Question

A spring is attached to the center of a frictionless horizontal turn table and at the other end a body of mass \(2 \mathrm{~kg}\) is attached. The length of the spring is \(35 \mathrm{~cm}\). Now when the turn table is rotated with an angular speed of \(10 \mathrm{rad} \mathrm{s}^{-1}\), the length of the spring becomes \(40 \mathrm{~cm}\) then the force constant of the spring is \(\ldots \ldots \mathrm{N} / \mathrm{m}\). (A) \(1.2 \times 10^{3}\) (B) \(1.6 \times 10^{3}\) (C) \(2.2 \times 10^{3}\) (D) \(2.6 \times 10^{3}\)

Short answer

The force constant of the spring is \(1.6 \times 10^3 N/m\), which corresponds to option (B).

Step-by-step solution

Calculate the centripetal force of the rotating body

To calculate the centripetal force, \(F_c\), we use the formula: \[F_c = m \cdot \omega^2 \cdot r\] where - \(m = 2 kg\) = mass of the body attached to the spring end - \(\omega = 10 rad/s\) = angular speed of the turntable - \(r = 0.4 m\) = stretched length of the spring (converted from 40 cm to meters by multiplying by 0.01) Plugging these values into the formula, we get: \(F_c = 2 \times (10)^2 \times 0.4\)

Apply Hooke's Law to find spring force

Hooke's Law states that the force exerted by a spring, \(F_s\), is proportional to its elongation/extension, \(\Delta x\): \[F_s = k \cdot \Delta x\] where - \(k\) = force constant of the spring (which we want to find) - \(\Delta x = 0.4 - 0.35 = 0.05 m\) = extension of the spring when the turntable is rotating (converted lengths to meters)

Relate the centripetal force to spring force

Since there's no other external force in the horizontal plane, the centripetal force equals the spring force: \[F_c = F_s\] So we can write the equation as: \[m \cdot \omega^2 \cdot r = k \cdot \Delta x\]

Solve for the force constant of the spring

Now we can plug in all the known values and solve for \(k\): \(2 \times (10)^2 \times 0.4 = k \times 0.05\) Solving for \(k\), we get: \[k = \frac{2 \times (10)^2 \times 0.4}{0.05} = 1.6 \times 10^3 N/m\] So the force constant of the spring is \(1.6 \times 10^3 N/m\), which corresponds to option (B).

Key concepts

Angular Speed

Angular speed is a way to quantify how fast something is rotating. Imagine you're spinning in a chair. Angular speed tells you how many times you spin around your axis in a given time. It is usually measured in radians per second (rad/s). One full rotation is equivalent to 2π radians.
In the given problem, the table is rotating at an angular speed of 10 rad/s. This indicates that any point on the rotating object travels 10 radians along the circular path every second. Angular speed helps us determine forces acting on rotating bodies, such as the centripetal force that keeps the rotating mass tethered on its circular path.
Knowing the angular speed helps us understand how quickly actions, like the spring stretching, happen as the disk spins.

Hooke's Law

Hooke's Law is fundamental to understanding how springs work. It states that the force exerted by a spring is directly proportional to the extent it is stretched or compressed from its original position.
This law is expressed by the formula: \( F_s = k \cdot \Delta x \), where:

• \( F_s \) is the spring force.
• \( k \) is the spring constant, which measures the spring's stiffness.
• \( \Delta x \) represents the change in the spring's length from its original position (extension or compression).

In our problem, Hooke's Law helps in determining how much force the spring exerts, which affects how the body remains in place on the turntable as it rotates.
Applying Hooke's Law allows us to connect the experimentally observed spring extension with forces keeping the spring-mass system stable during motion.

Spring Constant

The spring constant, denoted by \( k \), is a measure of a spring’s resistance to deformation. In simpler terms, it tells us how stiff or stretchy a spring is.
In our problem, the spring constant is what we find to understand how much stretch results from a given force. A higher \( k \) means a stiffer spring that barely extends even if high force is applied.
The formula derived from Hooke’s Law, \( k = \frac{F_s}{\Delta x} \), is used to compute \( k \) when the force and extension are known. The units of \( k \) are typically N/m (Newtons per meter).
In the exercise, we found \( k \) to be \( 1.6 \times 10^3 \) N/m, indicating a relatively stiff spring since it requires significant force for a small extension related to its original length.

Spring Extension

Spring extension refers to how far the spring stretches beyond its original, uncompressed length. It is an important aspect, as it determines the force the spring exerts according to Hooke's Law.
In the given scenario, the extension \( \Delta x \) is the difference between the spring’s original length (35 cm) and its length when stretched (40 cm). After converting units from centimeters to meters, the extension \( \Delta x = 0.05 \) m.
This extension plays a crucial role in calculating the force via Hooke's Law and determines how the body, attached to the spring’s end, reacts while the system spins on the turntable. Understanding \( \Delta x \) is essential for solving problems related to rotational dynamics with springs, as it directly affects calculations for determining the spring constant \( k \) and analyzing rotational systems.