between the unit vectors easily derived by inspection of the figure, we can derive the Rotation matrix between the coordinate systems. Defining $\beta$ as the angle between $\hat{j}$ and $\widehat{e_1}$, we have:

$\begin{array}{cccc}\widehat{e_1}\cdot \hat{i}& =\sin \beta & \widehat{e_1}\cdot \hat{j}& =\cos \beta \\ \widehat{e_2}\cdot \hat{i}& =-\cos \beta & \widehat{e_2}\cdot \hat{j}& =\sin \beta \end{array}$

Take the inner product of $\overrightarrow{J}$ with the unit vectors $\widehat{e_1}$ and $\widehat{e_2}$ to obtain:

$J_1=J_x\sin \beta +J_y\cos \beta$   and   $J_2=-J_x\cos \beta +J_y\sin \beta$

Four equations of motion.

The impulse torque on the block

Following the concepts developed at Wikipedia's Impulse we integrate the equations of motion under the assumption that the collision is so brief that the ball and block change velocity but not position during the collision. The block is assumed to be initially at rest, with a vertical wall facing the incoming ball. The first equation relates the impulse of torque to the angular velocity of the block about the axis, which is situated a distance R away from the point of impact:

$I\dot{\theta }=-R{J}_2,$

where I is the block's moment of inertia and the minus sign above is due to the "negative" orientation of the unit vector.

The impulse torque on the ball

For a ball that is a solid sphere, we use Wikipedia's list of moments of inertia relate the change in the ball's angular velocity to the appropriate component of the impulse:

$\frac{2}{5}m{r}^2({\omega }_f-{\omega }_i)=-r{J}_y,$

where here the minus sign occurs because the impulse to the ball is opposite that to the block.

The (questionable) assumption of brief stationary contact between ball and block

The third and fourth equation arise from a single (two-dimensional) vector equation that arises from the fact that for an instant in time, the ball and block are in contact. At that moment, the portion of the ball and block that are in contact are moving at the same velocity. It must be emphasized that this equality of both objects at the point of contact might not actually occur. We this caveat in mind, we use the concept of relative velocity and use the rotation of the ball's surface about the center of mass to find the velocity at the point of contact with the block.

${\overrightarrow{v}}_f+{\omega }_{f}r\hat{j}=-R\dot{\theta }\widehat{e_2}$

On nature of these equations

Before solving a set equations, it is important to first classify the type of equations they are. In this case, the four equations are linear, or first order, because in the four unknowns, no polynomial has degree higher than one. Here, the "unknowns" are the variables that describe the angular and linear velocities associated with the ball and ramp immediately after the collision, namely $(\dot{\theta },{\omega }_f,v_{fx},v_{fy})$. These "unknowns" can be deduced if the initial velocities $({\omega }_i,v_{ix},v_{iy})$ are known. This linear set of equations is also inhomogeneous , as discussed in Wikipedia's System of linear equations, where it is shown that such equations have either no solution, one unique solution, or an infinite number of solutions. We shall later see that one unique solution exists for this system.

Finding a solution

It would be interesting to see if Matlab or Mathematica can solve this problem symbolically. Meanwhile, we shall outline the steps required to obtain a solution that arrives at the final velocities (angular and linear) given the initial velocities. The first step is to write these equations as an inhomogeneous set of equations:

Algebra

The algebra is tedious. In a physics lab we will break down the steps and have individual groups verify two or three steps. We will also see if Matlab symbolic manipulation is up to the task.

{{cot|click to view the steps}}

$J_1=J_x\sin \beta +J_y\cos \beta =m{v}_{i1}-m{v}_{f1}$

$J_2=-J_x\cos \beta +J_y\sin \beta =m{v}_{i2}-m{v}_{f2}$

Step 0:Collect the equations into one place:

$I\dot{\theta }=-R{J}_2$

$m\frac{2}{5}{r}^2({\omega }_f-{\omega }_i)=-r{J}_y$

${\overrightarrow{v}}_f+{\omega }_{f}r\hat{j}=-R\dot{\theta }\widehat{e_2}$

Step 1: Eliminate J (after factoring)

$I\dot{\theta }=mR(v_{f2}-v_{i2})$

If everything above this Algebra section is true. Then step 0 and the first of Step 1 is confirmed:

$m\frac{2}{5}{r}^2{\omega }_f-m\frac{2}{5}{r}^2{\omega }_i=mr{v}_{fy}-mr{v}_{iy}$

${\overrightarrow{v}}_f\cdot \hat{i}+{\omega }_{f}r\hat{j}\cdot \hat{i}=-R\dot{\theta }\widehat{e_2}\cdot \hat{i}$

${\overrightarrow{v}}_f\cdot \hat{j}+{\omega }_{f}r\hat{j}\cdot \hat{j}=-R\dot{\theta }\widehat{e_2}\cdot \hat{j}$

Step 2: Move inhomogeneous terms to the RHS (right hand side) and simplify the vector equation.

$I\dot{\theta }-mR{v}_{f2}=-mR{v}_{i2}$

$\frac{2}{5}m{r}^2{\omega }_f-mr{v}_{fy}=\frac{2}{5}m{r}^2{\omega }_i-mr{v}_{iy}$

$v_{fx}+R\dot{\theta }\widehat{e_2}\cdot \hat{i}=0$

$v_{fy}+{\omega }_{f}r+R\dot{\theta }\widehat{e_2}\cdot \hat{j}=0$

Step 3: When working by hand it helps to simplify the notation with temporary variables:

$𝒜=-mR{v}_{i2}$

$ℬ=\frac{2}{5}r{\omega }_i-v_{iy}$

$\widehat{e_2}\cdot \hat{i}=-𝒞=-\cos \beta$

$\widehat{e_2}\cdot \hat{j}=𝒮=\sin \beta$

Step 4: Substitute

$I\dot{\theta }-mR{v}_{f2}=𝒜$

$\frac{2}{5}r{\omega }_f-v_{fy}=ℬ$

$v_{fx}-R\dot{\theta }𝒞=0$

$v_{fy}+r{\omega }_f+R\dot{\theta }𝒮=0$

Step 5: Substitute: $v_{f2}=-v_{fx}𝒞+v_{fy}𝒮$

$I\dot{\theta }+mR𝒞v_{fx}-mR𝒮v_{fy}=𝒜$

$\frac{2}{5}r{\omega }_f-v_{fy}=ℬ$

$v_{fx}-R𝒞\dot{\theta }=0$

$v_{fy}+r{\omega }_f+R𝒮\dot{\theta }=0$

Step 6: Reduce to three equations using $v_{fx}=R𝒞\dot{\theta }$

$I\dot{\theta }+m{R}^2{𝒞}^2\dot{\theta }-mR𝒮v_{fy}=𝒜$

$\frac{2}{5}r{\omega }_f-v_{fy}=ℬ$

$v_{fy}+r{\omega }_f+R𝒮\dot{\theta }=0$

Step 7: Substitute :$v_{fy}=-r{\omega }_f-R𝒮\dot{\theta }$

$I\dot{\theta }+m{R}^2{𝒞}^2\dot{\theta }+mR𝒮r{\omega }_f+m{R}^2{𝒮}^2\dot{\theta }=𝒜$

$\frac{2}{5}r{\omega }_f+r{\omega }_f+R𝒮\dot{\theta }=ℬ$

Step 8: Simplify

$I\dot{\theta }+m{R}^2\dot{\theta }-mR𝒮r{\omega }_f=𝒜$

$\frac{7}{5}r{\omega }_f+R𝒮\dot{\theta }=ℬ$

Step 9: Substitute $r{\omega }_f=-\frac{5}{7}R𝒮\dot{\theta }+\frac{5}{7}ℬ$

$I\dot{\theta }+m{R}^2\dot{\theta }-mR𝒮(-\frac{5}{7}R𝒮\dot{\theta }+\frac{5}{7}ℬ)=𝒜$

$I\dot{\theta }+m{R}^2\dot{\theta }+\frac{5}{7}m{R}^2{𝒮}^2\dot{\theta }=𝒜+\frac{5}{7}mR𝒮ℬ$

Step 10: Substitute $𝒜=-mR{v}_{i2}$   and   $ℬ=\frac{2}{5}r{\omega }_i-v_{iy}$

$I\dot{\theta }+m{R}^2\dot{\theta }+\frac{5}{7}m{R}^2{𝒮}^2\dot{\theta }=-mR{v}_{i2}+\frac{5}{7}mR𝒮(\frac{2}{5}r{\omega }_i-v_{iy})$

$I\dot{\theta }+m{R}^2\dot{\theta }+\frac{5}{7}m{R}^2{𝒮}^2\dot{\theta }=-mR{v}_{i2}+\frac{2}{7}mRr𝒮{\omega }_i-\frac{5}{7}mR𝒮v_{iy}$

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$I\dot{\theta }+m{R}^2\dot{\theta }+\frac{5}{7}m{R}^2{𝒮}^2\dot{\theta }=m{v}_{ix}R𝒞-m{v}_{iy}R𝒮+\frac{2}{7}mRr𝒮{\omega }_i-\frac{5}{7}mR𝒮v_{iy}$

Step 11: Make the small S approximation for a thin board $(𝒮=\sin \beta =a/R<<1)$

$(I+m{R}^2)\dot{\theta }\approx m{v}_{ix}R\cos (\beta )=m{v}_{ix}b$