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A thin elastic cord of mass m, natural length l, & stiffness k , is used to form a circular loop kept on a smooth horizontal. if the loop is made to rotate at const. angular vel. w, about a vertical axis passing through its centre, What is the radius of the loop ?

tushar pandey , 13 Years ago
Grade 12
anser 1 Answers
Askiitians Tutor Team

Last Activity: 7 Days ago

To determine the radius of a circular loop formed by a thin elastic cord rotating at a constant angular velocity, we need to consider the forces acting on the cord due to its elasticity and the centrifugal force resulting from the rotation. Let's break this down step by step.

Understanding the Forces at Play

When the loop rotates, each segment of the cord experiences a centrifugal force that pushes it outward. This force is balanced by the tension in the cord, which is a result of its elasticity. The relationship between the tension in the cord and its extension can be described using Hooke's Law.

Key Variables

  • m: Mass of the cord
  • l: Natural length of the cord
  • k: Stiffness (spring constant) of the cord
  • w: Angular velocity of the loop
  • R: Radius of the loop

Applying Hooke's Law

According to Hooke's Law, the tension (T) in the cord when it is stretched can be expressed as:

T = k * Δx

Here, Δx is the extension of the cord from its natural length. When the loop is rotating, the extension can be related to the radius of the loop:

Δx = R - l

Balancing Forces

The centrifugal force acting on a small segment of the cord can be expressed as:

F_c = m * R * w²

For a uniform loop, the total centrifugal force must equal the total tension in the cord. If we consider the entire length of the cord, the tension can be expressed as:

T = k * (R - l)

Since the loop is circular, we can relate the total tension to the centrifugal force acting on the entire mass of the cord:

T = m * R * w²

Setting Up the Equation

Now, we can set the total tension equal to the centrifugal force:

k * (R - l) = m * R * w²

Solving for the Radius

Rearranging this equation gives us:

kR - kl = mRw²

Now, we can isolate R:

R(k - mw²) = kl

Thus, the radius R can be expressed as:

R = kl / (k - mw²)

Final Thoughts

This formula gives us the radius of the loop in terms of the mass of the cord, its natural length, stiffness, and the angular velocity. It's essential to ensure that the denominator (k - mw²) is positive for the radius to be physically meaningful. If k is less than mw², the loop cannot maintain its shape under the given conditions.

In summary, by balancing the forces acting on the rotating loop and applying Hooke's Law, we can derive a clear relationship that defines the radius of the loop based on the properties of the elastic cord and the conditions of its rotation.

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