91Ó°ÊÓ

Linear momentum is a property of objects which are changing their position with respect to a reference point. Angular momentum is a property of objects which are changing the angle of their position vector with respect to a reference point.

Linear and angular momentum are conserved independently of each other. Being by writing the linear momentum equation as you normally would and use relations until you have an equation that can be solved for \({v_3}\) and \({\theta _3}\) in terms on known quantities. Then rite out the angular momentum equation and do the same until you can solve for \({\omega _3}\).

From the law of conservation of linear momentum,

\({\overrightarrow p _1} + {\overrightarrow p _2} = {\overrightarrow p _3} + {\overrightarrow p _4}\)

Here,\({\overrightarrow p _1}\)and \({\overrightarrow p _3}\)are the initial and final momenta of the dumbbell,\({\overrightarrow p _2}\)and \({\overrightarrow p _4}\)are the initial and final momenta of the particle.

Write this equation in terms of the mass and velocities of the system.

\(2{m_1}{\overrightarrow v _1} + {m_2}{\overrightarrow v _2} = 2{m_1}{\overrightarrow v _3} + 2{m_2}{\overrightarrow v _4}\)

Now separate the vector in to \(\left( x \right)\) and \(\left( y \right)\) component equations.

\(2{m_1}{v_{1x}} + {m_2}{v_{2x}} = 2{m_1}{v_{3x}} + {m_2}{v_{4x}}\) ….. (1)

\(2{m_1}{v_{1y}} + {m_2}{v_{2y}} = 2{m_1}{v_{3y}} + {m_2}{v_{4y}}\) ….. (2)

Now use the vector identify as given below.

\({v_x} = v\cos \theta \)

And \({v_y} = v\sin \theta \)

Hence, equation (1) and (2) becomes,

\(\begin{aligned}{l}2{m_1}{v_1} + {m_2}{v_2}\cos {\theta _2} = 2{m_1}{v_3}\cos {\theta _3} + {m_2}{v_4}\cos {\theta _4}\\{m_2}{v_2}\sin {\theta _2} = 2{m_1}{v_3}\sin {\theta _3} + {m_2}{v_4}\sin {\theta _4}\end{aligned}\)

Now, you have two equations with two unknowns. These equations can be used to solve for \({\theta _3}\) and \({v_3}\).

Solve for \({\omega _3}\)by using angular momentum conservation. From the law of conservation of angular momentum.

\({\overrightarrow L _1} + {\overrightarrow L _2} = {\overrightarrow L _3} + {\overrightarrow L _4}\)

Angular momentum vector is,

\(\begin{aligned}{c}\overrightarrow L = I\overrightarrow \omega \\ = \overrightarrow r \times \overrightarrow p \\ = m\overrightarrow r \times \overrightarrow v \end{aligned}\)

You can write the equation as follows:

\({I_1}{\overrightarrow \omega _1} + {m_2}{\overrightarrow r _2} \times {\overrightarrow v _2} = {I_3}{\overrightarrow \omega _3} + {m_2}{\overrightarrow r _4} \times {\overrightarrow v _4}\)

Now, you need to solve for the moments of inertia and the lever-arms.

The moment of inertia for a dumbbell shaped object is,

\(I = \frac{1}{2}m{L^2}\)

Here,\(L\)is the length of the rod.

The object does not change its shape or mass, so the moment of inertia before and after the collision is,

\(\begin{aligned}{l}{I_1} = {I_3}\\ = \frac{1}{2}{m_1}{L^2}\end{aligned}\)

The lever-arm of the incoming particle is the shortest distance from the particle to the axis of rotation. You can choose a lever-arm at any point along the particle’s path and the angular momentum of the particle will remain the same. Choose a possible lever-arm\({\overrightarrow r _4}\)that simplifies the calculations. You know that immediately before and after the collision the particle is at\(B\)distance away from the axis of rotation and the angle the momentum makes with the rod’s center-of-mass, s calculates the angular momentum at the collision point.

Substitute the moment of inertia and lever-arms into the angular momentum equations.

\(\frac{1}{2}{m_1}{L^2}{\overrightarrow \omega _1} + {m_2}\overrightarrow B \times {\overrightarrow v _2} = \frac{1}{2}{m_1}{L^2}{\overrightarrow \omega _3} + {m_2}\overrightarrow B \times {\overrightarrow v _4}\)

Now you only need to solve for the initial angular velocity of the dumbbell. The dumbbell was not spinning, so the initial angular velocity was zero.

\({m_2}\overrightarrow B \times {\overrightarrow v _2} = \frac{1}{2}{m_1}{L^2}{\overrightarrow \omega _3} + {m_2}\overrightarrow B \times {\overrightarrow v _4}\) ….. (3)

The vector \(\overrightarrow B \)is,

\(\overrightarrow B = \left\langle {0, - B,0} \right\rangle \)

The cross products are fairly easy to calculate and come out to be, \({m_2}\overrightarrow B \times {\overrightarrow v _2} = - mB{v_2}\widehat z\)and \({m_2}\overrightarrow B \times {\overrightarrow v _4} = - mB{v_4}\widehat z\).

Substitute these two values into the equation(3).

Therefore, the angular momentum equation is.

\( - mB{v_2}\widehat z = \frac{1}{2}{m_1}{L^2}{\overrightarrow \omega _3} - mB{v_4}\widehat z\)

This is a single equation with only one unknown \({\overrightarrow \omega _3}\).This can be used to solve for the angular velocity.