Circular Motion and Polar Coordinates

Centripetal acceleration, angular velocity, and motion in polar coordinates

Circular Motion and Polar Coordinates

Uniform Circular Motion

For motion in a circle of radius rr at constant speed vv:

Centripetal acceleration: ac=v2r=ω2ra_c = \frac{v^2}{r} = \omega^2 r

Angular velocity: ω=vr\omega = \frac{v}{r}

Period: T=2πrv=2πωT = \frac{2\pi r}{v} = \frac{2\pi}{\omega}

Polar Coordinates

Position in polar coordinates: (r,θ)(r, \theta)

Unit vectors: r^\hat{r} (radial), θ^\hat{\theta} (tangential)

Important: These unit vectors are not constant—they change direction as the particle moves.

Time derivatives of unit vectors: dr^dt=θ˙θ^=ωθ^\frac{d\hat{r}}{dt} = \dot{\theta}\hat{\theta} = \omega\hat{\theta}

dθ^dt=−θ˙r^=−ωr^\frac{d\hat{\theta}}{dt} = -\dot{\theta}\hat{r} = -\omega\hat{r}

Velocity in Polar Coordinates

Position vector: r⃗=rr^\vec{r} = r\hat{r}

Velocity: v⃗=dr⃗dt=drdtr^+rdr^dt\vec{v} = \frac{d\vec{r}}{dt} = \frac{dr}{dt}\hat{r} + r\frac{d\hat{r}}{dt}

v⃗=r˙r^+rθ˙θ^\vec{v} = \dot{r}\hat{r} + r\dot{\theta}\hat{\theta}

Radial component: vr=r˙v_r = \dot{r}

Tangential component: vθ=rθ˙=rωv_{\theta} = r\dot{\theta} = r\omega

Speed: v=vr2+vθ2=r˙2+r2θ˙2v = \sqrt{v_r^2 + v_{\theta}^2} = \sqrt{\dot{r}^2 + r^2\dot{\theta}^2}

Acceleration in Polar Coordinates

a⃗=dv⃗dt=ddt(r˙r^+rθ˙θ^)\vec{a} = \frac{d\vec{v}}{dt} = \frac{d}{dt}(\dot{r}\hat{r} + r\dot{\theta}\hat{\theta})

After applying product rule and substituting unit vector derivatives:

a⃗=(r¨−rθ˙2)r^+(rθ¨+2r˙θ˙)θ^\vec{a} = (\ddot{r} - r\dot{\theta}^2)\hat{r} + (r\ddot{\theta} + 2\dot{r}\dot{\theta})\hat{\theta}

Radial component: ar=r¨−rω2a_r = \ddot{r} - r\omega^2

Tangential component: aθ=rα+2r˙ωa_{\theta} = r\alpha + 2\dot{r}\omega

where α=θ¨\alpha = \ddot{\theta} is angular acceleration.

Circular Motion (r = constant)

When radius is constant (r˙=0\dot{r} = 0, r¨=0\ddot{r} = 0):

Radial acceleration (centripetal): ar=−rω2=−v2ra_r = -r\omega^2 = -\frac{v^2}{r}

Tangential acceleration: aθ=rα=rdωdta_{\theta} = r\alpha = r\frac{d\omega}{dt}

The negative sign in ara_r indicates the acceleration points toward the center (opposite to r^\hat{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=(v2r)2+(rα)2a = \sqrt{a_c^2 + a_t^2} = \sqrt{\left(\frac{v^2}{r}\right)^2 + (r\alpha)^2}

Example: Spiral Motion

A particle moves in a spiral: r(t)=btr(t) = bt, θ(t)=ωt\theta(t) = \omega t (both constant bb and ω\omega)

Position: r⃗=btr^\vec{r} = bt\hat{r}

Velocity: v⃗=br^+btωθ^\vec{v} = b\hat{r} + bt\omega\hat{\theta}

Acceleration: a⃗=−btω2r^+2bωθ^\vec{a} = -bt\omega^2\hat{r} + 2b\omega\hat{\theta}

Notice the 2r˙ω2\dot{r}\omega term (Coriolis-like term) appears even though α=0\alpha = 0.

📚 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.

💡 Show Solution

Given:

  • R = 200 m
  • v₀ = 20 m/s, v = 30 m/s
  • Δt = 10 s
  • At t = 5 s

(a) Tangential acceleration:

at=ΔvΔt=30−2010a_t = \frac{\Delta v}{\Delta t} = \frac{30 - 20}{10}

at=1.0 m/s2\boxed{a_t = 1.0 \text{ m/s}^2}

(Constant throughout)

(b) Centripetal acceleration at t = 5 s:

Speed at t = 5 s: v(5)=20+(1.0)(5)=25 m/sv(5) = 20 + (1.0)(5) = 25 \text{ m/s}

ac=v2R=(25)2200=625200a_c = \frac{v^2}{R} = \frac{(25)^2}{200} = \frac{625}{200}

ac=3.13 m/s2\boxed{a_c = 3.13 \text{ m/s}^2}

(c) Total acceleration:

atotal=at2+ac2=(1.0)2+(3.13)2a_{total} = \sqrt{a_t^2 + a_c^2} = \sqrt{(1.0)^2 + (3.13)^2}

atotal=1.0+9.8=10.8a_{total} = \sqrt{1.0 + 9.8} = \sqrt{10.8}

atotal=3.29 m/s2\boxed{a_{total} = 3.29 \text{ m/s}^2}

2Problem 2easy

❓ 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.

💡 Show Solution

Given:

  • R = 200 m
  • v₀ = 20 m/s, v = 30 m/s
  • Δt = 10 s
  • At t = 5 s

(a) Tangential acceleration:

at=ΔvΔt=30−2010a_t = \frac{\Delta v}{\Delta t} = \frac{30 - 20}{10}

at=1.0 m/s2\boxed{a_t = 1.0 \text{ m/s}^2}

(Constant throughout)

(b) Centripetal acceleration at t = 5 s:

Speed at t = 5 s: v(5)=20+(1.0)(5)=25 m/sv(5) = 20 + (1.0)(5) = 25 \text{ m/s}

ac=v2R=(25)2200=625200a_c = \frac{v^2}{R} = \frac{(25)^2}{200} = \frac{625}{200}

ac=3.13 m/s2\boxed{a_c = 3.13 \text{ m/s}^2}

(c) Total acceleration:

atotal=at2+ac2=(1.0)2+(3.13)2a_{total} = \sqrt{a_t^2 + a_c^2} = \sqrt{(1.0)^2 + (3.13)^2}

atotal=1.0+9.8=10.8a_{total} = \sqrt{1.0 + 9.8} = \sqrt{10.8}

atotal=3.29 m/s2\boxed{a_{total} = 3.29 \text{ m/s}^2}

3Problem 3easy

❓ 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.

💡 Show Solution

Given:

  • R = 200 m
  • v₀ = 20 m/s, v = 30 m/s
  • Δt = 10 s
  • At t = 5 s

(a) Tangential acceleration:

at=ΔvΔt=30−2010a_t = \frac{\Delta v}{\Delta t} = \frac{30 - 20}{10}

at=1.0 m/s2\boxed{a_t = 1.0 \text{ m/s}^2}

(Constant throughout)

(b) Centripetal acceleration at t = 5 s:

Speed at t = 5 s: v(5)=20+(1.0)(5)=25 m/sv(5) = 20 + (1.0)(5) = 25 \text{ m/s}

ac=v2R=(25)2200=625200a_c = \frac{v^2}{R} = \frac{(25)^2}{200} = \frac{625}{200}

ac=3.13 m/s2\boxed{a_c = 3.13 \text{ m/s}^2}

(c) Total acceleration:

atotal=at2+ac2=(1.0)2+(3.13)2a_{total} = \sqrt{a_t^2 + a_c^2} = \sqrt{(1.0)^2 + (3.13)^2}

atotal=1.0+9.8=10.8a_{total} = \sqrt{1.0 + 9.8} = \sqrt{10.8}

atotal=3.29 m/s2\boxed{a_{total} = 3.29 \text{ m/s}^2}

4Problem 4medium

❓ 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)=dθdt=6t rad/s\omega(t) = \frac{d\theta}{dt} = 6t \text{ rad/s}

At t = 2: ω(2)=12 rad/s\boxed{\omega(2) = 12 \text{ rad/s}}

α(t)=dωdt=6 rad/s2\alpha(t) = \frac{d\omega}{dt} = 6 \text{ rad/s}^2

α=6 rad/s2\boxed{\alpha = 6 \text{ rad/s}^2} (constant)

(b) Tangential and centripetal accelerations:

at=rα=(2.0)(6)a_t = r\alpha = (2.0)(6)

at=12 m/s2\boxed{a_t = 12 \text{ m/s}^2}

ac=ω2r=(12)2(2.0)=144×2.0a_c = \omega^2 r = (12)^2(2.0) = 144 \times 2.0

ac=288 m/s2\boxed{a_c = 288 \text{ m/s}^2}

(c) Total acceleration:

Tangential component: 12 m/s² (in direction of motion) Centripetal component: 288 m/s² (toward center)

∣a⃗∣=at2+ac2=144+82944|\vec{a}| = \sqrt{a_t^2 + a_c^2} = \sqrt{144 + 82944}

∣a⃗∣=288.2 m/s2\boxed{|\vec{a}| = 288.2 \text{ m/s}^2}

Angle from radial direction: tan⁡ϕ=atac=12288  ⟹  ϕ=2.4°\tan\phi = \frac{a_t}{a_c} = \frac{12}{288} \implies \phi = 2.4°

5Problem 5medium

❓ 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)=dθdt=6t rad/s\omega(t) = \frac{d\theta}{dt} = 6t \text{ rad/s}

At t = 2: ω(2)=12 rad/s\boxed{\omega(2) = 12 \text{ rad/s}}

α(t)=dωdt=6 rad/s2\alpha(t) = \frac{d\omega}{dt} = 6 \text{ rad/s}^2

α=6 rad/s2\boxed{\alpha = 6 \text{ rad/s}^2} (constant)

(b) Tangential and centripetal accelerations:

at=rα=(2.0)(6)a_t = r\alpha = (2.0)(6)

at=12 m/s2\boxed{a_t = 12 \text{ m/s}^2}

ac=ω2r=(12)2(2.0)=144×2.0a_c = \omega^2 r = (12)^2(2.0) = 144 \times 2.0

ac=288 m/s2\boxed{a_c = 288 \text{ m/s}^2}

(c) Total acceleration:

Tangential component: 12 m/s² (in direction of motion) Centripetal component: 288 m/s² (toward center)

∣a⃗∣=at2+ac2=144+82944|\vec{a}| = \sqrt{a_t^2 + a_c^2} = \sqrt{144 + 82944}

∣a⃗∣=288.2 m/s2\boxed{|\vec{a}| = 288.2 \text{ m/s}^2}

Angle from radial direction: tan⁡ϕ=atac=12288  ⟹  ϕ=2.4°\tan\phi = \frac{a_t}{a_c} = \frac{12}{288} \implies \phi = 2.4°

6Problem 6medium

❓ 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)=dθdt=6t rad/s\omega(t) = \frac{d\theta}{dt} = 6t \text{ rad/s}

At t = 2: ω(2)=12 rad/s\boxed{\omega(2) = 12 \text{ rad/s}}

α(t)=dωdt=6 rad/s2\alpha(t) = \frac{d\omega}{dt} = 6 \text{ rad/s}^2

α=6 rad/s2\boxed{\alpha = 6 \text{ rad/s}^2} (constant)

(b) Tangential and centripetal accelerations:

at=rα=(2.0)(6)a_t = r\alpha = (2.0)(6)

at=12 m/s2\boxed{a_t = 12 \text{ m/s}^2}

ac=ω2r=(12)2(2.0)=144×2.0a_c = \omega^2 r = (12)^2(2.0) = 144 \times 2.0

ac=288 m/s2\boxed{a_c = 288 \text{ m/s}^2}

(c) Total acceleration:

Tangential component: 12 m/s² (in direction of motion) Centripetal component: 288 m/s² (toward center)

∣a⃗∣=at2+ac2=144+82944|\vec{a}| = \sqrt{a_t^2 + a_c^2} = \sqrt{144 + 82944}

∣a⃗∣=288.2 m/s2\boxed{|\vec{a}| = 288.2 \text{ m/s}^2}

Angle from radial direction: tan⁡ϕ=atac=12288  ⟹  ϕ=2.4°\tan\phi = \frac{a_t}{a_c} = \frac{12}{288} \implies \phi = 2.4°

7Problem 7hard

❓ 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.

💡 Show Solution

Given:

  • R = 0.5 m
  • ω = 6 rad/s (constant rotation)
  • Mass m (general)
  • g = 9.8 m/s²

(a) Effective potential:

In rotating frame, forces on bead:

  • Weight: mg (downward)
  • Normal force: N (perpendicular to hoop)
  • Centrifugal force: mω²r = mω²(R sin θ) (outward)

Effective potential energy: Ueff(θ)=mgR(1−cos⁡θ)−12mω2R2sin⁡2θU_{eff}(\theta) = mgR(1 - \cos\theta) - \frac{1}{2}m\omega^2 R^2 \sin^2\theta

Ueff=mgR(1−cos⁡θ)−12mω2R2sin⁡2θ\boxed{U_{eff} = mgR(1 - \cos\theta) - \frac{1}{2}m\omega^2 R^2 \sin^2\theta}

(b) Equilibrium angle:

At equilibrium: dUeffdθ=0\frac{dU_{eff}}{d\theta} = 0

mgRsin⁡θ−mω2R2sin⁡θcos⁡θ=0mgR\sin\theta - m\omega^2 R^2 \sin\theta\cos\theta = 0

mgRsin⁡θ=mω2R2sin⁡θcos⁡θmgR\sin\theta = m\omega^2 R^2 \sin\theta\cos\theta

If θ ≠ 0: g=ω2Rcos⁡θg = \omega^2 R\cos\theta

cos⁡θ=gω2R=9.8(6)2(0.5)=9.818\cos\theta = \frac{g}{\omega^2 R} = \frac{9.8}{(6)^2(0.5)} = \frac{9.8}{18}

cos⁡θ=0.544\cos\theta = 0.544

θ=57.0°\boxed{\theta = 57.0°}

(c) Normal force at equilibrium:

Force balance perpendicular to hoop: N=mω2Rsin⁡θ+mgcos⁡θN = m\omega^2 R\sin\theta + mg\cos\theta

Actually, force balance along radius: Nsin⁡θ=mω2Rsin⁡θN\sin\theta = m\omega^2 R\sin\theta Ncos⁡θ=mgN\cos\theta = mg

From second equation: N=mgcos⁡θ=mg0.544N = \frac{mg}{\cos\theta} = \frac{mg}{0.544}

N=1.84mg\boxed{N = 1.84mg}

Or with numbers (for m = 1 kg): N=1.84(1)(9.8)=18.0 NN = 1.84(1)(9.8) = 18.0 \text{ N}

8Problem 8hard

❓ 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.

💡 Show Solution

Given:

  • R = 0.5 m
  • ω = 6 rad/s (constant rotation)
  • Mass m (general)
  • g = 9.8 m/s²

(a) Effective potential:

In rotating frame, forces on bead:

  • Weight: mg (downward)
  • Normal force: N (perpendicular to hoop)
  • Centrifugal force: mω²r = mω²(R sin θ) (outward)

Effective potential energy: Ueff(θ)=mgR(1−cos⁡θ)−12mω2R2sin⁡2θU_{eff}(\theta) = mgR(1 - \cos\theta) - \frac{1}{2}m\omega^2 R^2 \sin^2\theta

Ueff=mgR(1−cos⁡θ)−12mω2R2sin⁡2θ\boxed{U_{eff} = mgR(1 - \cos\theta) - \frac{1}{2}m\omega^2 R^2 \sin^2\theta}

(b) Equilibrium angle:

At equilibrium: dUeffdθ=0\frac{dU_{eff}}{d\theta} = 0

mgRsin⁡θ−mω2R2sin⁡θcos⁡θ=0mgR\sin\theta - m\omega^2 R^2 \sin\theta\cos\theta = 0

mgRsin⁡θ=mω2R2sin⁡θcos⁡θmgR\sin\theta = m\omega^2 R^2 \sin\theta\cos\theta

If θ ≠ 0: g=ω2Rcos⁡θg = \omega^2 R\cos\theta

cos⁡θ=gω2R=9.8(6)2(0.5)=9.818\cos\theta = \frac{g}{\omega^2 R} = \frac{9.8}{(6)^2(0.5)} = \frac{9.8}{18}

cos⁡θ=0.544\cos\theta = 0.544

θ=57.0°\boxed{\theta = 57.0°}

(c) Normal force at equilibrium:

Force balance perpendicular to hoop: N=mω2Rsin⁡θ+mgcos⁡θN = m\omega^2 R\sin\theta + mg\cos\theta

Actually, force balance along radius: Nsin⁡θ=mω2Rsin⁡θN\sin\theta = m\omega^2 R\sin\theta Ncos⁡θ=mgN\cos\theta = mg

From second equation: N=mgcos⁡θ=mg0.544N = \frac{mg}{\cos\theta} = \frac{mg}{0.544}

N=1.84mg\boxed{N = 1.84mg}

Or with numbers (for m = 1 kg): N=1.84(1)(9.8)=18.0 NN = 1.84(1)(9.8) = 18.0 \text{ N}

9Problem 9hard

❓ 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.

💡 Show Solution

Given:

  • R = 0.5 m
  • ω = 6 rad/s (constant rotation)
  • Mass m (general)
  • g = 9.8 m/s²

(a) Effective potential:

In rotating frame, forces on bead:

  • Weight: mg (downward)
  • Normal force: N (perpendicular to hoop)
  • Centrifugal force: mω²r = mω²(R sin θ) (outward)

Effective potential energy: Ueff(θ)=mgR(1−cos⁡θ)−12mω2R2sin⁡2θU_{eff}(\theta) = mgR(1 - \cos\theta) - \frac{1}{2}m\omega^2 R^2 \sin^2\theta

Ueff=mgR(1−cos⁡θ)−12mω2R2sin⁡2θ\boxed{U_{eff} = mgR(1 - \cos\theta) - \frac{1}{2}m\omega^2 R^2 \sin^2\theta}

(b) Equilibrium angle:

At equilibrium: dUeffdθ=0\frac{dU_{eff}}{d\theta} = 0

mgRsin⁡θ−mω2R2sin⁡θcos⁡θ=0mgR\sin\theta - m\omega^2 R^2 \sin\theta\cos\theta = 0

mgRsin⁡θ=mω2R2sin⁡θcos⁡θmgR\sin\theta = m\omega^2 R^2 \sin\theta\cos\theta

If θ ≠ 0: g=ω2Rcos⁡θg = \omega^2 R\cos\theta

cos⁡θ=gω2R=9.8(6)2(0.5)=9.818\cos\theta = \frac{g}{\omega^2 R} = \frac{9.8}{(6)^2(0.5)} = \frac{9.8}{18}

cos⁡θ=0.544\cos\theta = 0.544

θ=57.0°\boxed{\theta = 57.0°}

(c) Normal force at equilibrium:

Force balance perpendicular to hoop: N=mω2Rsin⁡θ+mgcos⁡θN = m\omega^2 R\sin\theta + mg\cos\theta

Actually, force balance along radius: Nsin⁡θ=mω2Rsin⁡θN\sin\theta = m\omega^2 R\sin\theta Ncos⁡θ=mgN\cos\theta = mg

From second equation: N=mgcos⁡θ=mg0.544N = \frac{mg}{\cos\theta} = \frac{mg}{0.544}

N=1.84mg\boxed{N = 1.84mg}

Or with numbers (for m = 1 kg): N=1.84(1)(9.8)=18.0 NN = 1.84(1)(9.8) = 18.0 \text{ N}

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