Why does the yo-yo fall slower than a freely falling object?

Gravity must do two jobs: accelerate the center of mass downward and spin the yo-yo up. Some of the gravitational potential energy is converted into rotational kinetic energy, leaving less for translational motion. The rotational inertia (I) resists angular acceleration, effectively reducing the linear acceleration below g.

How does changing the axle radius affect the motion?

A smaller axle radius (r) decreases the torque exerted by the string tension (since τ = Tr). For a given tension, this reduces the angular acceleration. However, the kinematic constraint a = rα also changes. The net effect, seen in the equation a = g/(1+I/(mr²)), is that a smaller r in the denominator (mr²) makes the fractional term I/(mr²) larger, further reducing the linear acceleration.

What does the optional friction torque represent?

It models energy dissipation in the real system, such as bearing friction at the axle or air drag on the spinning body. This torque opposes the rotation, requiring a higher string tension to achieve the same angular acceleration. It causes the yo-yo to accelerate downward more quickly than in the frictionless case, as more gravitational force is 'used up' overcoming friction rather than creating rotational kinetic energy.

Is the yo-yo's mass important, or just its shape?

Both are crucial. Mass (m) appears directly in the translational equation (F=ma). The shape, through the moment of inertia (I), determines how mass is distributed relative to the axis. For a given mass, a shape with mass concentrated far from the axis (like a hoop, I=mR²) has a larger I than one with mass near the axis (like a disk, I=½mR²), resulting in a slower descent for the hoop.

Yo-Yo Dynamics

A yo-yo descending and ascending on its string is a classic demonstration of combined translational and rotational motion. This simulator models the dynamics of a yo-yo, represented as a solid disk or cylinder of mass m and moment of inertia I, unwinding from a string of negligible mass wrapped around an axle of radius r. The core physics involves applying Newton's second law for linear motion (ΣF = ma) to the center of mass and for rotational motion (Στ = Iα) about the center of mass. The primary forces are gravity (mg downward) and string tension (T upward). The key kinematic constraint is that the linear acceleration a of the center of mass and the angular acceleration α are linked by a = rα, assuming the string unwinds without slipping. Combining these equations yields the characteristic result for the linear acceleration: a = g / (1 + I/(mr²)). This shows the acceleration is less than gravitational free-fall (g) due to the rotational inertia; the term (1 + I/(mr²)) acts as an effective mass increase. The simulator allows exploration of parameters like mass distribution (changing I), axle radius, and the inclusion of an optional constant friction torque τ on the axle, which modifies the net torque equation. Students can learn how rotational inertia governs the descent speed, how tension relates to the system's parameters (T = mg - ma), and how energy is partitioned between translational and rotational kinetic energy. The model simplifies real yo-yo motion by treating the string as massless and inextensible, ignoring air resistance, and assuming a pure vertical descent without wobble, focusing on the fundamental coupled equations of motion.

Who it's for: Undergraduate physics students studying rigid body dynamics in a calculus-based mechanics course, and advanced high school students exploring beyond simple translational motion.

Key terms

Rotational Inertia
Newton's Second Law for Rotation
Constraint Equation (a = rα)
Translational-Rotational Coupling
String Tension
Angular Acceleration
Torque
Kinetic Energy (Translational and Rotational)

Live graphs

How it works

A yo-yo falls while string unwinds from the small axle of radius r. If the string does not slip on the axle, the center-of-mass speed and angular speed stay locked by v = ωr. Newton’s laws give mg − T = ma and Tr − τ_friction = Iα with α = a/r, so a = g/(1 + I/(mr²)) when τ = 0 — much less than free fall because translational kinetic energy is shared with spin. Larger I or smaller r (larger I/(mr²)) makes the descent slower. A constant opposing friction torque τ at the axle reduces α and a until, if τ ≥ mgr, the yo-yo can hang without accelerating.

Key equations

mg − T = ma · Tr − τ = Iα · a = αr

τ = 0: a = g/(1 + I/(mr²)) · α = (mgr − τ)/(I + mr²)

Frequently asked questions

Why does the yo-yo fall slower than a freely falling object?: Gravity must do two jobs: accelerate the center of mass downward and spin the yo-yo up. Some of the gravitational potential energy is converted into rotational kinetic energy, leaving less for translational motion. The rotational inertia (I) resists angular acceleration, effectively reducing the linear acceleration below g.
How does changing the axle radius affect the motion?: A smaller axle radius (r) decreases the torque exerted by the string tension (since τ = Tr). For a given tension, this reduces the angular acceleration. However, the kinematic constraint a = rα also changes. The net effect, seen in the equation a = g/(1+I/(mr²)), is that a smaller r in the denominator (mr²) makes the fractional term I/(mr²) larger, further reducing the linear acceleration.
What does the optional friction torque represent?: It models energy dissipation in the real system, such as bearing friction at the axle or air drag on the spinning body. This torque opposes the rotation, requiring a higher string tension to achieve the same angular acceleration. It causes the yo-yo to accelerate downward more quickly than in the frictionless case, as more gravitational force is 'used up' overcoming friction rather than creating rotational kinetic energy.
Is the yo-yo's mass important, or just its shape?: Both are crucial. Mass (m) appears directly in the translational equation (F=ma). The shape, through the moment of inertia (I), determines how mass is distributed relative to the axis. For a given mass, a shape with mass concentrated far from the axis (like a hoop, I=mR²) has a larger I than one with mass near the axis (like a disk, I=½mR²), resulting in a slower descent for the hoop.