For motion in a circle of radius r at constant speed v:
Centripetal acceleration:ac
📚 Practice Problems
1Problem 1easy
❓ Question:
A car travels around a circular track of radius R = 200 m. The speed increases uniformly from v₀ = 20 m/s to v = 30 m/s over 10 seconds. Find at t = 5 s: (a) the tangential acceleration, (b) the centripetal acceleration, and (c) the magnitude of total acceleration.
Centripetal acceleration, angular velocity, and motion in polar coordinates
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Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 3 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
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Circular Motion and Polar Coordinates is part of the AP Physics C: Mechanics course on Study Mondo, specifically in the Kinematics section. You can explore the full course for more related topics and practice resources.
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=
rv2=
ω2r
Angular velocity:ω=rv
Period:T=v2πr=ω2π
Polar Coordinates
Position in polar coordinates: (r,θ)
Unit vectors: r^ (radial), θ^ (tangential)
Important: These unit vectors are not constant—they change direction as the particle moves.
Time derivatives of unit vectors:dtdr^=θ˙θ^=ωθ^
dtdθ^=−θ˙r^=−ωr^
Velocity in Polar Coordinates
Position vector: r=rr^
Velocity:
v=dtdr=dtdrr^+rdtdr^
v=r˙r^+rθ˙θ^
Radial component:vr=r˙
Tangential component:vθ=rθ˙=rω
Speed:v=vr2+vθ2=r˙2+r2θ˙2
Acceleration in Polar Coordinates
a=dtdv=dtd(r˙r^+rθ˙θ^)
After applying product rule and substituting unit vector derivatives:
a=(r¨−rθ˙2)r^+(rθ¨+2r˙θ˙)θ^
Radial component:ar=r¨−rω2
Tangential component:aθ=rα+2r˙ω
where α=θ¨ is angular acceleration.
Circular Motion (r = constant)
When radius is constant (r˙=0, r¨=0):
Radial acceleration (centripetal):ar=−rω2=−rv2
Tangential acceleration:aθ=rα=rdtdω
The negative sign in ar indicates the acceleration points toward the center (opposite to r^).
Non-Uniform Circular Motion
When speed changes: both centripetal and tangential acceleration exist.
Centripetal: changes direction of velocity
Tangential: changes magnitude of velocity
Total acceleration magnitude:
a=ac2+at2=(rv2)2+(rα)
Example: Spiral Motion
A particle moves in a spiral: r(t)=bt, θ(t)=ωt (both constant b and ω)
Position:r=btr^
Velocity:v=br^+btωθ^
Acceleration:a=−btω2r^+2bωθ^
Notice the 2r˙ω term (Coriolis-like term) appears even though α=0.
at=ΔtΔv=1030−20
at=1.0 m/s2
(Constant throughout)
(b) Centripetal acceleration at t = 5 s:
Speed at t = 5 s:
v(5)=20+(1.0)(5)=25 m/s
ac=Rv2=200(25)2=200625
ac=3.13 m/s2
(c) Total acceleration:
atotal=at2+ac2=(1.0)2+(3.13)2
atotal=1.0+9.8=10.8
atotal=3.29 m/s2
2Problem 2medium
❓ Question:
A particle moves in a circle of radius r = 2.0 m with angular position θ(t) = 3t² rad (where t is in seconds). Find at t = 2 s: (a) the angular velocity and angular acceleration, (b) the tangential and centripetal accelerations, and (c) the total acceleration vector.
💡 Show Solution
Given:
r = 2.0 m
θ(t) = 3t² rad
At t = 2 s
(a) Angular velocity and acceleration:
ω(t)=dtdθ=6t rad/s
At t = 2:
ω(2)=12 rad/s
α(t)=dtdω=6 rad/s
α=6 rad/s2 (constant)
(b) Tangential and centripetal accelerations:
at=rα=(2.0)(6)
at=12 m/s2
ac=ω2r=(12)2
ac=288 m/s2
(c) Total acceleration:
Tangential component: 12 m/s² (in direction of motion)
Centripetal component: 288 m/s² (toward center)
∣a∣=a
∣a∣=288.2 m/s
Angle from radial direction:
tanϕ=aca
3Problem 3hard
❓ Question:
A bead slides without friction on a horizontal circular hoop of radius R = 0.5 m that rotates about a vertical axis at constant angular velocity ω = 6 rad/s. The bead is at angle θ from the vertical. Using calculus and force analysis, find: (a) the effective potential energy, (b) the equilibrium angle, and (c) the normal force at equilibrium.
Force balance perpendicular to hoop:
N=mω2Rsinθ+mgcosθ
Actually, force balance along radius:
Nsinθ=mω2RsinθNcosθ=
From second equation:
N=cosθmg=
N=1.84mg
Or with numbers (for m = 1 kg):
N=1.84(1)(9.8)=18.0 N
▾
Yes, this page includes 3 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.