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## Instructions:

Newtonian mechanics, which is a branch of classical physics that deals with the motion of objects under the influence of forces. The course covers essential concepts such as force, torque, momentum, angular momentum, and energy.

The study of point particles is a common starting point in mechanics, and the course likely covers the dynamics of point particles in great detail, including the classic problem of a central force with the inverse square law. Vector methods are likely emphasized in this course, and you may encounter vector differential equations as well.

Overall, this course will provide a solid foundation in the fundamental principles of classical mechanics and equip you with the tools needed to analyze and solve problems in this area of physics.

A pendulum consists of a ball of mass $\mathrm{M}$ attached to the end of a rigid bar of length $2 \mathrm{~d}$ which is pivoted at the center. At the other end of the bar is a container (“catch”). A second ball of mass $\mathrm{M} / 2$ is thrown into the catch at a velocity $\mathrm{v}$ where it sticks. For this problem, ignore the mass of the pendulum bar and catch, and treat the balls as if they were point masses (i.e., neglect their individual moments of inertia).

(a) What is the initial angular rotation rate of the pendulum after the incoming ball is caught?

(a) The angular momentum of the system relative to the pivot point just prior to the upper ball being caught is:

$$

L=\frac{M}{2} v d

$$

As there are no external torques acting on the system relative to the pivot point during the time of the collision, the angular momentum can be determined as:

$$

L=I \omega=\left(M d^2+\frac{M}{2} d^2\right) \omega=\frac{3}{2} M d^2 \omega

$$

hence,

$$

\omega=\frac{v}{3 d}

$$

(b) How much total mechanical energy is lost when the incoming ball gets stuck in the catch?

(b) The energy of the upper ball prior to sticking to the catch is:

$$

E_i=\frac{1}{2} \frac{M}{2} v^2=\frac{1}{4} M v^2

$$

After the collision, all of the energy can be expressed as pure rotation about the stationary pivot point, hence:

$$

E_f=\frac{1}{2} I \omega^2=\frac{1}{2}\left(\frac{3}{2} M d^2\right)\left(\frac{v}{3 d}\right)^2=\frac{1}{12} M v^2

$$

so

$$

\Delta E=E_f-E_i=\frac{1}{6} M v^2

$$

\begin{prob}

(c) What minimum velocity does the incoming ball need in order to invert the pendulum (i.e., rotate it by $180^{\circ}$ )?

\end{prob}

\begin{proof}

(c) Because there are only conservative forces acting (the pivot does not move so it

does no work) total mechanical energy is conserved. Assuming that the bottom of

the swing sets the zeropoint for gravitational potential, at the point that the ball

sticks the total energy is:

$$

E_i=\frac{M}{2}(2 d) g+\frac{1}{12} M v^2

$$

The point at which the pendulum just comes to rest inverted, the bottom mass is at height $2 \mathrm{~d}$ :

$$

E_f=M(2 d) g

$$

With $\Delta \mathrm{E}=0$ this gives:

$$

\begin{aligned}

& 2 M d g=M d g+\frac{1}{12} M v^2 \

& \Rightarrow v=\sqrt{12 d g}

\end{aligned}

$$