Physics Solution Manual for 1100 and 2101

67 10 11 n m 2 kg 2 h598 10 c 24 kg h 7690 m

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Unformatted text preview: 4 m g cos 50.0° ( 9.80 m/s 2 ) cos 50.0° 2 ______________________________________________________________________________ 260 DYNAMICS OF UNIFORM CIRCULAR MOTION 30. REASONING The centripetal force Fc required to keep an object of mass m that moves with speed v on a circle of radius r is Fc = mv 2 / r (Equation 5.3). From Equation 5.1, we know that v = 2 π r / T , where T is the period or the time for the suitcase to go around once. Therefore, the centripetal force can be written as m ( 2 π r / T ) 2 4 mπ 2 r Fc = = r T2 (1) This expression can be solved for T. However, we must first find the centripetal force that acts on the suitcase. SOLUTION Three forces act on the suitcase. They are the weight mg of the suitcase, the force of static friction f sMAX , and the normal force FN exerted on the suitcase by the surface of the carousel. The following figure shows the free body diagram for the suitcase. In this +y diagram, the y axis is along the vertical direction. MAX The force of gravity acts, then, in the –y direction. FN fs The centripetal force that causes the suitcase to θ move on its circular path is provided by the net θ +x force in the +x direction in the diagram. From the diagram, we can see that only the forces FN and f sMAX have horizontal components. Thus, we have θ Fc = f sMAX cosθ – FN sin θ , where the minus sign mg indicates that the x component of FN points to the left in the diagram. Using Equation 4.7 for the maximum static frictional force, we can write this result as in equation (2). Fc = µ s FN cosθ – FN sin θ = FN ( µ s cosθ – sin θ ) (2) If we apply Newton's second law in the y direction, we see from the diagram that FN cosθ + f s MAX sinθ – mg = ma y = 0 or FN cosθ + µ s FN sinθ – mg = 0 where we again have used Equation 4.7 for the maximum static frictional force. Solving for the normal force, we find mg FN = cosθ + µ s sinθ Using this result in equation (2), we obtain the magnitude of the centripetal force that acts on the suitcase: mg ( µ s cosθ – sin θ ) Fc = FN ( µ s cosθ – sin θ ) = cosθ + µ s sinθ With this expression for the centripetal force, equation (1) becomes Chapter 5 Problems mg ( µ s cosθ – sin θ ) cosθ + µ s sinθ = 261 4 mπ 2 r T2 Solving for the period T, we find T= 4π 2 r ( cos θ + µ s sinθ ) g ( µs cos θ – sin θ ) = 4π 2 (11.0 m) ( cos 36.0° + 0.760 sin 36.0° ) (9.80 m/s2 ) ( 0.760 cos 36.0° – sin 36.0° ) = 45 s 31. REASONING The speed v of a satellite in circular orbit about the earth is given by v = GM E / r (Equation 5.5), where G is the universal gravitational constant, ME is the mass of the earth, and r is the radius of the orbit. The radius is measured from the center of the earth, not the surface of the earth, to the satellite. Therefore, the radius is found by adding the height of the satellite above the surface of the earth to the radius of the earth 6 (6.38 × 10 m). SOLUTION First we add the orbital heights to the radius of the earth to obtain the orbital radii. Then we use Equation 5.5 to calculate the speeds. Satellite A rA = 6.38 × 10 6 m + 360 × 10 3 m = 6.74 × 10 6 m v= GM E rA = 6 c.67 × 10 −11 N ⋅ m 2 / kg 2 h5.98 × 10 c 24 kg h= 7690 m / s kg h= 7500 m / s 6.74 × 10 m 6 Satellite B rA = 6.38 × 10 6 m + 720 × 10 3 m = 7.10 × 10 6 m v= GM E rA = 6 c.67 × 10 −11 N ⋅ m 2 / kg 2 c h5.98 × 10 24 7 .10 × 10 6 m _____________________________________________________________________________________________ 32. REASONING AND SOLUTION 2 We have for Jupiter v = GMJ/r, where 5 7 7 r = 6.00 × 10 m + 7.14 × 10 m = 7.20 × 10 m Thus, v= ( 6.67 ×10−11 N ⋅ m 2 / kg 2 )(1.90 ×1027 kg ) = 7.20 × 10 m 7 4.20 × 104 m/s ____________________________________________________________________________________________ 262 DYNAMICS OF UNIFORM CIRCULAR MOTION 33. SSM WWW REASONING Equation 5.5 gives the orbital speed for a satellite in a circular orbit around the earth. It can be modified to determine the orbital speed around any planet P by replacing the mass of the earth M E by the mass of the planet M P : v= GM P / r . SOLUTION The ratio of the orbital speeds is, therefore, v2 v1 = GM P / r2 GM P / r1 = r1 r2 Solving for v2 gives v 2 = v1 r1 r2 = (1.70 × 10 4 m / s) 5.25 × 10 6 m = 1.33 × 10 4 m / s 8.60 × 10 6 m 34. REASONING AND SOLUTION The normal force exerted by the wall on each astronaut is 2 the centripetal force needed to keep him in the circular path, i.e., Fc = mv /r. Rearranging and letting Fc = (1/2)mg yields 2 2 2 r = 2v /g = 2(35.8 m/s) /(9.80 m/s ) = 262 m 35. REASONING In Section 5.5 it is shown that the period T of a satellite in a circular orbit about the earth is given by (see Equation 5.6) T= 2π r 3 / 2 GM E where r is the radius of the orbit, G is the universal gravitational constant, and ME is the mass of the earth. The ratio of the periods of satellites A and B is, then, 3 2π rA/2 TA TB = GM E 3/2 2π rB = 3/2 rA 3/2 rB GM E We do not know the radii rA and rB...
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