Physics Solution Manual for 1100 and 2101

It is certainly true that greater chapter 8 problems

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ctly to find the angular displacement, because the final angular velocity (not the initial angular velocity), the acceleration, and the time are known. We will combine two of the equations, Equations 8.4 and 8.6 to obtain an expression for the angular displacement that contains the three known variables. 406 ROTATIONAL KINEMATICS SOLUTION The angular displacement of each wheel is equal to the average angular velocity multiplied by the time θ= 1 2 ( ω0 + ω ) t 14243 (8.6) ω The initial angular velocity ω0 is not known, but it can be found in terms of the angular acceleration and time, which are known. The angular acceleration is defined as (with t0 = 0 s) α= ω − ω0 or t ω0 = ω − α t (8.4) Substituting this expression for ω0 into Equation 8.6 gives θ = 1 ( ω − α t ) + ω t = ω t − 1 α t 2 2 2 43 1 24 ω0 = ( +74.5 rad /s )( 4.50 s ) − 1 2 ( +6.70 rad /s ) ( 4.50 s ) 2 2 = +267 rad ______________________________________________________________________________ 28. REASONING AND SOLUTION circumstance. The angular acceleration is found for the first 2 ω 2 − ω 0 ( 3.14 × 104 rad/s ) − (1.05 × 104 rad/s ) 4 2 = = 2.33 × 10 rad/s α= 4 2θ 2 (1.88 × 10 rad ) 2 2 For the second circumstance t= ω − ω 0 7.85 × 104 rad/s − 0 rad/s = = 3.37 s 4 2 α 2.33 × 10 rad/s ____________________________________________________________________________________________ 29. REASONING There are three segments to the propeller’s angular motion, and we will calculate the angular displacement for each separately. In these calculations we will remember that the final angular velocity for one segment is the initial velocity for the next segment. Then, we will add the separate displacements to obtain the total. Chapter 8 Problems 407 SOLUTION For the first segment the initial angular velocity is ω0 = 0 rad/s, since the propeller starts from rest. Its acceleration is α = 2.90 × 10−3 rad/s2 for a time t = 2.10 × 103 s. Therefore, we can obtain the angular displacement θ1 from Equation 8.7 of the equations of rotational kinematics as follows: [First segment] θ1 = ω 0t + 1 α t 2 = ( 0 rad/s ) ( 2.10 ×103 s ) + 1 ( 2.90 × 10−3 rad/s 2 ) ( 2.10 ×103 s ) 2 2 2 = 6.39 ×103 rad The initial angular velocity for the second segment is the final velocity for the first segment, and according to Equation 8.4, we have ω = ω 0 + α t = 0 rad/s + ( 2.90 ×10−3 rad/s 2 ) ( 2.10 × 103 s ) = 6.09 rad/s Thus, during the second segment, the initial angular velocity is ω0 = 6.09 rad/s and remains constant at this value for a time of t = 1.40 × 103 s. Since the velocity is constant, the angular acceleration is zero, and Equation 8.7 gives the angular displacement θ2 as [Second segment] θ 2 = ω 0t + 1 α t 2 = ( 6.09 rad/s ) (1.40 ×103 s ) + 1 ( 0 rad/s 2 ) (1.40 ×103 s ) = 8.53 ×103 rad 2 2 2 During the third segment, the initial angular velocity is ω0 = 6.09 rad/s, the final velocity is ω = 4.00 rad/s, and the angular acceleration is α = −2.30 × 10−3 rad/s2. When the propeller picked up speed in segment one, we assigned positive values to the acceleration and subsequent velocity. Therefore, the deceleration or loss in speed here in segment three 2 means that the acceleration has a negative value. Equation 8.8 (ω 2 = ω 0 + 2αθ 3 ) can be used to find the angular displacement θ3. Solving this equation for θ3 gives [Third segment] 2 ω 2 − ω 0 ( 4.00 rad/s ) − ( 6.09 rad/s ) = = 4.58 ×103 rad θ3 = −3 2 2α 2 ( −2.30 ×10 rad/s ) 2 2 The total angular displacement, then, is θ Total = θ1 + θ 2 + θ 3 = 6.39 ×103 rad + 8.53 × 103 rad + 4.58 ×103 rad = 1.95 × 104 rad 30. REASONING Since the time t and angular acceleration α are known, we will begin by using Equation 8.7 from the equations of kinematics to determine the angular displacement θ : 408 ROTATIONAL KINEMATICS θ = ω0 t + 1 α t 2 2 However, the initial angular velocity ω0 is not given. We can determine it by resorting to another equation of kinematics, ω = ω0 + α t (Equation 8.4), which relates ω0 to the final angular velocity ω, the angular acceleration, and the time, all of which are known. SOLUTION Solving Equation 8.4 for ω0 gives ω0 = ω − α t . Substituting this result into θ = ω0t + 1 α t 2 gives 2 θ = ω0t + 1 α t 2 = (ω − α t ) t + 1 α t 2 = ω t − 1 α t 2 2 2 2 = ( +1.88 rad/s )(10.0 s ) − 1 ( −5.04 rad/s 2 ) (10.0 s ) = +2.71× 102 rad 2 2 31. REASONING According to Equation 3.5b, the time required for the diver to reach the water, assuming free-fall conditions, is t = 2 y / a y . If we assume that the "ball" formed by the diver is rotating at the instant that she begins falling vertically, we can use Equation 8.2 to calculate the number of revolutions made on the way down. SOLUTION Taking upward as the positive direction, the time required for the diver to reach the water is 2(–8.3 m) t= 2 = 1.3 s –9.80 m/s Solving Equation 8.2 for ∆ θ , we find ∆θ = ω ∆t = (1.6 rev/s)(1.3 s)= 2.1 rev 32....
View Full Document

Ask a homework question - tutors are online