Quantity Symbol SI Unit Relation to Linear Angular displacement θ \theta θ rad s = r θ s = r\theta s = r θ Angular velocity ω \omega ω rad/s v = r ω v = r\omega v = r ω Angular acceleration α \alpha α rad/s2 ^2 2 a t = r α a_t = r\alpha a t = r α
ω = Δ θ Δ t \omega = \frac{\Delta\theta}{\Delta t} ω = Δ t Δ θ For uniform circular motion, ω \omega ω
T = 2 π ω , f = 1 T = ω 2 π T = \frac{2\pi}{\omega}, \quad f = \frac{1}{T} = \frac{\omega}{2\pi} T = ω 2 π , f = T 1 = 2 π ω Where T T T f f f
V = r ω = 2 π r T = 2 π r f V = r\omega = \frac{2\pi r}{T} = 2\pi rf V = r ω = T 2 π r = 2 π r f
Example A CD rotates at 200 r p m 200\mathrm{ rpm} 200 rpm 5 c m 5\mathrm{ cm} 5 cm
ω = 200 × 2 π 60 = 400 π 60 = 20.9 r a d / s \omega = 200 \times \frac{2\pi}{60} = \frac{400\pi}{60} = 20.9\mathrm{ rad/s} ω = 200 × 60 2 π = 60 400 π = 20.9 rad/s V = r ω = 0.05 × 20.9 = 1.05 m / s V = r\omega = 0.05 \times 20.9 = 1.05\mathrm{ m/s} V = r ω = 0.05 × 20.9 = 1.05 m/s An object in uniform circular motion has a constantly changing velocity (direction changes), so it Is always accelerating toward the centre of the circle.
A c = v 2 r = ω 2 r = 4 π 2 r T 2 A_c = \frac{v^2}{r} = \omega^2 r = \frac{4\pi^2 r}{T^2} A c = r v 2 = ω 2 r = T 2 4 π 2 r Always directed toward the centre of the circular path (radially inward).
Centripetal acceleration changes the direction of velocity, not its magnitude. If the centripetal force is removed, the object moves in a straight line (tangent to the circle) by Newton”s first law. The word “centripetal” means “centre-seeking.”
Exam Tip Centripetal force is NOT a new force — it is the NET force toward the centre provided by existing Forces (gravity, tension, friction, normal force, etc.). Never include “centripetal force” as a Separate force on a free-body diagram.
F c = m a c = m v 2 r = m ω 2 r F_c = ma_c = \frac{mv^2}{r} = m\omega^2 r F c = m a c = r m v 2 = m ω 2 r Always directed toward the centre of the circle.
Situation Centripetal Force Provided By Car turning on a flat road Friction between tyres and road Car on a banked curve Horizontal component of normal force Satellite in orbit Gravitational force Object on a string (horizontal circle) Tension in the string Conical pendulum Horizontal component of tension Motorcyclist in vertical circle Combination of weight and normal reaction
For an object of mass m m m r r r v v v
T = m v 2 r T = \frac{mv^2}{r} T = r m v 2 Where T T T
A mass m m m L L L r r r θ \theta θ
Vertical : T cos θ = m g T\cos\theta = mg T cos θ = m g
Horizontal : T sin θ = m v 2 r T\sin\theta = \dfrac{mv^2}{r} T sin θ = r m v 2
Dividing: tan θ = v 2 r g \tan\theta = \dfrac{v^2}{rg} tan θ = r g v 2
The radius: r = L sin θ r = L\sin\theta r = L sin θ
The period: T p = 2 π L cos θ g T_p = 2\pi\sqrt{\dfrac{L\cos\theta}{g}} T p = 2 π g L cos θ
Example A 0.5 k g 0.5\mathrm{ kg} 0.5 kg 1 m 1\mathrm{ m} 1 m 3 m / s 3\mathrm{ m/s} 3 m/s
\tan\theta = \frac{v^2}`\{rg}` = \frac{9}{r \times 9.81} Also r = L sin θ = sin θ r = L\sin\theta = \sin\theta r = L sin θ = sin θ
tan θ = 9 9.81 sin θ \tan\theta = \frac{9}{9.81\sin\theta} tan θ = 9.81 sin θ 9 sin θ cos θ = 0.917 sin θ \frac{\sin\theta}{\cos\theta} = \frac{0.917}{\sin\theta} cos θ sin θ = sin θ 0.917 sin 2 θ = 0.917 cos θ \sin^2\theta = 0.917\cos\theta sin 2 θ = 0.917 cos θ 1 − cos 2 θ = 0.917 cos θ 1 - \cos^2\theta = 0.917\cos\theta 1 − cos 2 θ = 0.917 cos θ Let u = cos θ u = \cos\theta u = cos θ u 2 + 0.917 u − 1 = 0 u^2 + 0.917u - 1 = 0 u 2 + 0.917 u − 1 = 0
U = − 0.917 + 0.841 + 4 2 = − 0.917 + 2.200 2 = 0.642 U = \frac{-0.917 + \sqrt{0.841 + 4}}{2} = \frac{-0.917 + 2.200}{2} = 0.642 U = 2 − 0.917 + 0.841 + 4 = 2 − 0.917 + 2.200 = 0.642 θ = arccos ( 0.642 ) = 50.1 ° \theta = \arccos(0.642) = 50.1\degree θ = arccos ( 0.642 ) = 50.1°
T = \frac`\{mg}`{\cos\theta} = \frac{0.5 \times 9.81}{0.642} = 7.64\mathrm{ N} For an object moving in a vertical circle, the speed varies (it is fastest at the bottom, slowest at The top) because gravity does work.
T − m g = m v b o t t o m 2 r ⟹ T = m g + m v b o t t o m 2 r T - mg = \frac{mv_{\mathrm{bottom}}^2}{r} \implies T = mg + \frac{mv_{\mathrm{bottom}}^2}{r} T − m g = r m v bottom 2 ⟹ T = m g + r m v bottom 2 T + m g = m v t o p 2 r ⟹ T = m v t o p 2 r − m g T + mg = \frac{mv_{\mathrm{top}}^2}{r} \implies T = \frac{mv_{\mathrm{top}}^2}{r} - mg T + m g = r m v top 2 ⟹ T = r m v top 2 − m g For the object to complete the full circle, T ≥ 0 T \ge 0 T ≥ 0
\frac{mv_{\mathrm{top}}^2}{r} \ge mg \implies v_{\mathrm{top}} \ge \sqrt`\{gr}` 1 2 m v b o t t o m 2 = 1 2 m v t o p 2 + m g ( 2 r ) \frac{1}{2}mv_{\mathrm{bottom}}^2 = \frac{1}{2}mv_{\mathrm{top}}^2 + mg(2r) 2 1 m v bottom 2 = 2 1 m v top 2 + m g ( 2 r ) V b o t t o m 2 = v t o p 2 + 4 g r V_{\mathrm{bottom}}^2 = v_{\mathrm{top}}^2 + 4gr V bottom 2 = v top 2 + 4 g r For minimum speed at the top (v t o p = g r v_{\mathrm{top}} = \sqrt{gr} v top = g r
V b o t t o m = 5 g r V_{\mathrm{bottom}} = \sqrt{5gr} V bottom = 5 g r
Example A 0.3 k g 0.3\mathrm{ kg} 0.3 kg 0.8 m 0.8\mathrm{ m} 0.8 m
V b o t t o m = 5 g r = 5 ( 9.81 ) ( 0.8 ) = 39.24 = 6.26 m / s V_{\mathrm{bottom}} = \sqrt{5gr} = \sqrt{5(9.81)(0.8)} = \sqrt{39.24} = 6.26\mathrm{ m/s} V bottom = 5 g r = 5 ( 9.81 ) ( 0.8 ) = 39.24 = 6.26 m/s For a car on a banked curve at angle θ \theta θ r r r v v v
Vertical : N cos θ = m g N\cos\theta = mg N cos θ = m g
Horizontal : N sin θ = m v 2 r N\sin\theta = \dfrac{mv^2}{r} N sin θ = r m v 2
Dividing: tan θ = v 2 r g \tan\theta = \dfrac{v^2}{rg} tan θ = r g v 2
The ideal (no friction needed) speed:
V = r g tan θ V = \sqrt{rg\tan\theta} V = r g tan θ When friction is present, the car can travel at speeds above or below the ideal speed. Friction acts Up the slope (to prevent sliding down) or down the slope (to prevent sliding up).
Torque (moment of force) is the rotational equivalent of force:
τ = F d sin θ = F r ⊥ \tau = Fd\sin\theta = Fr_\perp τ = F d sin θ = F r ⊥ Where:
F F F d d d θ \theta θ r ⊥ = d sin θ r_\perp = d\sin\theta r ⊥ = d sin θ The unit of torque is N ⋅ m \mathrm{N}\cdot\mathrm{m} N ⋅ m
Newton’s second law for rotation:
τ n e t = I α \tau_{\mathrm{net}} = I\alpha τ net = I α Where I I I α \alpha α
For an object in static equilibrium :
Translational : ∑ F ⃗ = 0 \sum \vec{F} = 0 ∑ F = 0 Rotational : ∑ τ = 0 \sum \tau = 0 ∑ τ = 0 The second condition must hold about ANY axis.
Example A uniform beam of mass 10 k g 10\mathrm{ kg} 10 kg 4 m 4\mathrm{ m} 4 m 20 k g 20\mathrm{ kg} 20 kg 1 m 1\mathrm{ m} 1 m
Taking moments about the left end (clockwise positive):
R r i g h t × 4 − 10 g × 2 − 20 g × 1 = 0 R_{\mathrm{right}} \times 4 - 10g \times 2 - 20g \times 1 = 0 R right × 4 − 10 g × 2 − 20 g × 1 = 0 4 R r i g h t = 196.2 + 196.2 = 392.4 4R_{\mathrm{right}} = 196.2 + 196.2 = 392.4 4 R right = 196.2 + 196.2 = 392.4 R r i g h t = 98.1 N R_{\mathrm{right}} = 98.1\mathrm{ N} R right = 98.1 N By vertical equilibrium:
R l e f t + R r i g h t = 10 g + 20 g = 294.3 N R_{\mathrm{left}} + R_{\mathrm{right}} = 10g + 20g = 294.3\mathrm{ N} R left + R right = 10 g + 20 g = 294.3 N R l e f t = 294.3 − 98.1 = 196.2 N R_{\mathrm{left}} = 294.3 - 98.1 = 196.2\mathrm{ N} R left = 294.3 − 98.1 = 196.2 N The moment of inertia I I I
I = ∑ m i r i 2 I = \sum m_i r_i^2 I = ∑ m i r i 2 For a continuous body:
I = ∫ r 2 d m I = \int r^2\,dm I = ∫ r 2 d m Object Axis I I I Solid cylinder/disk Central axis 1 2 M R 2 \dfrac{1}{2}MR^2 2 1 M R 2 Hollow cylinder Central axis M R 2 MR^2 M R 2 Solid sphere Diameter 2 5 M R 2 \dfrac{2}{5}MR^2 5 2 M R 2 Hollow sphere Diameter 2 3 M R 2 \dfrac{2}{3}MR^2 3 2 M R 2 Thin rod (centre) Perpendicular through centre 1 12 M L 2 \dfrac{1}{12}ML^2 12 1 M L 2 Thin rod (end) Perpendicular through end 1 3 M L 2 \dfrac{1}{3}ML^2 3 1 M L 2 Point mass At distance r r r M r 2 Mr^2 M r 2
For a body of mass M M M I c m I_{\mathrm{cm}} I cm
I = I c m + M d 2 I = I_{\mathrm{cm}} + Md^2 I = I cm + M d 2 Where d d d
L ⃗ = I ω ⃗ \vec{L} = I\vec{\omega} L = I ω For a point mass: L = m v r = m r 2 ω L = mvr = mr^2\omega L = m v r = m r 2 ω
k g ⋅ m 2 / s \mathrm{kg}\cdot\mathrm{m}^2/\mathrm{s} kg ⋅ m 2 / s
In a closed system with no external torques:
I 1 ω 1 = I 2 ω 2 I_1\omega_1 = I_2\omega_2 I 1 ω 1 = I 2 ω 2 Ice skater spinning : Pulling arms in reduces I I I ω \omega ω Diving : Tucking reduces I I I Figure skater : Extending arms increases I I I ω \omega ω
Example A figure skater with arms extended has I = 4.5 k g ⋅ m 2 I = 4.5\mathrm{ kg}\cdot\mathrm{m}^2 I = 4.5 kg ⋅ m 2 2 r a d / s 2\mathrm{ rad/s} 2 rad/s I I I 1.5 k g ⋅ m 2 1.5\mathrm{ kg}\cdot\mathrm{m}^2 1.5 kg ⋅ m 2
I 1 ω 1 = I 2 ω 2 I_1\omega_1 = I_2\omega_2 I 1 ω 1 = I 2 ω 2 4.5 × 2 = 1.5 × ω 2 4.5 \times 2 = 1.5 \times \omega_2 4.5 × 2 = 1.5 × ω 2 ω 2 = 6 r a d / s \omega_2 = 6\mathrm{ rad/s} ω 2 = 6 rad/s Her angular velocity triples.
Δ L = τ ⋅ Δ t \Delta L = \tau \cdot \Delta t Δ L = τ ⋅ Δ t This is analogous to linear impulse: Δ p = F ⋅ Δ t \Delta p = F \cdot \Delta t Δ p = F ⋅ Δ t
E k , r o t = 1 2 I ω 2 E_{k,\mathrm{rot}} = \frac{1}{2}I\omega^2 E k , rot = 2 1 I ω 2 For an object that rolls without slipping:
E k = 1 2 m v 2 + 1 2 I ω 2 E_k = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 E k = 2 1 m v 2 + 2 1 I ω 2 Since v = r ω v = r\omega v = r ω
E k = 1 2 m v 2 + 1 2 I v 2 r 2 = 1 2 ( m + I r 2 ) v 2 E_k = \frac{1}{2}mv^2 + \frac{1}{2}I\frac{v^2}{r^2} = \frac{1}{2}\left(m + \frac{I}{r^2}\right)v^2 E k = 2 1 m v 2 + 2 1 I r 2 v 2 = 2 1 ( m + r 2 I ) v 2 For an object rolling down a frictionless-free incline (rolling without slipping):
M g h = 1 2 m v 2 + 1 2 I ω 2 Mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 M g h = 2 1 m v 2 + 2 1 I ω 2 V = 2 g h 1 + I M r 2 V = \sqrt{\frac{2gh}{1 + \frac{I}{Mr^2}}} V = 1 + M r 2 I 2 g h
Example Compare the speeds of a solid sphere, a hollow sphere, and a solid cylinder rolling down the same Incline from the same height.
Solid sphere: I = 2 5 M r 2 ⟹ v = 10 g h 7 I = \dfrac{2}{5}Mr^2 \implies v = \sqrt{\dfrac{10gh}{7}} I = 5 2 M r 2 ⟹ v = 7 10 g h Hollow sphere: I = 2 3 M r 2 ⟹ v = 6 g h 5 I = \dfrac{2}{3}Mr^2 \implies v = \sqrt{\dfrac{6gh}{5}} I = 3 2 M r 2 ⟹ v = 5 6 g h Solid cylinder: I = 1 2 M r 2 ⟹ v = 4 g h 3 I = \dfrac{1}{2}Mr^2 \implies v = \sqrt{\dfrac{4gh}{3}} I = 2 1 M r 2 ⟹ v = 3 4 g h The solid sphere is fastest, followed by the solid cylinder, then the hollow sphere. Objects with More mass concentrated near the centre (smaller I I I
Linear Quantity Rotational Equivalent Displacement s s s Angular displacement θ \theta θ Velocity v v v Angular velocity ω \omega ω Acceleration a a a Angular acceleration α \alpha α Mass m m m Moment of inertia I I I Force F F F Torque τ \tau τ Momentum p = m v p = mv p = m v Angular momentum L = I ω L = I\omega L = I ω F = m a F = ma F = ma τ = I α \tau = I\alpha τ = I α E k = 1 2 m v 2 E_k = \frac{1}{2}mv^2 E k = 2 1 m v 2 E k = 1 2 I ω 2 E_k = \frac{1}{2}I\omega^2 E k = 2 1 I ω 2 W = F s W = Fs W = F s W = τ θ W = \tau\theta W = τ θ P = F v P = Fv P = F v P = τ ω P = \tau\omega P = τ ω
A car of mass 1200 k g 1200\mathrm{ kg} 1200 kg 50 m 50\mathrm{ m} 50 m 15 m / s 15\mathrm{ m/s} 15 m/s
m v 2 r ≤ μ s m g \frac{mv^2}{r} \le \mu_s mg r m v 2 ≤ μ s m g \mu_s \ge \frac{v^2}`\{rg}` = \frac{225}{50 \times 9.81} = \frac{225}{490.5} = 0.459 A satellite orbits Earth at an altitude of 400 k m 400\mathrm{ km} 400 km
(a) Calculate the orbital period.
(M E = 5.97 × 10 24 k g M_E = 5.97 \times 10^{24}\mathrm{ kg} M E = 5.97 × 1 0 24 kg R E = 6.37 × 10 6 m R_E = 6.37 \times 10^6\mathrm{ m} R E = 6.37 × 1 0 6 m
R = 6.77 × 10 6 m R = 6.77 \times 10^6\mathrm{ m} R = 6.77 × 1 0 6 m \frac`\{GMm}`{r^2} = \frac{mv^2}{r} \implies v = \sqrt{\frac`\{GM}`{r}} = \sqrt{\frac{3.98 \times 10^{14}}{6.77 \times 10^6}} = 7669\mathrm{ m/s} T = 2 π r v = 2 π × 6.77 × 10 6 7669 = 5544 s ≈ 92.4 m i n T = \frac{2\pi r}{v} = \frac{2\pi \times 6.77 \times 10^6}{7669} = 5544\mathrm{ s} \approx 92.4\mathrm{ min} T = v 2 π r = 7669 2 π × 6.77 × 1 0 6 = 5544 s ≈ 92.4 min (b) Calculate the centripetal acceleration.
A c = v 2 r = 5.88 × 10 7 6.77 × 10 6 = 8.68 m / s 2 A_c = \frac{v^2}{r} = \frac{5.88 \times 10^7}{6.77 \times 10^6} = 8.68\mathrm{ m/s}^2 A c = r v 2 = 6.77 × 1 0 6 5.88 × 1 0 7 = 8.68 m/s 2 A 0.2 k g 0.2\mathrm{ kg} 0.2 kg 0.8 m 0.8\mathrm{ m} 0.8 m
(a) Find the minimum speed at the top of the circle for the ball to maintain contact.
V_{\mathrm{min}} = \sqrt`\{gr}` = \sqrt{9.81 \times 0.8} = \sqrt{7.85} = 2.80\mathrm{ m/s} (b) If the speed at the bottom is 8 m / s 8\mathrm{ m/s} 8 m/s
T b o t t o m = m g + m v 2 r = 0.2 ( 9.81 ) + 0.2 ( 64 ) 0.8 = 1.962 + 16 = 17.96 N T_{\mathrm{bottom}} = mg + \frac{mv^2}{r} = 0.2(9.81) + \frac{0.2(64)}{0.8} = 1.962 + 16 = 17.96\mathrm{ N} T bottom = m g + r m v 2 = 0.2 ( 9.81 ) + 0.8 0.2 ( 64 ) = 1.962 + 16 = 17.96 N (c) Find the tension at the top.
First find v t o p v_{\mathrm{top}} v top
1 2 m v b o t t o m 2 = 1 2 m v t o p 2 + m g ( 2 r ) \frac{1}{2}mv_{\mathrm{bottom}}^2 = \frac{1}{2}mv_{\mathrm{top}}^2 + mg(2r) 2 1 m v bottom 2 = 2 1 m v top 2 + m g ( 2 r ) 64 = v t o p 2 + 9.81 ( 1.6 ) = v t o p 2 + 15.70 64 = v_{\mathrm{top}}^2 + 9.81(1.6) = v_{\mathrm{top}}^2 + 15.70 64 = v top 2 + 9.81 ( 1.6 ) = v top 2 + 15.70 V t o p 2 = 48.30 ⟹ v t o p = 6.95 m / s V_{\mathrm{top}}^2 = 48.30 \implies v_{\mathrm{top}} = 6.95\mathrm{ m/s} V top 2 = 48.30 ⟹ v top = 6.95 m/s T t o p = m v t o p 2 r − m g = 0.2 ( 48.30 ) 0.8 − 1.962 = 12.075 − 1.962 = 10.11 N T_{\mathrm{top}} = \frac{mv_{\mathrm{top}}^2}{r} - mg = \frac{0.2(48.30)}{0.8} - 1.962 = 12.075 - 1.962 = 10.11\mathrm{ N} T top = r m v top 2 − m g = 0.8 0.2 ( 48.30 ) − 1.962 = 12.075 − 1.962 = 10.11 N A disc of mass 2 k g 2\mathrm{ kg} 2 kg 0.3 m 0.3\mathrm{ m} 0.3 m 10 r a d / s 10\mathrm{ rad/s} 10 rad/s
I = 1 2 M R 2 = 1 2 ( 2 ) ( 0.09 ) = 0.09 k g ⋅ m 2 I = \frac{1}{2}MR^2 = \frac{1}{2}(2)(0.09) = 0.09\mathrm{ kg}\cdot\mathrm{m}^2 I = 2 1 M R 2 = 2 1 ( 2 ) ( 0.09 ) = 0.09 kg ⋅ m 2 E k = 1 2 I ω 2 = 1 2 ( 0.09 ) ( 100 ) = 4.5 J E_k = \frac{1}{2}I\omega^2 = \frac{1}{2}(0.09)(100) = 4.5\mathrm{ J} E k = 2 1 I ω 2 = 2 1 ( 0.09 ) ( 100 ) = 4.5 J A diver has moment of inertia 15 k g ⋅ m 2 15\mathrm{ kg}\cdot\mathrm{m}^2 15 kg ⋅ m 2 4 k g ⋅ m 2 4\mathrm{ kg}\cdot\mathrm{m}^2 4 kg ⋅ m 2 2 r a d / s 2\mathrm{ rad/s} 2 rad/s
(a) Find her angular velocity when tucked.
I 1 ω 1 = I 2 ω 2 I_1\omega_1 = I_2\omega_2 I 1 ω 1 = I 2 ω 2 15 × 2 = 4 × ω 2 ⟹ ω 2 = 7.5 r a d / s 15 \times 2 = 4 \times \omega_2 \implies \omega_2 = 7.5\mathrm{ rad/s} 15 × 2 = 4 × ω 2 ⟹ ω 2 = 7.5 rad/s (b) How many complete somersaults can she perform in 1.2 s 1.2\mathrm{ s} 1.2 s
θ = ω 2 × t = 7.5 × 1.2 = 9 r a d \theta = \omega_2 \times t = 7.5 \times 1.2 = 9\mathrm{ rad} θ = ω 2 × t = 7.5 × 1.2 = 9 rad Number of somersaults = 9 2 π = 1.43 = \dfrac{9}{2\pi} = 1.43 = 2 π 9 = 1.43
She can complete 1 full somersault and is partway through a second.
Rotational motion is linear motion wearing a disguise. Every linear equation has a rotational twin: force becomes torque, mass becomes moment of inertia, and velocity becomes angular velocity. Moment of inertia is like rotational mass, but the answer varies based on on how far the mass is spread from the axis, which is why a figure skater spins faster with arms tucked in. Centripetal acceleration is the constant inward tug that keeps objects on circular paths, like a ball on a string that must always be pulled toward the center or the string breaks. Energy conservation links the top and bottom of a vertical circle, telling you exactly how fast you must go.
flowchart TD
A[4_Rotational Motion] --> B[Key Concepts]
A --> C[Core Principles]
A --> D[Practical Applications]
B --> E[Fundamental definitions]
C --> F[Design patterns]
D --> G[Real-world usage] Quantity Formula Angular velocity ω = v r = 2 π T \omega = \dfrac{v}{r} = \dfrac{2\pi}{T} ω = r v = T 2 π Centripetal acceleration a c = v 2 r = ω 2 r a_c = \dfrac{v^2}{r} = \omega^2 r a c = r v 2 = ω 2 r Centripetal force F c = m v 2 r = m ω 2 r F_c = \dfrac{mv^2}{r} = m\omega^2 r F c = r m v 2 = m ω 2 r Torque τ = F r ⊥ \tau = Fr_\perp τ = F r ⊥ Newton’s second law (rotation) τ = I α \tau = I\alpha τ = I α Angular momentum L = I ω L = I\omega L = I ω Rotational kinetic energy E k = 1 2 I ω 2 E_k = \dfrac{1}{2}I\omega^2 E k = 2 1 I ω 2 Conservation of angular momentum I 1 ω 1 = I 2 ω 2 I_1\omega_1 = I_2\omega_2 I 1 ω 1 = I 2 ω 2
Exam Strategy For circular motion problems, always draw a free-body diagram and identify which force(s) provide The centripetal force. For vertical circle problems, use energy conservation to relate speeds at Different points. For torque problems, identify the pivot and calculate the moment arm.
For constant angular acceleration α \alpha α
ω = ω 0 + α t \omega = \omega_0 + \alpha t ω = ω 0 + α t θ = ω 0 t + 1 2 α t 2 \theta = \omega_0 t + \frac{1}{2}\alpha t^2 θ = ω 0 t + 2 1 α t 2 ω 2 = ω 0 2 + 2 α θ \omega^2 = \omega_0^2 + 2\alpha\theta ω 2 = ω 0 2 + 2 α θ θ = 1 2 ( ω 0 + ω ) t \theta = \frac{1}{2}(\omega_0 + \omega)t θ = 2 1 ( ω 0 + ω ) t
Example A flywheel starts from rest and accelerates uniformly at 2 r a d / s 2 2\mathrm{ rad/s}^2 2 rad/s 2 5 s 5\mathrm{ s} 5 s
(a) Find the angular velocity after 5 s 5\mathrm{ s} 5 s
ω = 0 + 2 × 5 = 10 r a d / s \omega = 0 + 2 \times 5 = 10\mathrm{ rad/s} ω = 0 + 2 × 5 = 10 rad/s (b) Find the number of revolutions made.
θ = 0 + 1 2 ( 2 ) ( 25 ) = 25 r a d \theta = 0 + \frac{1}{2}(2)(25) = 25\mathrm{ rad} θ = 0 + 2 1 ( 2 ) ( 25 ) = 25 rad R e v o l u t i o n s = 25 2 π = 3.98 \mathrm{Revolutions} = \frac{25}{2\pi} = 3.98 Revolutions = 2 π 25 = 3.98 (c) Find the linear speed of a point 0.3 m 0.3\mathrm{ m} 0.3 m
V = r ω = 0.3 × 10 = 3.0 m / s V = r\omega = 0.3 \times 10 = 3.0\mathrm{ m/s} V = r ω = 0.3 × 10 = 3.0 m/s For a satellite of mass m m m r r r M M M
E_{\mathrm{total}} = -\frac`\{GMm}`{2r} E_k = \frac`\{GMm}`{2r} E_p = -\frac`\{GMm}`{r} V = \sqrt{\frac`\{GM}`{r}} T^2 = \frac{4\pi^2}`\{GM}`R^3 T 2 T^2 T 2 r 3 r^3 r 3
A geostationary satellite:
Orbits above the equator. Has a period of 24 hours (matches Earth’s rotation). Remains above the same point on Earth’s surface. Orbital radius ≈ 42200 k m \approx 42200\mathrm{ km} ≈ 42200 km A spinning gyroscope resists changes to its axis of rotation due to conservation of angular Momentum. This principle is used in:
Navigation systems (gyrocompasses). Stabilisation of ships and aircraft. Bicycle stability. Smartphone orientation sensors. When a torque is applied to a spinning object, instead of tipping over, the axis of rotation moves Perpendicular to the applied torque. This is called precession.
The precession angular velocity:
ω p = τ L \omega_p = \frac{\tau}{L} ω p = L τ A disc of mass 5 k g 5\mathrm{ kg} 5 kg 0.2 m 0.2\mathrm{ m} 0.2 m 0.5 N ⋅ m 0.5\mathrm{ N}\cdot\mathrm{m} 0.5 N ⋅ m 4 s 4\mathrm{ s} 4 s
(a) Find the angular acceleration.
α = τ I = 0.5 1 2 ( 5 ) ( 0.04 ) = 0.5 0.1 = 5 r a d / s 2 \alpha = \frac{\tau}{I} = \frac{0.5}{\frac{1}{2}(5)(0.04)} = \frac{0.5}{0.1} = 5\mathrm{ rad/s}^2 α = I τ = 2 1 ( 5 ) ( 0.04 ) 0.5 = 0.1 0.5 = 5 rad/s 2 (b) Find the angular velocity after 4 s 4\mathrm{ s} 4 s
ω = 0 + 5 × 4 = 20 r a d / s \omega = 0 + 5 \times 4 = 20\mathrm{ rad/s} ω = 0 + 5 × 4 = 20 rad/s (c) Find the rotational kinetic energy after 4 s 4\mathrm{ s} 4 s
E k = 1 2 I ω 2 = 1 2 ( 0.1 ) ( 400 ) = 20 J E_k = \frac{1}{2}I\omega^2 = \frac{1}{2}(0.1)(400) = 20\mathrm{ J} E k = 2 1 I ω 2 = 2 1 ( 0.1 ) ( 400 ) = 20 J (d) Find the work done by the torque.
W = τ θ = τ ⋅ 1 2 α t 2 = 0.5 × 1 2 ( 5 ) ( 16 ) = 20 J W = \tau\theta = \tau \cdot \frac{1}{2}\alpha t^2 = 0.5 \times \frac{1}{2}(5)(16) = 20\mathrm{ J} W = τ θ = τ ⋅ 2 1 α t 2 = 0.5 × 2 1 ( 5 ) ( 16 ) = 20 J This equals the change in rotational kinetic energy, confirming the work-energy theorem.
A thin rod of mass 2 k g 2\mathrm{ kg} 2 kg 1 m 1\mathrm{ m} 1 m
(a) Find the moment of inertia about the pivot.
I = 1 3 M L 2 = 1 3 ( 2 ) ( 1 ) = 0.667 k g ⋅ m 2 I = \frac{1}{3}ML^2 = \frac{1}{3}(2)(1) = 0.667\mathrm{ kg}\cdot\mathrm{m}^2 I = 3 1 M L 2 = 3 1 ( 2 ) ( 1 ) = 0.667 kg ⋅ m 2 (b) Find the initial angular acceleration.
The torque about the pivot: τ = m g × L 2 = 2 ( 9.81 ) ( 0.5 ) = 9.81 N ⋅ m \tau = mg \times \dfrac{L}{2} = 2(9.81)(0.5) = 9.81\mathrm{ N}\cdot\mathrm{m} τ = m g × 2 L = 2 ( 9.81 ) ( 0.5 ) = 9.81 N ⋅ m
α = τ I = 9.81 0.667 = 14.7 r a d / s 2 \alpha = \frac{\tau}{I} = \frac{9.81}{0.667} = 14.7\mathrm{ rad/s}^2 α = I τ = 0.667 9.81 = 14.7 rad/s 2 (c) Find the angular velocity as the rod passes through the vertical.
Using conservation of energy (taking the pivot as reference):
Loss of E p E_p E p L 2 = 0.5 m \dfrac{L}{2} = 0.5\mathrm{ m} 2 L = 0.5 m
M g L 2 = 1 2 I ω 2 Mg\frac{L}{2} = \frac{1}{2}I\omega^2 M g 2 L = 2 1 I ω 2 2 ( 9.81 ) ( 0.5 ) = 1 2 ( 0.667 ) ω 2 2(9.81)(0.5) = \frac{1}{2}(0.667)\omega^2 2 ( 9.81 ) ( 0.5 ) = 2 1 ( 0.667 ) ω 2 9.81 = 0.333 ω 2 ⟹ ω = 29.4 = 5.42 r a d / s 9.81 = 0.333\omega^2 \implies \omega = \sqrt{29.4} = 5.42\mathrm{ rad/s} 9.81 = 0.333 ω 2 ⟹ ω = 29.4 = 5.42 rad/s A horizontal turntable of radius 0.5 m 0.5\mathrm{ m} 0.5 m 3 r a d / s 3\mathrm{ rad/s} 3 rad/s 0.3 m 0.3\mathrm{ m} 0.3 m 0.4 0.4 0.4
A c = ω 2 r = 9 × 0.3 = 2.7 m / s 2 A_c = \omega^2 r = 9 \times 0.3 = 2.7\mathrm{ m/s}^2 A c = ω 2 r = 9 × 0.3 = 2.7 m/s 2 Required friction: f = m a c = m × 2.7 f = ma_c = m \times 2.7 f = m a c = m × 2.7
Maximum available friction: f max = μ s m g = 0.4 m ( 9.81 ) = 3.924 m f_{\max} = \mu_s mg = 0.4m(9.81) = 3.924m f m a x = μ s m g = 0.4 m ( 9.81 ) = 3.924 m
Since 2.7 m < 3.924 m 2.7m \lt 3.924m 2.7 m < 3.924 m
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