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Two cylinders of same material and of same length are joined to end as shown in figure. The upper end of \(A\) is rigidly fixed. Their radii are in ratio of \(1: 2\). If the lower end of \(B\) is twisted by an angle \(\theta\), the angle of twist of cylinder \(A\) is (a) \(\frac{15}{16} \theta\) (b) \(\frac{16}{15} \theta\) (c) \(\frac{16}{17} \theta\) (d) \(\frac{17}{16} \theta\)

Short Answer

Expert verified
The angle of twist of cylinder A is \( \frac{16}{17} \theta \) (option c).

Step by step solution

01

Understanding the Problem

We have two cylinders, A and B, of the same length and material, with radii in the ratio 1:2. When the lower end of cylinder B is twisted by angle \( \theta \), we need to find the angle of twist of cylinder A.
02

Relationship Between Torque and Angle of Twist

For a cylinder, the angle of twist \( \phi \) is given by \( \phi = \frac{TL}{JG} \), where \( T \) is the torque, \( L \) is the length, \( J \) is the polar moment of inertia, and \( G \) is the shear modulus. Since the cylinders are of the same material and length, their angle of twist is proportional to \( \frac{T}{J} \).
03

Calculating the Polar Moment of Inertia

The polar moment of inertia \( J \) for a cylinder is \( J = \frac{\pi r^4}{2} \). Let the radius of cylinder A be \( r \) and that of cylinder B be \( 2r \). Thus, the polar moment of inertia for cylinder A is \( \frac{\pi r^4}{2} \) and for cylinder B, it is \( \frac{\pi (2r)^4}{2} = 16\frac{\pi r^4}{2} \).
04

Relation of Torque and Angle of Twist in the System

Since both the cylinders are in series, the same torque \( T \) is applied to both. The angle of twist in cylinder B can be expressed in terms of its torque \( \phi_B = \frac{TL}{JG_B} \), and similarly for A, \( \phi_A = \frac{TL}{JG_A} \). But since \( G \) and \( L \) are constant across the same material and length, we find \( \frac{\phi_B}{\phi_A} = \frac{J_A}{J_B} = \frac{1}{16} \).
05

Determine the Angle of Twist for Cylinder A

The total twist \( T_B \) at B is given as \( \theta = \phi_A + \phi_B \). Substituting the relation \( \phi_B = 16\phi_A \), we get \( \theta = \phi_A + 16\phi_A = 17\phi_A \). Therefore, \( \phi_A = \frac{\theta}{17} \), relating \( \phi_A = \frac{\theta}{17/16} = \frac{16}{17} \theta \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cylinder Torque
Torque is a measure of the rotational force applied to an object, causing it to rotate around an axis. In the case of cylinders, understanding torque is crucial because it directly affects how the material twists under the applied force. Torque is represented by the symbol \( T \) and is measured in units like Newton-meters (Nm). When dealing with rotational systems like cylinders, torque is crucial because:
  • It determines the amount of force applied to rotate the cylinder.
  • In a system where multiple cylinders are joined, like in the exercise, torque remains constant across the system if they are in series.
In the problem, since both cylinders A and B experience the same torque, it is pivotal to understand that while the force stays the same, it is distributed differently across each cylinder due to their differing geometry, particularly their radii ratio of 1:2.
Polar Moment of Inertia
The polar moment of inertia is an important geometric property when assessing how objects resist torsion, or twisting. For a cylinder, the polar moment of inertia \( J \) is expressed through the formula \( J = \frac{\pi r^4}{2} \), where \( r \) is the radius of the cylinder.In our problem:
  • Cylinder A has radius \( r \) and a polar moment of inertia \( \frac{\pi r^4}{2} \).
  • Cylinder B, being of twice the radius (2r), has a significantly larger polar moment of inertia \( 16 \times \frac{\pi r^4}{2} \).
The polar moment of inertia indicates how strongly a cross-sectional area of a cylinder resists twisting. Therefore, cylinder B, having a much larger polar moment of inertia compared to cylinder A, resists twisting more effectively. This contrast affects the angle of twist each cylinder experiences, as a smaller \( J \) means a greater angle of twist for the same torque.
Shear Modulus
Shear modulus, often denoted as \( G \), measures a material's ability to resist deformation under shear stress. It is a fundamental property of materials that describes their elasticity in response to applied torque. The higher the shear modulus, the more resistant a material is to deformation. The units are typically Pascals (Pa).In the context of rotating cylinders subjected to torque:
  • Shear modulus contributes to the formula for angle of twist: \( \phi = \frac{TL}{JG} \).
  • Both cylinders A and B are made of the same material, meaning they share the same shear modulus \( G \).
This uniformity in shear modulus across both cylinders implies that any difference in the angle of twist is solely due to variations in their geometry (radii) and not because of the material's resistance to deformation. Understanding shear modulus helps predict how a given material will react under specific forces, which is crucial for calculating how much a cylinder will twist when torque is applied.

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Most popular questions from this chapter

When a weight of \(5 \mathrm{~kg}\) is suspended from a copper wire of length \(30 \mathrm{~m}\) and diameter \(0.5 \mathrm{~mm}\), the length of the wire increases by \(2.4 \mathrm{~cm}\). If the diameter is doubled, the extension produced is (a) \(1.2 \mathrm{~cm}\) (b) \(0.6 \mathrm{~cm}\) (c) \(0.3 \mathrm{~cm}\) (d) \(0.15 \mathrm{~cm}\)

A wire \(3 \mathrm{~m}\) in length and \(1 \mathrm{~mm}\) in diameter at \(30^{\circ} \mathrm{C}\) is kept in a low temperature at \(-170^{\circ} \mathrm{C}\) and is stretched by hanging a weight of \(10 \mathrm{~kg}\) at one end. The change in length of the wire is \(\left(Y=2 \times 10^{11} \mathrm{Nm}^{-2}, g=10 \mathrm{~ms}^{-2}\right.\) and \(\left.\alpha=1.2 \times 10^{-50} \mathrm{C}^{-1}\right) \quad\) IUP SEE 2006] (a) \(5.2 \mathrm{~mm}\) (b) \(2.5 \mathrm{~mm}\) (c) \(52 \mathrm{~mm}\) (d) \(25 \mathrm{~mm}\)

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A steel wire of length \(4.7 \mathrm{~m}\) and cross-sectional area \(3.0 \times 10^{-5} \mathrm{~m}^{2}\) stretches by the same amount as a copper wire of length \(3.5 \mathrm{~m}\) and cross-sectional area \(4.0 \times 10^{-5} \mathrm{~m}^{2}\) under a given load. The ratio of the Young's modulus of steel to that of copper is \([\) NCERT] (a) \(1.2\) (b) \(1.8\) (c) \(1.5\) (d) \(1.19\)

When a \(4 \mathrm{~kg}\) mass is hung vertically on a light spring that obeys Hooke's law, the spring stretches by \(2 \mathrm{~cm}\). The work required to be done by an external agent in stretching this spring by \(5 \mathrm{~cm}\) will be (a) \(4.9 \mathrm{~J}\) (b) \(2.45 \mathrm{~J}\) (c) \(0.495 \mathrm{~J}\) (d) \(0.245 \mathrm{~J}\)

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