Problem 549 - JEE Physics

Question

A body of mass \(\mathrm{m}\) is tied to one end of spring and whirled round in a horizontal plane with a constant angular velocity. The elongation in the spring is one centimetre. If the angular velocity is doubted, the elongation in the spring is \(5 \mathrm{~cm}\). The original length of spring is... \(\\{\mathrm{A}\\} 16 \mathrm{~cm}\) \(\\{B\\} 15 \mathrm{~cm}\) \(\\{\mathrm{C}\\} 14 \mathrm{~cm}\) \(\\{\mathrm{D}\\} 13 \mathrm{~cm}\)

Step-by-Step Solution

Answer

The original length of the spring is \(\frac{1}{3} \mathrm{~cm}\).

Step 1: Write down the centripetal force equation

The centripetal force acting on the body is given by the equation: $$F_c = m \omega^2 r$$ where \(F_c\) is the centripetal force, \(m\) is the mass of the body, \(\omega\) is the angular velocity, and \(r\) is the distance from the center of rotation to the body.

Step 2: Write down the Hooke's law equation

The force exerted by the spring is given by Hooke's law: $$F_s = kx$$ where \(F_s\) is the force exerted by the spring, \(k\) is the spring constant, and \(x\) is the elongation of the spring.

Step 3: Setup the relationship between centripetal force and Hooke's law

Since the centripetal force and the Hooke's law for the spring force should be equal: $$m \omega^2 r = kx$$ The relationship between \(kx\) and \(\omega^2 r\) can also be written as: $$k(x_0 + x) = m \omega^2 r$$ In both scenarios, we are given the elongations of the spring \((x)\) and the angular velocities \((\omega)\). With this information, we can create equations for both scenarios.

Step 4: Write equations for both scenarios

In the first scenario: 1. Elongation \(x_1 = 1 \mathrm{~cm}\) 2. Angular velocity \(\omega_1\) In the second scenario: 1. Elongation \(x_2 = 5 \mathrm{~cm}\) 2. Angular velocity \(\omega_2 = 2\omega_1\) We can write the equations for both scenarios: 1. \(k(x_0 + 1) = m \omega_1^2 r\) 2. \(k(x_0 + 5) = m (2\omega_1)^2 r\)

Step 5: Eliminate \(k\) and solve for \(x_0\)

Divide equation 2 by equation 1: $$\frac{x_0 + 5}{x_0 + 1} = \frac{m (2\omega_1)^2 r}{m \omega_1^2 r}$$ Simplifying the equation: $$\frac{x_0 + 5}{x_0 + 1} = 4$$ Cross-multiply to solve for \(x_0\): $$4(x_0 + 1) = x_0 + 5$$ $$4x_0 + 4 = x_0 + 5$$ $$3x_0 = 1$$ $$x_0 = \frac{1}{3} \mathrm{~cm}$$

Step 6: Find the original length of the spring

We are given that the elongation \(x_1\) in the first scenario is 1cm. To find the original length of the spring, we subtract the elongation from the total length: Original Length of Spring = \((x_0 + x_1) - x_1\) $$= \frac{1}{3} + 1 - 1$$ $$= \frac{1}{3} \mathrm{~cm}$$ The original length of the spring is not among the given options A, B, C, and D. There must be an error or typo in the problem statement or the given choices.

Key Concepts

Hooke's Law

Hooke's Law is a fundamental principle in physics that describes the behavior of springs under tension or compression. It states that the force required to extend or compress a spring by a certain distance is directly proportional to that distance. This relationship can be mathematically expressed as:

• \( F_s = kx \)
• Where \( F_s \) is the force exerted by the spring, \( k \) is the spring constant, and \( x \) is the displacement of the spring's natural length, known as elongation in this context.

A larger spring constant \( k \) indicates a stiffer spring, which requires more force to produce the same amount of displacement.
Understanding how Hooke's Law works is important when dealing with problems involving elastic materials. It helps to predict how much an object will stretch or compress under a given load. In this exercise, Hooke's Law helps determine how much the spring elongates under different angular velocities as the forces acting on it change.
When the spring elongation changes, the values of the force change according to Hooke's Law, allowing you to solve for unknown variables like the spring's original length.

Angular Velocity

Angular velocity is a measure of how quickly an object rotates or revolves around a point or axis. It is commonly used in situations involving circular motion, where objects move along a curved path.

• It is denoted by the Greek letter \( \omega \) and is expressed in radians per second.
• For a constant angular velocity, the object moves at a steady speed in a circular path, meaning all regions of the path are covered in equal intervals of time.

Angular velocity is pivotal in understanding motion because an increase in \( \omega \) means the object covers the circular path more rapidly.
In our exercise, when the angular velocity is doubled, it significantly affects the centripetal force required to keep an object in circular motion, consequently altering the spring elongation. The centripetal force is calculated using the formula:

• \( F_c = m \omega^2 r \)
• Here, \( m \) is the mass of the object, \( \omega \) is the angular velocity, and \( r \) is the radius of the circular path.

Understanding how angular velocity influences these quantities helps in analyzing how different speeds impact forces in circular motion.

Spring Elongation

Spring elongation refers to the increase in length of a spring from its natural or original length when a force is applied. This concept is directly tied to the forces at play in circular motion and Hooke's Law.

• The original length of the spring is its length with no external forces applied, and elongation is the additional length added when it is stretched.
• The total length of the spring when force is applied is equal to its original length plus elongation: \( L_{total} = L_{original} + x \), where \( x \) is the elongation.

In our context, an increase in angular velocity resulted in increased elongation. This is because the force exerted on the spring by the rotating object increased, requiring the spring to stretch further.
The relationship between angular velocity and spring elongation is essential in determining the physical behavior of systems involving elastic potential energy and rotational dynamics.
By understanding how these factors relate, you can predict how changes in rotational speed affect spring behavior, aiding in solutions like finding the original length of springs based on varying conditions.