Physics Solution Manual for 1100 and 2101

3 fc mv0 r we are given the values for m and r however

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Unformatted text preview: ________________________________________________ 56. REASONING The speed of the satellite is given by Equation 5.1 as v = 2π r / T . Since we are given that the period is T = 1.20 × 104 s, it will be possible to determine the speed from Equation 5.1 if we can determine the radius r of the orbit. To find the radius, we will use Equation 5.6, which relates the period to the radius according to T = 2π r 3 / 2 / GM E , where G is the universal gravitational constant and ME is the mass of the earth. SOLUTION According to Equation 5.1, the orbital speed is v= 2π r T To find a value for the radius, we begin with Equation 5.6: 2π r 3 / 2 T= GM E r 3/ 2 = or T GM E 2π Next, we square both sides of the result for r3/2: ( r 3/ 2 ) 2 T GM E = 2π 2 or r3 = T 2GM E 4π 2 276 DYNAMICS OF UNIFORM CIRCULAR MOTION We can now take the cube root of both sides of the expression for r3 in order to determine r: r= T 2GM E 3 4π 2 1.20 × 10 4 s ) ( 6.67 × 10 −11 N ⋅ m 2 /kg 2 )( 5.98 × 1024 kg ) 3( = = 1.13 × 107 m 2 4π 2 With this value for the radius, we can use Equation 5.1 to obtain the speed: ( ) 7 2π r 2π 1.13 × 10 m v= = = 5.92 × 103 m/s 4 T 1.20 × 10 s 57. SSM REASONING AND SOLUTION The centripetal acceleration for any point on the blade a distance r from center of the circle, according to Equation 5.2, is ac = v 2 / r . From Equation 5.1, we know that v = 2 π r / T where T is the period of the motion. Combining these two equations, we obtain ac = ( 2π r / T ) 2 4π 2 r = r T2 a. Since the turbine blades rotate at 617 rev/s, all points on the blades rotate with a period of T = (1/617) s = 1.62 × 10–3 s . Therefore, for a point with r = 0.020 m , the magnitude of the centripetal acceleration is ac = 4 π 2 ( 0.020 m) = 3.0 × 10 5 m / s 2 –3 2 (1.62 × 10 s) b. Expressed as a multiple of g, this centripetal acceleration is c a c = 3.0 × 10 5 m / s 2 hF 1.00 gs I = G m/ J H K 9.80 2 3.1 × 10 4 g _____________________________________________________________________________________________ 58. REASONING The centripetal acceleration for any point that is a distance r from the center of the disc is, according to Equation 5.2, ac = v 2 / r . From Equation 5.1, we know that v = 2 π r / T where T is the period of the motion. Combining these two equations, we obtain ac = ( 2π r / T ) 2 4π 2 r = r T2 Chapter 5 Problems 277 SOLUTION Using the above expression for ac , the ratio of the centripetal accelerations of the two points in question is a 2 4 π 2 r2 / T22 r2 / T22 = = a 1 4 π 2 r1 / T12 r1 / T12 Since the disc is rigid, all points on the disc must move with the same period, so T1 = T2 . Making this cancellation and solving for a2 , we obtain a 2 = a1 r2 r1 c = 120 m / s 2 FJ G mK h0.050 m I = H 0.030 2.0 × 10 2 m / s 2 Note that even though T1 = T2 , it is not true that v1 = v 2 . Thus, the simplest way to approach this problem is to express the centripetal acceleration in terms of the period T which cancels in the final step. _____________________________________________________________________________________________ 59. SSM WWW REASONING Let v0 be the initial speed of the ball as it begins its 2 projectile motion. Then, the centripetal force is given by Equation 5.3: FC = mv0 / r . We are given the values for m and r; however, we must determine the value of v0 from the details of the projectile motion after the ball is released. In the absence of air resistance, the x component of the projectile motion has zero acceleration, while the y component of the motion is subject to the acceleration due to gravity. The horizontal distance traveled by the ball is given by Equation 3.5a (with a x = 0 m/s2): x = v 0 x t = ( v 0 cosθ ) t with t equal to the flight time of the ball while it exhibits projectile motion. The time t can be found by considering the vertical motion. From Equation 3.3b, v y = v0 y + a y t After a time t, v y = − v0 y . Assuming that up and to the right are the positive directions, we have −2 v 0 y −2 v 0 sin θ t= = ay ay and x = ( v 0 cosθ ) F2 v sinθ I − Ga J H K 0 y Using the fact that 2 sin θ cosθ = sin 2θ , we have 278 DYNAMICS OF UNIFORM CIRCULAR MOTION x=− 2 2 v 0 cosθ sin θ ay =− 2 v 0 sin 2θ (1) ay Equation (1) (with upward and to the right chosen as the positive directions) can be used to determine the speed v0 with which the ball begins its projectile motion. Then Equation 5.3 can be used to find the centripetal force. SOLUTION Solving equation (1) for v0 , we have v0 = −x ay sin 2θ = − ( 86.75 m)(–9.80 m / s 2 ) = 29 . 3 m / s sin 2(41° ) Then, from Equation 5.3, 2 mv 0 (7.3 kg)(29.3 m / s) 2 FC = = = 3500 N r 1.8 m _____________________________________________________________________________________________ 60. REASONING AND SOLUTION a. The centripetal force is provided by the normal force exerted on the rider by the wall . b. Newton's second law applied in the horizontal direction gives 2 2 FN = mv /r = (55.0 kg)(10.0 m/s) /(3.30 m) = 1670 N c....
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