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What is the extension of a spring mass rotating in a horizontal circle?
When a mass m tied to a spring (with spring constant k and natural length L₀) rotates in a horizontal circle at a steady angular velocity ω, the spring stretches by a distance x. Because the block moves in a circle, it needs a centripetal force directed toward the centre. On a frictionless table, the only force acting inward is the spring force. By equating Hooke's Law to the centripetal force formula, you can solve for the exact extension.
The spring's total length becomes L₀ + x, which is also the radius of the circular path. Setting the spring force equal to the centripetal force derived from OpenStax gives the governing equation:
k x = m ω² (L₀ + x)
You can rearrange this to find x:
k x - m ω² x = m ω² L₀
x (k - m ω²) = m ω² L₀
x = m ω² L₀ / (k - m ω²)
This formula shows that as the rotation speeds up, the spring stretches more. As with any application of Hooke's Law, this relationship holds as long as the spring stays within its elastic limit.
Why does the spring break at the critical angular velocity?
Look at the denominator of the extension formula: k - m ω². As the angular velocity ω increases, m ω² gets closer to k. When m ω² equals k, the denominator becomes zero, meaning the theoretical extension x shoots to infinity.
In reality, the spring cannot stretch forever. It will either permanently deform or break. The speed at which this happens, ωc = √(k/m), is the critical angular velocity, derived from the same centripetal force principles. At speeds equal to or greater than this, the spring simply cannot provide enough inward pull to keep the mass moving in a circle. The required centripetal force grows faster with distance than the spring's restoring force does.
If you struggle to picture how forces change as an object moves, Fermi's AI tutor runs the simulation of a rotating mass, showing you the exact moment the centripetal requirement overtakes the spring's limits.
What happens if the table is not frictionless?
In standard exam problems, you often assume a frictionless table. But what if friction is present? When a block rotates on a rough horizontal surface, static friction can also point toward or away from the centre, depending on the spring's extension. If the block is moving slowly, the spring might be highly stretched, pulling inward, while static friction acts outward to prevent the block from sliding closer to the centre.
As the angular velocity increases, the required centripetal force increases. Once the spring alone cannot provide enough inward force, static friction will point inward to help. The maximum speed before the block slips further out occurs when static friction reaches its maximum inward value, f = μ N, according to the friction models provided by OpenStax. Setting up the equation then becomes k x + μ m g = m ω² (L₀ + x). This variation is a frequent trap in advanced exams, testing whether you can correctly identify the direction of the friction vector.
How do you solve circular motion spring problems?
When tackling these problems for JEE or NEET, follow this step-by-step framework:
Identify the forces: Draw a free-body diagram of the block. On a horizontal surface, gravity acts downwards and the normal force pushes upwards. Since there is no vertical motion, these two forces perfectly cancel each other out. This leaves only the horizontal forces to consider. On a frictionless table, the sole horizontal force is the spring force pointing directly to the centre of rotation.
Find the correct radius: The radius of the circular path is a common source of errors. It is never just the natural length of the spring; it is always the natural length plus the extension, so r = L₀ + x. Forgetting to add the extension is the most frequent mistake students make under exam pressure.
Set up the dynamic equation: Write Spring Force = Centripetal Force, which translates to k x = m ω² r. Always double-check that your units are consistent before substituting numbers into this equation.
Solve for the target unknown: Substitute r = L₀ + x into your equation. You can then isolate the specific variable you need, whether the question asks for the extension x, the spring constant k, the mass m, or the total rotating length.
If you make a mistake while setting up the radius, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than just printing an answer. For instance, if you forget to add x to L₀, it works out what you know and walks you back to the correct radius.
Practising this method also helps with related mechanics setups, such as finding the tension in a simple pendulum string or finding the accelerations at the instant a string is cut in a spring-mass system.
