Selected Hw Problems for Ch. 10

Selected Hw Problems for Ch. 10 - 10.2 What is the angular...

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10.2 What is the angular speed of (a) the second hand, (b) the minute hand, and (c) the hour hand of a smoothly running analog watch? Answer in radians per second. a) b) 1rev 2 rad min rad 0.1047 min rev 60s s π = c) 3 1rev 2 rad min rad 1.745x10 60min rev 60s s π = 4 1rev 2 rad hr min rad 1.454x10 12hr rev 60min 60s s π =
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10.16 A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is rotating at 10 rev/s; 60 revolutions later, its angular speed is 15 rev/s. Calculate (a) the angular acceleration, (b) the time required to complete the 60 revolutions, (c) the time required to reach the 10 rev/s angular speed, and (d) the number of revolutions from rest until the time the disk reaches the 10 rev/s angular speed. 22 oo 2( ) ω=ω+αθ−θ o 2 o rev rev (15 ) (10 ) rev ss 1.042 2( ) 2(60 rev) s ω−ω α= = = θ−θ a) By analogy with, vv 2 a ( x x ) =+ we have Solving for α, b) Since we’re told the angular speeds at the start and end of the 60 rev rotation we use (by analogy with ), o t ω=ω +α o a t = + o 2 rev rev 15 10 t 4.80 s rev 1.042 s == = α Solving for t ,
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2 oo 1 xx v t a t 2 =+ + d) Using (by analogy with ) 2 1 tt 2 θ=θ +ω + α Since the disk starts from rest, 22 2 11 r e v t (1.042 )(9.60s) 48.0 rev s θ= α = = o 2 rev rev 10 0 ss t9 . 6 0 s rad 1.042 s ω−ω == = α c) The time from starting at rest to going 62.83 rad/s ,
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10.22 If an airplane propeller rotates at 2000 rev/min while the airplane flies at a speed of 480 km/h relative to the ground, what is the linear speed of a point on the tip of the propeller, at radius 1.5 m, as seen by (a) the pilot and (b) an observer on the ground? The plane's velocity is parallel to the propeller's axis of rotation. θ s r The identity relating linear to circular measure is, sr = θ For constant r, differentiating wrt time gives, ds d r dt dt θ = vr = ω Then for the pilot, rev 2 rad m v r (1.5m)(2000 ) 314.16 60s rev s π =ω= =
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For an observer on the ground, the propeller tip slices a spiral path through the air. Its speed perpendicular to the direction of the planes motion is what we calculated on the previous page. We must add to this the planes speed as vector quantities. I.e. if looked at from above while the propeller is vertical we have two perpendicular velocity vectors to be added, tip velocity plane’s velocity net velocity Then, speed of point = 22 km 1000 m hr m m (480 ) (314.16 ) 341.28 hr 3600 km s s s +=
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10. 33
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Selected Hw Problems for Ch. 10 - 10.2 What is the angular...

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