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These are the fundamental points in the following Lecture Slides : Orbits, Diameter, Car Traveling, Hour, Constant, Total Path, Velocity, Total Time, Total Distance, Period
Typology: Slides
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Example: Your car’s wheels are 65 cm in diameter and are spinning at ω = 101 rads/sec. How fast in km/hour is the car traveling, assuming no slipping?
v = total distance total time = ( 2 π^ r ) N (^ T ) N^ = 2 π r T = ω r = (^) ( 1 01 rads/sec) ( 32. 5 cm) = 3. 28 × 10 3 cm/sec = 118 km/hr Given: d = 65 cm = 0.65 m ω = 101 rads/sec Find: v =? Idea: since the car is moving with constant velocity, divide total path to total time!
The time it takes to go one time around a closed path is called the period (T). T r v 2 π total time total distance av = = Comparing to v = rω: f T π π ω 2 2 = = f is called the frequency , the number of revolutions (or cycles) per second. Period and Frequency
Escape Speed The escape speed is the speed needed for an object to soar off into space and not return For the earth, v
is about 11.2 km/s Note, v is independent of the mass of the object v esc = 2 GM E R E
Example: How high above the surface of the Earth does a satellite need to be so that it has an orbit period of 24 hours? r GM v e Previously: (^) = Also need, T r v 2 π = Combine these and solve for r: 3 1 2 2 4 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = T GM r e π ( )( ) ( )
Kepler’s First Law All planets move in elliptical orbits with the Sun at one focus.
Kepler’s Second Law A line drawn from the Sun to any planet will sweep out equal areas in equal times
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Given: h = 613 km = 6.13 × 105 m Find: T =? ∑^ F =^ F g = GmHs Me r 2 = m HS a r = m HS v 2 r Need v! To find it, write 2 nd Newton’s law: T = 2 π r v Idea: period T is the time needed to complete one revolution, i.e. a path of length s = 2πr! So, ⇒ T = 2 π r 3 GM e = 2 π ( 613 × 10 3 m + 6371 × 10 6 m ) 3 ( 6. 674 × 10 − 11 Nm 2 / kg 2 )( 5. 974 × 10 24 kg ) T = 5808 s = 1. 613 h Or with numbers (and remembering that r = RE+h):
13 Tangential and Angular Acceleration
av t t f i Δ Δ = Δ − = ω ω ω α
2
t (^) Δ t Δ = Δ→ ω α 0 lim
tangential component of velocity: v r ω t = α ω a r t t r t v a t t t = Δ → Δ Δ = Δ Δ = (in thelimit 0 )
v v a x x v t a t x v v t v v v a t fx ix x ix x fx ix x fx ix x − = Δ Δ = Δ + Δ Δ = + Δ Δ = − = Δ 2 2 1 ( ) 2 1 2 2 2 v (^) t = r ω , at = r α ω ω α θ θ ω α θ ω ω ω ω ω α − = Δ Δ = Δ + Δ Δ = + Δ Δ = − = Δ 2 2 1 ( ) 2 1 2 2 2 f i i f i f i t t t t Solution: given final and initial angular velocity and angle. Find acceleration.
f 2
i 2
f 2
i 2
2
2
Example: A disk rotates with constant angular acceleration. The initial angular speed of the disk is
the magnitude of the angular acceleration? (b) How much time did it take for the disk to rotate through
of a point located at a distance of 5.0 cm from the center of the disk? v v a x x v t a t x v v t v v v a t fx ix x ix x fx ix x fx ix x − = Δ Δ = Δ + Δ Δ = + Δ Δ = − = Δ 2 2 1 ( ) 2 1 2 2 2 v (^) t = r ω , at = r α ω ω α θ θ ω α θ ω ω ω ω ω α − = Δ Δ = Δ + Δ Δ = + Δ Δ = − = Δ 2 2 1 ( ) 2 1 2 2 2 f i i f i f i t t t t (^2 2 2 ) 2 2 f i^2 f i
(a) (b) f i f i 1 2 2(10 rad) ( ) , so 2.2 s. 2 7 rad s 2 rad s t t θ π θ ω ω ω ω π π Δ Δ = + Δ Δ = = =
(c) 2 2 2 t (7 rad s) (2 rad s) (0.050 m) 0.35 m s 2(10 rad) a r π π α π − = = =
x y
x y
FBD for person at the bottom position FBD for person at the top position
y r 2
y r 2
Apply Newton’s 2nd^ Law to each:
Example: A space station is shaped like a ring and rotates to simulate gravity. If the radius of the space station is 120m, at what frequency must it rotate so that it simulates Earth’s gravity? Using the result from the previous slide:
Example: What is the minimum speed for the car so that it maintains contact with the loop when it is in the top (pictured) position?
Example: What is the minimum speed for the car so that it maintains contact with the loop when it is in the top (pictured) position? FBD for the car at the top of the loop:
y x Apply Newton’s 2nd^ Law: r v N w m r v F N w ma m y r 2 2
