A thin, uniform rod is hinged at its midpoint. To begin...

A thin, uniform rod is hinged at its midpoint. To begin with, one-half of the rod is bent upward and is perpendicular to the other half. This bent object is rotating at an angular velocity of $9.0 \mathrm{rad} / \mathrm{s}$ about an axis that is perpendicular to the left end of the rod and parallel to the rod's upward half (see the drawing). Without the aid of external torques, the rod suddenly assumes its straight shape. What is the angular velocity of the straight rod?

Key Concepts

Moment of Inertia

The moment of inertia quantifies an object's resistance to changes in its rotational motion and depends on the mass distribution relative to the rotation axis. When the shape of a body is altered, its moment of inertia changes, affecting the rotational speed needed to conserve angular momentum. This concept is essential when analyzing systems that undergo reconfiguration, as in the transition from the bent rod to the straight rod.

Rigid Body Reconfiguration

Reconfiguration of a rigid body involves a change in its shape or mass distribution without any external torques acting on it. In such cases, the redistribution of mass alters the moment of inertia, and therefore, the angular velocity adjusts to maintain the conserved angular momentum. This concept is key when dealing with problems where a system’s geometry, and hence its rotational characteristics, changes suddenly.

Conservation of Angular Momentum

This principle states that if no external torques act on a system, the total angular momentum remains constant throughout any process. In rotational dynamics problems like this one, it allows the initial and final angular momenta of the system to be set equal, providing a critical relationship for solving unknown quantities such as the final angular velocity when the configuration changes.

Transcript

00:01 Hello, my name is david.

00:02 In this video, we'll cover the conservation of angular momentum.

00:06 So for this problem, we're given that a uniform thin rock is hinge at its midpoint and is rotating with an initial angular velocity of 9 .0 radiance per second.

00:17 And with the presence of no external torques, it assumes its straight shape.

00:22 So we want to find the angular velocity when it does that.

00:28 So now before it assumes a straight shape, the initial moment of inertia is going to be equal to the moment of of the vertical portion plus the moment of inertia of the horizontal portion.

00:44 Now the moment of inertia for the vertical portion, you will apply the parallel axis theorem is going to be equal to the moment of inertia about the center of mass plus the mass times the distance from the rod to the axis of rotation square.

01:03 So the moment of inertia for the vertical portion is going to be equal to one half m r square plus m d square.

01:17 Now since this is a thin rod, that means that the radius is going to be negligible or equal to zero.

01:23 So the moment of inertia about its center of mass is going to be equal to zero.

01:28 So the moment of inertia for the vertical portion is going to be equal to use md square, where the mass is equal to the entire mass of the rod divided by two.

01:44 And the distance from the rod to the axis of rotation is going to be the length of the rod divided by two.

01:50 So we plug this to n we get that the moment of inertia for the vertical portion is equal to m divided by 2 times l divided by 2 square which gives us m over 2 times l square over 4 so the moment of inertia for the vertical portion is equal to m l square over 8 now for the horizontal rod the moment of inertia about one end is given by one third and m l square so again lowercase m is going to be equal to the total moment of inert that mean the total mass of the rot divided by two and l is going to be the length of the horizontal portion of the rod which is just the total length of the rot divided by two so we plug this end we get that the moment of inertia for the horizontal portion is equal to one -third times m over two times l over two squared.

03:17 So this gives us m over 6 times l square over 4.

03:25 So the moment of inertia for the horizontal portion is going to be equal to nl square divided by 24...

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