Physics 2414 Assignment 6 Solutions: Circular Motion and Gravitation - Prof. Braden Abbott, Study notes of Physics

Download Physics 2414 Assignment 6 Solutions: Circular Motion and Gravitation - Prof. Braden Abbott and more Study notes Physics in PDF only on Docsity! Phys. 2414 Solutions for Assignment 6 October 16, 2010 REMEMBER, when going through the solutions, some of your numbers may be different than the ones provided here. # 1. A 0.075 kg toy airplane is tied to the ceiling with a string. When the airplane’s motor is started, it moves with a constant speed of 1.13 m/s in a horizontal circle of radius 0.39 m. We begin by summing forces in the x and y directions. Defining x as hori- zontal, y as vertical, and T is the tension in the string, we have: ∑ Fx = mv2 r = T sin(θ) (1) and ∑ Fy = mg − T cos(θ) = 0→ mg = T cos(θ). (2) Solving the above equations for the tension T, we are left with T = mv2 r sin(θ) (3) and T = mg cos(θ) . (4) Combining these equations gives mv2 r sin(θ) = mg cos(θ) . (5) 1 Solving Eq. (5) for θ and using the identity tan(θ) = sin(θ) cos(θ) , It is seen that θ = tan−1 ( v2 rg ) = 18.5◦. (6) Now that we have the angle, we can use Eq. (4) above to solve the tension: T = mg cos(θ) = 0.775 N (7) # 2. Four masses are positioned at the corners of a rectangle as shown below. For simplicity, lets call the force of the 1 kg mass on the 2 kg mass F12, for the 3 kg mass and 2 kg mass F32, etc. In this problem we will be using Newton’s law of gravitation, F12 = G m1m2 r2 (8) where G = 6.67× 10−11 and r is the distance between the two masses. Now, we have F12 = G m1m2 x2 F32 = G m3m2 y2 F42 = G m4m2 x2+y2 It is simple to see that F12 is in the x-direction and F32 is in the y-direction. So we have to break F42 into it’s x and y components. Now we have F42x = F42 cos(θ) F42y = F42 sin(θ) 2 # 5. A curve of radius 75 m is banked for a design speed of 110 km/h. If the coefficient of static friction is 0.30 (wet pavement), at what range of speeds can a car safely make the curve? The first step is to find the angle at which the curve is banked. This is given by tan(θ) = v2 gr (20) Now, to find the maximum speed that a vehicle can have, we must first sum forces in the x and y directions. In the y-direction we have∑ Fy = FN cos(θ)− µsFN sin(θ)−mg = 0 (21) And in the x-direction, ∑ Fx = m v2 r = FN sin(θ) + µsFN cos(θ) (22) Solving for the normal force, we have FN = mg cos(θ)− µs sin(θ) (23) We can plug this back into the equation for the sum of forces in the x-direction and solve for velocity. We get v2 = gr sin(θ) + µs cos(θ) cos(θ)− µs sin(θ) (24) Now for the minimum speed, we follow the same steps, except that the sign on the frictional force is switched. So the minimum speed is v2 = gr sin(θ)− µs cos(θ) cos(θ) + µs sin(θ) (25) # 6. You are an astronaut in the space shuttle pursuing a satellite in need of repair. You are in a circular orbit of the same radius as the satellite (380 km above the Earth), but 20 km behind it.Constants used in this problem: G = 6.67× 10−11 5 RE = 6.38× 106 m ME = 5.98× 1024 kg It is also VERY important to keep track of your units in this problem. Throughout the problem, it is assumed that all distances are measure in METERS. (a) How long will it take to overtake the satellite if you reduce your or- bital radius by 1.0 km? For this problem, we will use the relation mv2 r′ = G mME r′2 ⇒ v = √ GME r′ (26) where r′ = r + RE, r being the original orbital radius is the orbital radius being considered. Now we can find the velocity of the satellite vs directly: vs = √ GME r′ = 7681.4 m/s (27) The speed of the astronaut va will be different due to the fact that the radius is decreased by 1000 m. So we have va = √ GME r′ − 1000 = 7681.96 m/s (28) The difference between this two, vd = va − vs = 0.56 m/s is the speed with which the astronaut will catch the satellite. So the time to catch is given by t = x vd = 20000 0.56 = 35714.3 s = 9.92 hrs. (b) By how much must you reduce your orbital radius to catch up in 7 hours? First, we must find with what velocity the astronaut has to travel to catch the satellite in 7 hours. This is found by v = x t = 20000 25200 = 0.79365 m/s (29) 6 Now the new velocity of the astronaut is va = v + vs = 7682.19 m/s. Using the equation v = √ GME r ⇒ r = GME v2 = 6.7586 m (30) Subtracting off the radius of the earth and the original radius of orbit, we are left how much the radius should be shortened, r = 1400 m. # 7. A 310 kg piano slides 4.6 m down a 30◦ incline and is kept from accelerating by a man who is pushing back on it parallel to the incline. The effective coefficient of kinetic friction is 0.40. (a) Calculate the force exerted by the man. First, we set the x-axis along the motion of the piano. Now, using the notation that Fp is the force with which the man pushes, and the fact that since velocity is constant so there is no acceleration, the forces are∑ Fx = Fmg − Ff − Fp = 0⇒ Fp = Fmg − Ff (31) Since Fmg = mg sin θ and Ff = µkmg cos θ, we now have Fp = mg(sin θ − µk cos θ = 310(9.8)(sin(30)− 0.4 cos(30)) = 466.6 N (32) (b) Calculate the work done by the man on the piano. The work done by the man, Wm is given by Wm = Fpd cos(θ) = 466.6(4.6) cos(180) = −2146.4 J (33) (c) Calculate the work done by the friction force. Wf = Ffd cos(θ) = 1052.4(4.6) cos(180) = −4841 J (34) (d) What is the work done by the force of gravity? Wmg = Fmgd cos θ = 1519(4.6) cos(0) = 6987.4 J (35) (e) What is the net work done on the piano? 7