Physics Solution Manual for 1100 and 2101

4 with t0 0 s as 0 t if has the same sign as 0 then

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Unformatted text preview: than θ0, then ω is negative. SOLUTION The average angular velocity is given by Equation 8.2 as ω = t – t0 = 2.0 s is the elapsed time: (a ) (b) ω= (c) ω= (d ) 3. ω= ω= θ − θ0 t − t0 θ − θ0 t − t0 θ − θ0 t − t0 θ − θ0 t − t0 = 4.2 rad − 5.4 rad = −0.60 rad /s 2.0 s = , where 0.54 rad − 0.94 rad = −0.20 rad /s 2.0 s = t − t0 0.75 rad − 0.45 rad = +0.15 rad /s 2.0 s = θ − θ0 3.8 rad − 3.0 rad = +0.4 rad /s 2.0 s REASONING The average angular velocity ω is defined as the angular displacement ∆θ divided by the elapsed time ∆t during which the displacement occurs: ω = ∆θ / ∆t (Equation 8.2). This relation can be used to find the average angular velocity of the earth as it spins on its axis and as it orbits the sun. 394 ROTATIONAL KINEMATICS SOLUTION a. As the earth spins on its axis, it makes 1 revolution (2π rad) in a day. Assuming that the positive direction for the angular displacement is the same as the direction of the earth’s rotation, the angular displacement of the earth in one day is ( ∆θ )spin = +2π rad . The average angular velocity is (converting 1 day to seconds): ω= ( ∆θ )spin = ( ∆t )spin +2π rad = +7.3 × 10−5 rad/s 24 h 3600 s 1 day 1 day 1 h ( ) b. As the earth orbits the sun, the earth makes 1 revolution (2π rad) in one year. Taking the positive direction for the angular displacement to be the direction of the earth’s orbital motion, the angular displacement in one year is ( ∆θ )orbit = +2π rad . The average angular velocity is (converting 365¼ days to seconds): ω= ( ∆θ )orbit = ( ∆t )orbit ( +2π rad 24 h 365 1 days 4 1 day ) 3600 s 1 h = +2.0 ×10−7 rad/s ____________________________________________________________________________________________ 4. REASONING The average angular velocity of either mandible is given by ω = ∆θ ∆t (Equation 8.2), where ∆θ is the angular displacement of the mandible and ∆t is the elapsed time. In order to calculate the average angular velocity in radians per second, we will first convert the angular displacement ∆θ from degrees to radians. SOLUTION Converting an angular displacement of 90° into radians, we find that the angular displacement of the mandible is 1 rev ∆θ = 90 degrees 360 degrees ( ) 2π rad π = rad 1 rev 2 The average angular velocity of the mandible is π rad ∆θ = 2 −4 = 1.2 × 10 4 rad/s ω= ∆t 1.3 × 10 s (8.2) Chapter 8 Problems 5. 395 SSM REASONING The average angular velocity is equal to the angular displacement divided by the elapsed time (Equation 8.2). Thus, the angular displacement of the baseball is equal to the product of the average angular velocity and the elapsed time. However, the problem gives the travel time in seconds and asks for the displacement in radians, while the angular velocity is given in revolutions per minute. Thus, we will begin by converting the angular velocity into radians per second. SOLUTION Since 2π rad = 1 rev and 1 min = 60 s, the average angular velocity ω (in rad/s) of the baseball is 330 rev 2 π rad 1 min ω = = 35 rad/s min 1 rev 60 s Since the average angular velocity of the baseball is equal to the angular displacement ∆θ divided by the elapsed time ∆t, the angular displacement is ∆θ = ω ∆t = ( 35 rad/s )( 0.60 s ) = 21 rad 6. (8.2) REASONING The jet is maintaining a distance of r = 18.0 km from the air traffic control tower by flying in a circle. The angle that the jet’s path subtends while its nose crosses over the moon is the same as the angular width θ of the moon. The corresponding distance the jet travels is the length of arc s subtended by the moon’s diameter. We will use the relation s = rθ (Equation 8.1) to determine the distance s. SOLUTION In order to use the relation s = rθ (Equation 8.1), the angle θ must be expressed in radians, as it is. The result will have the same units as r. Because s is required in meters, we first convert r to meters: 1000 m 4 r = 18.0 km = 1.8 × 10 m 1 km ( ) Therefore, the distance that the jet travels while crossing in front of the moon is ( )( ) s = rθ = 1.80 ×104 m 9.04 ×10−3 rad = 163 m 7. REASONING α= ω − ω0 t − t0 The average angular acceleration has the same direction as ω − ω0, because , according to Equation 8.4. If ω is greater than ω0, α is positive. If ω is less than ω0, α is negative. 396 ROTATIONAL KINEMATICS SOLUTION The average angular acceleration is given by Equation 8.4 as α = ω − ω0 t − t0 , where t – t0 = 4.0 s is the elapsed time. (a ) (b) α= (c) α= (d ) 8. α= α= ω − ω0 t − t0 ω − ω0 t − t0 ω − ω0 t − t0 ω − ω0 t − t0 = +5.0 rad /s − 2.0 rad /s 2 = +0.75 rad /s 4.0 s = +2.0 rad /s − 5.0 rad /s = −0.75 rad /s 2 4.0 s = = −3.0 rad /s − ( −7.0 rad /s ) 4.0 s −4.0 rad /s − ( +4.0 rad /s ) 4.0 s = +1.0 rad /s 2 = −2.0 rad /s 2 REASONING The relation between the final angular velocity ω, the initial angular velocity ω0, and the angular acceleration α is given by Equati...
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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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