KErot = ½Iω²
where:
KEtotal = KEtrans + KErot = ½mv² + ½Iω²
For pure rolling: v = rω
A solid sphere (I = 2/5MR²) of mass 2 kg and radius 0.1 m rolls without slipping at 3 m/s. Calculate its total kinetic energy.
Solution:
v = 3 m/s, ω = v/r = 30 rad/s
KEtrans = ½(2)(3²) = 9 J
KErot = ½(2/5)(2)(0.1²)(30²) = 3.6 J
KEtotal = 12.6 J
W = τθ
where:
P = τω
where ω = angular velocity (rad/s)
L = Iω
L = r × p (for a point mass)
where:
Angular momentum is a vector quantity perpendicular to the plane of rotation!
If τnet = 0, then L = constant
I₁ω₁ = I₂ω₂
A figure skater spinning with arms extended (I = 3.0 kg⋅m²) at 2.0 rad/s pulls their arms in, reducing I to 0.5 kg⋅m². Find the new angular velocity.
Solution:
I₁ω₁ = I₂ω₂
(3.0)(2.0) = (0.5)ω₂
ω₂ = 12 rad/s
v = rω (no slipping)
acm = rα (center of mass acceleration)
For a solid sphere:
KEtotal = ½mv² + ½(2/5mR²)ω² = 7/10mv²
Solid Sphere: I = 2/5MR²
Hollow Sphere: I = 2/3MR²
Solid Cylinder: I = ½MR²
Hollow Cylinder: I = MR²
Rod (about end): I = 1/3ML²
I = Icm + Md²
where d = distance from rotation axis to CM
ΔPE + ΔKEtrans + ΔKErot = Wnonconservative
A solid cylinder rolls down a 3m incline. Find its final velocity if it starts from rest.
Solution:
mgh = ½mv² + ½Iω²
mgh = ½mv² + ½(½mR²)(v²/R²)
v = √(4gh/3)
Ω = mgR/Iω
where:
Key Steps:
AP Exam Tips:
