Circular Motion Tangential & Angular Acceleration v • Tangential Acceleration: t The arc length s is related to the angle θ (in radians = rad) as follows: s = rθ

The tangential velocity vt is related to the angular velocity ω as follows: Tangential vt = rω Velocity

The tangential acceleration at is related to the angular acceleration α as follows:

dv dω vt = rω Δω dω t a at = = r = rα α = lim = (radians/s2) t dt dt Δt→0 Δt dt • Overall Acceleration: Radial Axis ar r r r ˆ Tangential r atot = aradial + at = −aradial rˆ + atθ Acceleration

r 2 2 atot = atot = aradial + at Radial Acceleration Rick Field 2/6/2014 PHY 2053 Page 1 University of Florida Angular Equations of Motion

• Angular Equations of Motion (constant α): If the angular acceleration α is constant then α(t) = α (radians/s2) a (t) = rα (m/s2) θ t ω(t) =θω +αt (radians/s) v (t) = v + a t (m/s) ω 0 ω t t0 t ω 2 2 (t) = + t + 1 αt (radians) s(t) = s + v t + 1 a t (m) 0 0α θ2 0 t0 2 t 2 2 2 2 v t (t) − v = 2a (s(t) − s ) (t) − 0 = 2 () (t) −θ0 t0 t 0 at 2 2 2 2 a (t) = rω (t) (m/s ) aradial (t) = v t (t) / r (m/s ) radial Radial Axis

ar

Tangential r Acceleration Radial Acceleration

Rick Field 2/6/2014 PHY 2053 Page 2 University of Florida Angular Equations of Motion

• Angular Equations of Motion (constant α): θ (t) Let N = Number of revolutions (rev) N(t) = 2π ω(t) Let f = Number of revolutions per second f (t) = (frequency) 2π

α (rev/s2) α(t) = α (rad/s2) 2π θ α ω(t) =θω +αt (rad/s) f (t) = f + ( )t (rev/s) ω 0 ω 0 2π ω 2 1 α 2 (t) = + t + 1 αt (rad) N(t) = N + f t + ( )t (rev) 0 0α θ2 0 0 2 2π 2 2 2 2 α (t) − 0 = 2 ()(t) −θ0 f (t) − f0 = 2( 2π )(N(t) − N0 )

Rick Field 2/6/2014 PHY 2053 Page 3 University of Florida Angular Equations: Examples

• A disk rotates about its central axis starting from rest at t = 0 and accelerates with constant angular acceleration. At one time it is rotating at 4 rev/s; 60 revolutions later, its angular speed is 16 rev/s. Starting at t = 0, what is the time required to complete 64 revolutions? α f 2 (t) − f 2 (16rev / s)2 − (4rev / s)2 Answer: t = 8 seconds = 0 = = 2rev / s2 2π 2()N(t) − N0 2()60rev 2(N(t) − N ) 2(64rev) 1 α 2 t = 0 = = 8s N(t) − N0 = 2 ( 2π )t α 2 ( 2π ) (2rev / s )

• An astronaut is being tested in a centrifuge. The centrifuge has a radius R and, in starting from rest at t = 0, rotates with a constant angular acceleration α = 0.25 rad/s2 . At what time t > 0 is the ω magnitude of the tangential acceleration equal to the magnitude of the radial accelerationα (i.e. centripetal acceleration)? 2 2 2 aradial (t) = R (t) = R t = at = Rα Answer: t = 2 seconds 1 1 t = = = 2s α 0.25rad / s2 Rick Field 2/6/2014 PHY 2053 Page 4 University of Florida Exam 2 Spring 2011: Problem 2

• A race car accelerates uniformly from a speed of 40 m/s to a speed of 58 m/s in 6 seconds while traveling around a circular track of radius 625 m. When the car reaches a speed of 50 m/s what is the magnitude of its total acceleration (in m/s2)? Answer: 5 % Right: 49% v2 − v1 (58m / s) − (40m / s) at = = = 3m / s t2 − t1 6s v2 (50m / s)2 a = = = 4m / s r R 625m 2 2 atot = at + ar = 5m / s

Rick Field 2/6/2014 PHY 2053 Page 5 University of Florida Gravitation: Circular Orbits (M >> m)

For circular orbits the gravitational force is perpendicular to the velocity and hence the speed of the mass m is constant. The force

Fg is equal to the mass times the radial (i.e. centripetal) M v acceleration as follows: r Fg m GmM v2 GM F = = ma = m = mrω 2 r = (radius of the orbit, constant) g r 2 radial r v2 GM Assume M >> m so that the v = (speed, constant) position of M is fixed! r

GM For circular orbits r, v, and ω = π (angular velocity, constant) 3 ω are also constant. r π 2 r r r 3 T = = 2 r = 2π In general both masses rotate v GM GM (period of rotation) about the center-of-mass and the formulas are more complicated! • Kepler’s Third Law: v 2 3 M m 2 4π r The period squared is CM Fg T = proportional to the radius cubed. × m GM rM rm VM Rick Field 2/6/2014 PHY 2053 Page 6 University of Florida Circular Orbits: Example

• Two satellites are in circular orbit around the Earth. The first satellite

has mass M1 and is travelling in a circular orbit of radius R1. The second satellite has mass M2 = M1 is travelling in a circular orbit of radius R2 = 4R1. If the first satellite completes one revolution of the Earth in time T, how long doesπ it take the second satellite to make one revolution of the Earth? π Answer: 8T

2 3 2 3 2 3 2 3 2 4π R1 2 4 R2 4 (4R1) 4π R1 2 T1 = T2 = = = 64 = 64T1 GM1 GM 2 GM1 GM1

T2 = 8T1 = 8T

Rick Field 2/6/2014 PHY 2053 Page 7 University of Florida Circular Orbits: Example

• Two diametrically opposed masses m revolve around a M = 2m circle of radius R. A third mass M = 2m is located at the m m center of the circle. What is the period T of rotation for R this system of three masses? 4π R3 Answer: T = 3 Gm

2 2 π GmM Gm Gm v F M = 2m 1 m m Fgrav = 2 + 2 = 2 ()M + 4 m = maradial = m R π (2R) R R R

G(M + 1 m) v = π4 R π 2 R R R 4R 4π R3 T = = 2 R = 2 R = 2 R = 1 1 v G(M + 4 m) G(2m + 4 m) 9Gm 3 Gm

Rick Field 2/6/2014 PHY 2053 Page 8 University of Florida