1.1. Solution

In the previous lecture, we had derived that the end-to-end rotation Ω is given by (1) but it holds only when the cross-section is uniform over the entire length. For the current case, a different form of the above equation is useful as shown below: (2) twist(κ) where κ denotes twist or the rate of change of rotation of the cross-section.

This chunk introduces the foundational equations related to the twisting of the cylinder. The first point is that the standard equation for end-to-end rotation, labeled as (1), is applicable only when the cross-section of the cylinder is consistent throughout its length. In this case, due to variable radius, a modified equation needs to be applied, indicated in formula (2). Here, twist (κ) is defined as the measure of how much the cross-section of the cylinder rotates along its length.

Imagine twisting a rubber band. If you twist it uniformly, it turns in a predictable manner. However, if one end is thicker and the other thinner and you twist it, the thicker part will twist less than the thinner part. This is similar to the problem we're addressing, where the radius of the cylinder changes, influencing how much it twists.

As the same torque T acts in every cross-section (as shown below), the above equation implies that twist varies along the length as the polar moment of area J is varying due to varying radius.

In this chunk, we learn about the consistency of torque applied across the entire cylinder. Even though the radius changes at different points, the torque (T) remains the same at every cross-section, indicating the importance of torque in determining how the twist (κ) varies. This is affected by the changing polar moment of area, which influences how much each section twists as you move down the length of the cylinder.

Think of holding a towel and twisting it. The point of grip (where you apply torque) dictates how the towel twists throughout its length, regardless of whether it is thicker or thinner at certain points. So as you twist it, the same force affects the entire towel, but the twist may vary depending on how thick the fabric is at different spots.

The variation of T along the axis of the cylinder can be found by making a free body diagram. We cut a section of the cylinder at a distance x from the clamped end. The two cut parts are shown in Figure 2.

This chunk discusses creating a free body diagram of the cylinder to analyze the internal mechanics. By slicing through the cylinder at a distance x from where it is clamped, we can study the forces and torques acting on that section. This allows us to visualize how external and internal torques relate and confirms that they must be balanced for the system to stay in equilibrium.

Imagine slicing a straw in half and looking at one of the halves. The pressure on one side of the cut needs to be balanced with the other half to keep the juice from spilling out. Similarly, the torque balance ensures that the cylinder's internal forces correctly counterbalance the applied external forces.

We then do the moment balance for the right section. The total moment on this section should add up to zero, i.e., T − T(x) = 0 ⇒ T(x) = T. This means that the torque at every cross-section is the same and equal to applied external torque T even though cross-sectional radius is changing here.

In this step, the moment balance equation is established, which states that the torque value at any cross-section of the cylinder (T(x)) must equal the applied torque (T). This reinforces that despite changes in the cross-section, the torque remains constant along the length of the cylinder. It’s a crucial insight that helps in further calculations of how the material deforms under the application of torque.

Think about a train moving on a track. No matter how many turns the track takes, the engine (representing the applied torque) maintains the same pulling force throughout. Similarly, the 'constant torque' concept illustrates how the twisting effect is uniformly applied across varying cross-sections, allowing the entire cylinder to work as a cohesive unit.

To find J(x) for a cross-section of radius r(x), we integrate over the area of the cross section.

Here, we focus on determining the polar moment of inertia (J) related to the cross-sectional area of the cylinder based on its varying radius. The formula entails that to compute the polar moment accurately, you must consider the entire area of the cross-section. This step is vital because J is essential for calculating how much the material will twist under the applied torque.

Consider how force is distributed across a trampoline. Depending on how stretched or compressed its surface is, each section can handle varying amounts of force. Similarly, as the radius of the cylinder changes, we need to account for that fluctuation to grasp how effective the cross-section is at resisting twist.

Using equation (4), the end-to-end rotation can be found by integrating twist along the axis of the cylinder, i.e., (left end of the tube is clamped).

This chunk explains how to compute the total end-to-end rotation (Ω) by integrating the previously deduced twist (κ) along the entire length of the cylinder. Given that one end is clamped, the integration considers how twist accumulates from that fixed point to the free end, yielding a comprehensive view of the cylinder's deformation due to the applied torque.

Think of pulling on a licorice stick. As you pull on one end, the twist you create at the other end accumulates based on how you pull it. The longer the licorice, the more twist happens from that initial pull until the other end. Similarly, integration of twist gives us the total rotation experienced by the cylinder from end to end.

--