2 concentric cylinders with radius r and 2r having a common axis are present . if the bottom part is given a twist 0 then the twist at the middle part is ?

1. 0
2. 0/2
3. 160/17
4. 170/16

To solve the problem of determining the twist at the middle part of two concentric cylinders when the bottom part is twisted, we need to consider how the twist propagates through the material of the cylinders. The key here is to understand the relationship between the twist applied to the inner cylinder and how it affects the outer cylinder.

Understanding the Setup

We have two cylinders: the inner cylinder with radius \( r \) and the outer cylinder with radius \( 2r \). Both cylinders share a common axis, and when the bottom part of the inner cylinder is twisted, this twist will influence the outer cylinder due to their physical connection.

Analyzing the Twist Propagation

When a twist is applied to the inner cylinder, it creates a shear strain that affects the outer cylinder. The amount of twist experienced by the outer cylinder depends on the ratio of the radii of the two cylinders. Since the outer cylinder is larger, it will experience a smaller twist compared to the inner cylinder.

Mathematical Representation

Let’s denote the twist applied to the inner cylinder as \( \theta \) (in degrees). The twist at the outer cylinder can be calculated using the following relationship:

• The twist at the outer cylinder, \( \theta_{outer} \), can be expressed as:
• \( \theta_{outer} = \frac{r_{inner}}{r_{outer}} \times \theta_{inner} \)

Substituting the values of the radii:

• Inner radius \( r_{inner} = r \)
• Outer radius \( r_{outer} = 2r \)

Thus, the equation becomes:

\( \theta_{outer} = \frac{r}{2r} \times \theta_{inner} = \frac{1}{2} \times \theta_{inner} \)

Applying the Given Twist

If the bottom part of the inner cylinder is twisted by \( 0^\circ \), then:

• Let’s say \( \theta_{inner} = 0^\circ \)
• Then, substituting this into our equation gives:
• \( \theta_{outer} = \frac{1}{2} \times 0^\circ = 0^\circ \)

This means that if the inner cylinder is not twisted at all, the outer cylinder will also not experience any twist. Therefore, the twist at the middle part of the outer cylinder remains \( 0^\circ \).

Final Thoughts

In summary, when the bottom part of the inner cylinder is given a twist of \( 0^\circ \), the twist at the middle part of the outer cylinder is also \( 0^\circ \). This illustrates how the mechanical properties of materials and their geometrical relationships influence the transmission of rotational motion.