Physics Solution Manual for 1100 and 2101

00 rads400 s 1 250 rads 2 400 s 2 400 101 rad 2 2

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Unformatted text preview: are equal. SOLUTION The time during which some part of the arrow remains in the open angular space is the time it takes the arrow to travel at a speed v through a distance equal to its own length L. This time is tArrow = L/v. The time it takes for the edge to rotate at an angular speed ω through the angle θ between the blades is tBlade = θ/ω. The maximum angular speed is the angular speed such that these two times are equal. Therefore, we have L θ = v ω { { Arrow Blade In this expression we note that the value of the angular opening is θ = 60.0º, which is θ = 1 ( 2π ) rad = 1 π rad . Solving the expression for ω gives 6 3 ω= θv L = πv 3L Substituting the given values for v and L into this result, we find that a. ω= b. ω= c. ω= πv 3L πv 3L πv 3L = = = π ( 75.0 m/s ) 3 ( 0.71 m ) π ( 91.0 m/s ) 3 ( 0.71 m ) π ( 91.0 m/s ) 3 ( 0.81 m ) = 111 rad/s = 134 rad/s = 118 rad/s 15. REASONING AND SOLUTION The baton will make four revolutions in a time t given by t= θ ω Half of this time is required for the baton to reach its highest point. The magnitude of the initial vertical velocity of the baton is then 400 ROTATIONAL KINEMATICS v0 = g ( 1 t ) = g 2θω 2 With this initial velocity the baton can reach a height of 2 h= v0 2g = gθ 2 8ω 2 (9.80 m/s ) (8π rad ) 2 = 2 rev 2π rad 8 1.80 s 1 rev 2 = 6.05 m 16. REASONING The time required for the bullet to travel the distance d is equal to the time required for the discs to undergo an angular displacement of 0.240 rad. The time can be found from Equation 8.2; once the time is known, the speed of the bullet can be found using Equation 2.2. SOLUTION From the definition of average angular velocity: ω= the required time is ∆t = ∆θ ω = ∆θ ∆t 0.240 rad = 2.53 × 10 −3 s 95.0 rad/s Note that ω = ω because the angular speed is constant. The (constant) speed of the bullet can then be determined from the definition of average speed: v= 17. ∆x d 0.850 m = = = ∆t ∆t 2.53 × 10−3 s 336 m/s SSM REASONING AND SOLUTION a. If the propeller is to appear stationary, each blade must move through an angle of 120° or 2π / 3 rad between flashes. The time required is t= θ = ω (2π / 3) rad –2 = 2.00 ×10 s 2π rad (16.7 rev/s) 1 rev b. The next shortest time occurs when each blade moves through an angle of 240°, or 4π / 3 rad, between successive flashes. This time is twice the value that we found in part a, or 4.00 × 10 −2 s. Chapter 8 Problems 18. REASONING AND SOLUTION The figure at the right shows the relevant angles and dimensions for either one of the celestial bodies under consideration. 401 r θ s person on earth celestial body a. Using the figure above θ moon = θ sun = s moon 3.48 × 10 6 m –3 = = 9.04 × 10 rad 8 rmoon 3.85 × 10 m s sun 1.39 × 109 m –3 = = 9.27 × 10 rad rsun 1.50 × 1011 m b. Since the sun subtends a slightly larger angle than the moon, as measured by a person standing on the earth, the sun cannot be completely blocked by the moon. Therefore, a "total" eclipse of the sun is not really total . c. The relevant geometry is shown below. r sun R sun s sun R r moon b s b θ sun moon θ moon person on earth The apparent circular area of the sun as measured by a person standing on the earth is given 2 by: Asun = π Rsun , where Rsun is the radius of the sun. The apparent circular area of the sun 2 that is blocked by the moon is Ablocked = π Rb , where Rb is shown in the figure above. Also from the figure above, it follows that Rsun = (1/2) ssun and Rb = (1/2) sb Therefore, the fraction of the apparent circular area of the sun that is blocked by the moon is 402 ROTATIONAL KINEMATICS Ablocked Asun = 2 π Rb 2 π Rsun = θ = moon θ sun π ( sb / 2)2 π ( ssun s = b 2 s / 2) sun 2 θ moon rsun = θ sun rsun 2 2 2 9.04 × 10−3 rad = = 0.951 9.27 × 10−3 rad The moon blocks out 95.1 percent of the apparent circular area of the sun. 19. REASONING AND SOLUTION Since the ball spins at 7.7 rev/s, it makes (7.7 rev/s)t revolutions while in flight, where t is the time of flight and must be determined. The ball’s vertical displacement is y = 0 m since the ball returns to its launch point. The vertical component of the ball’s initial velocity is v0y = (19 m/s) sin 55°, assuming upward to be the positive direction. The acceleration due to gravity is ay = −9.80 m/s2. With these three 1 2 pieces of information at hand, we use y = v0 y t + a y t 2 (Equation 3.5b) to determine the time of flight. Noting that y = 0 m, we can solve this expression for t and find that t=− 2v0 y ay =− 2 (19 m/s ) sin 55° −9.80 m/s 2 =3.2 s and Number of = 7.7 rev/s 3.2 s = 25 rev )( ) revolutions ( ____________________________________________________________________________________________ 20. REASONING The angular displacement is given as θ = 0.500 rev, while the initial angular velocity is given as ω0 = 3.00 rev/s and the final angular velocity as ω = 5.00 rev/s. Since we seek the time t, we can use Equation 8.6 θ = 1 (ω0 + ω ) t from the equations of 2 rotational kinematics to obtain it. SOLUTION Solving Equation 8.6 for the time t, we find that t= 21. 2 ( 0.500 rev...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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