Kinetic energy and momentum in circular paths

[Like Tony Stark]

Homework Statement

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Relevant Equations

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I have some doubts about ballistic pendulums.
First, we say that if a bullet hits the pendulum, the linear momentum is conserved. But when we consider a rod attached to a pivot at one of its ends instead of a pendulum we say that the linear momentum isn't conserved because the rod can't move freely. What's the difference?
That also makes me wonder "if a body moves with constant speed in a circular path, is momentum conserved?", like a ball attached to a string.
And lastly, what about kinetic energy? Because there's rotational kinetic energy and there's translational kinetic energy. When do we have to distinguish them? Let's suppose a ball rolling and not sliding, it has both. But what about a pendulum? It rotates and its centre of mass moves too, so does it have both of them? Or a rod leaning against a wall and falling, what type of kinetic energy does it have?

 

[kuruman]

The argument about the ballistic pendulum is that linear momentum is conserved during the infinitesimally small time interval over which the projectile becomes embedded in the pendulum bob. In other words, the pendulum is not sufficiently displaced while the projectile sticks to it.

Linear momentum is a vector. It changes direction as the object moves in a circular path, therefore linear momentum is not conserved in that case.

Rotational kinetic energy is derived from linear kinetic energy therefore it is the same kind of beast. If you consider a rigid body consisting of many pieces $m_i$ each with its own speed $v_i$, the total kinetic energy is $K=\frac{1}{2}\sum_i m_iv_i^2$. If it so happens that the body is constrained to rotate about a fixed axis with angular speed $\omega,$ then $v_i=\omega r_i$ where $r_i$ is the distance of $m_i$ from the axis. Thus, $K=\frac{1}{2}\omega^2\sum_i m_ir_i^2$. With the moment of inertia about said axis $I=\sum_i m_ir_i^2$, you have $K=\frac{1}{2}I\omega^2$ also known as the rotational energy. This expression for the kinetic energy is always correct as long as the moment of inertia $I$ is about the axis of rotation, wherever that is. It is often customary but not necessary to write the kinetic energy as having two parts, translational kinetic energy of the center of mass plus rotational kinetic energy about the center of mass, $$K=\frac{1}{2}MV_{cm}^2+\frac{1}{2}I_{cm}\omega^2.$$

 

[haruspex]

we say that if a bullet hits the pendulum, the linear momentum is conserved. But when we consider a rod attached to a pivot at one of its ends instead of a pendulum we say that the linear momentum isn't conserved because the rod can't move freely.

The earliest ballistic pendulum had a relatively massive arm, so the rotational inertia was taken into account. (Angular momentum about the pivot would be conserved.)
In later ones the arm was very light, and in some the target was suspended fore and aft so that it did not rotate on its own axis.
https://en.m.wikipedia.org/wiki/Ballistic_pendulum

 

[jbriggs444]

But when we consider a rod attached to a pivot at one of its ends instead of a pendulum we say that the linear momentum isn't conserved because the rod can't move freely. What's the difference?

The collision is such that negligible horizontal impulse is required at the pivot point.

The horizontal linear force at the pivot point has to be enough to make the pendulum arm to rotate so that it stays attached to the pivot. If the pendulum arm is of negligible mass and if the bullet hits the block nicely centered then the required horizontal force is negligible.

 

[haruspex]

The collision is such that negligible horizontal impulse is required at the pivot point.

The horizontal linear force at the pivot point has to be enough to make the pendulum arm to rotate so that it stays attached to the pivot. If the pendulum arm is of negligible mass and if the bullet hits the block nicely centered then the required horizontal force is negligible.

... but you also need the mass to be pivoted where it joins the arm or the moment of inertia of the mass becomes an issue. See the reference to a parallelogram arrangement at the link I posted.