A solid disk (M=2 kg, R=0.5 m) has α=2 rad/s². What torque is applied?

I = ½MR² = ½(2)(0.25) = 0.25 kg·m². τ = Iα = 0.25 × 2 = 0.5 N·m.

Starting from rest with α=2 rad/s², what is ω after 5 seconds?

Ω = ω₀ + αt = 0 + 2(5) = 10 rad/s.

A solid sphere and a ring of equal mass and radius roll down an incline. Which reaches the bottom first?

The sphere (I=⅖MR²) has a = gsinθ/(1+2/5) = 5gsinθ/7. The ring (I=MR²) has a = gsinθ/2. The sphere is faster.

Pro 🔒~25 min

Rotational Kinematics (Advanced)

Angular velocity, acceleration, and rolling motion with calculus

How it works

Rotational kinematics parallels translational kinematics: angular displacement θ replaces x, angular velocity ω replaces v, and angular acceleration α replaces a. For constant α: ω(t) = ω₀ + αt and θ(t) = θ₀ + ω₀t + ½αt². The moment of inertia I depends on mass distribution: I = ½MR² (solid disk/cylinder), I = MR² (ring/thin cylinder), I = ⅖MR² (solid sphere). Newton's second law for rotation is τ = Iα. For rolling without slipping, the contact point has zero velocity: v_cm = Rω. The total kinetic energy is K = ½Mv²_cm + ½Iω² = ½(M + I/R²)v²_cm. When rolling down an incline of angle θ, a = gsinθ/(1 + I/(MR²)), meaning objects with larger I/MR² roll more slowly.

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Step-by-step

Select an object shape and set its mass and radius.
Apply an angular acceleration and watch it spin.
The canvas shows the rotating object with velocity vectors, while graphs display θ(t), ω(t), and the tangential/centripetal acceleration components.
Press Play to start the animation.

Key formulas

$\theta(t) = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2$ Angular displacement as a function of time with constant angular acceleration
$\tau = I\alpha$ Newton's second law for rotation: net torque equals moment of inertia times angular acceleration
$v_{\text{cm}} = R\omega, \quad a_{\text{cm}} = R\alpha$ Rolling without slipping: the center-of-mass velocity and acceleration relate to angular quantities by the radius

Frequently asked questions

A solid disk (M=2 kg, R=0.5 m) has α=2 rad/s². What torque is applied?: I = ½MR² = ½(2)(0.25) = 0.25 kg·m². τ = Iα = 0.25 × 2 = 0.5 N·m.
Starting from rest with α=2 rad/s², what is ω after 5 seconds?: Ω = ω₀ + αt = 0 + 2(5) = 10 rad/s.
A solid sphere and a ring of equal mass and radius roll down an incline. Which reaches the bottom first?: The sphere (I=⅖MR²) has a = gsinθ/(1+2/5) = 5gsinθ/7. The ring (I=MR²) has a = gsinθ/2. The sphere is faster.