Relate mass distribution to angular acceleration

Move the mass before you start

Two identical masses can respond differently to the same turning effort depending on their distance from the axis. Predict what happens when both masses move twice as far out. The total mass is unchanged, but each mass must follow a larger circle for the same angular motion.

The investigation prepares two 1 kg point masses on opposite ends of an ideal massless connector. It rotates horizontally about a fixed vertical axis. Each trial starts from rest, keeps its chosen radius fixed and applies a constant net torque for two seconds. Changing radius prepares a new trial; it does not move masses during a spin. The diagram shows the initial geometry, not the final orientation.

Describe a turn in radians

An angle of one radian sweeps an arc whose length equals its radius. A complete turn is 2π radians. If a point at radius r moves through angle θ, its signed arc displacement is rθ. Here r is measured in metres and θ in radians; substituting degrees directly gives an incorrect distance.

Angular velocity ω measures how quickly angle changes, in radians per second. Angular acceleration α measures how quickly angular velocity changes, in radians per second squared. Counterclockwise viewed from above is positive. Clockwise angular velocity is negative, but speed is never negative.

Both masses share the same angular velocity because the connector is rigid. Their speed is v = r|ω|. A point twice as far from the axis would move twice as fast at the same angular velocity. That comparison holds angular velocity fixed; the control experiment instead holds torque and time fixed.

Calculate the resistance to angular acceleration

Rotational inertia I adds each mass multiplied by its squared perpendicular distance from the axis: I = Σmr². The symbol Σ means add the contributions. I is measured in kg·m². For the two masses at radius 1 m, I = 1 × 1² + 1 × 1² = 2 kg·m². At radius 2 m, I = 1 × 2² + 1 × 2² = 8 kg·m².

The fixed-axis equation is τ = Iα, where τ is net external torque in N·m. Read it as: the required turning effort equals rotational inertia times angular acceleration. For +2 N·m and I = 2 kg·m², α = 2/2 = 1 rad/s². With I = 8 kg·m², the same torque produces α = 2/8 = 0.25 rad/s².

The radius doubled, inertia quadrupled and angular acceleration fell to one quarter. Total mass alone cannot predict rotational response. The specified axis matters too. These definitions are developed in OpenStax rotational inertia.

Predict the graph and compare trials

Starting from rest with constant α, angular velocity is ω = αt and angle is θ = ½αt², with time t in seconds. At radius 1 m and torque +2 N·m, two seconds gives ω = 2 rad/s and θ = 2 rad. Each mass moves at 2 m/s. At radius 2 m, the same torque and duration give ω = 0.5 rad/s, θ = 0.5 rad and speed 1 m/s.

Capture both settings. Compare the straight-line slopes on the angular-velocity graph. The graph keeps the same axes between trials so a flatter line represents smaller angular acceleration. Use the time ledger to check intermediate values; doubling time doubles ω but quadruples θ in these rest-start trials.

Now reverse torque to −2 N·m. Predict which columns change sign before reading them. Angular acceleration, angular velocity and angle reverse; speed remains nonnegative. Zero torque leaves this initially resting system at rest. It would not stop a rotor that already had a nonzero initial angular velocity; that is a different initial condition.

Relate angular and linear acceleration

Tangential acceleration changes speed along a circle: a_t = rα, with its sign measured along the positive local rotation direction. At radius 1 m in the default trial, it is 1 m/s². Radial acceleration changes the direction of velocity and has inward magnitude a_r = rω². At two seconds it is 4 m/s². Angular acceleration is neither of these linear accelerations; its units differ.

The actuator is modeled as a pure couple: equal tangential forces in opposite spatial directions, producing the same turning sense. Each force has magnitude |τ|/(2r). At +2 N·m, each force is 1 N for radius 1 m and 0.5 N for radius 2 m. Holding torque fixed therefore changes the required forces. The connector provides the radial constraint. Its strength and deformation are not calculated.

These ideal predictions use exact settings, not measured observations. A physical investigation would need a measured inertia or mass distribution, time resolution and an estimate of friction. A mismatch might reflect an omitted connector mass or bearing torque rather than failure of the torque law.

Transfer to a different mass distribution

Two 2 kg point masses lie 0.5 m from a fixed axis. Starting from rest, a constant +3 N·m net torque acts for two seconds. Predict inertia, angular acceleration, angular velocity and each mass speed. Explain why using the total 4 kg as if it were I would be dimensionally wrong.

Check the inertia transfer

I = 2 × 2 × 0.5² = 1 kg·m². Angular acceleration is 3 rad/s², angular velocity after two seconds is 6 rad/s and each mass speed is 0.5 × 6 = 3 m/s. Mass in kg is not rotational inertia in kg·m²; the distance from the specified axis is essential.