Physics 106 Group Problems Summer 2015 Circular motion and Newton's law of gravity Name: TA: 1. (5 points) One of the many planets found outside our solar system is Kepler-23C. Its radius is 3.2 times greater than Earths and it has a mass 860 times greater than Earths. What would be the gravitational acceleration on Kepler-23Cs surface in gs, where a g = 9.8 m=s2? Solution: ME g = G 2 RE M23C g23C = G 2 R23C 860ME g23C = G 2 (3:2RE) ME g23C = 84G 2 RE g23C = 84g 2. (5 points) Two masses m1 and m2 are held a distance d apart and the gravitational force between the two of them is measured. They are then moved a distance 4d apart. By what factor has the gravitational attraction between the two changed? Solution: m m F = G 1 2 NEW (4d)2 1 F = F NEW 16 SOLUTIONS 1 3. (5 points) Suppose one night the radius of the earth doubled but its mass stayed the same. What would be an approximate new value for the free-fall acceleration at the surface of the earth? Solution: ME g = G 2 RE ME gNEW = G 2 2RE 1 ME gNEW = G 2 4 RE 1 g = g NEW 4 4. (5 points) Two identical stars, at a fixed distance 2R apart, revolve in a circle about their mutual center of mass, as shown to the right. Each star has mass M and speed v. What is the relation among M,R and v? Solution: Fc = Fgrav v2 M 2 M = G R (2R)2 GM v2 = 4R SOLUTIONS 2 5. (5 points) A flywheel is initially rotating at !0 =20 rad=s and has a constant angular acceleration. After 9.0 s it has rotated through θ=450 rad. What is the angular acceleration α of the wheel? Solution: 2 ∆θ = !0t + 1=2αt ∆θ − ! t α = 2 0 t2 α = 6:7rad=s2 6. (5 points) A wheel of radius R=0.5 m rolls without sliding on a horizontal surface. Starting from rest, the wheel moves with constant angular acceleration α =6 rad=s2. What is the distance traveled by the center of the wheel from t =0 to t =3 s ? Solution: Whenever the problem mentions rolling without sliding or slipping that is a hint for us to use the relation between angular and linear acceleration a = αr and velocity v = !r. a = αr 1 ∆x = x + v t + at2 0 0 2 1 ∆x = 0 + 0 + at2 2 ∆x = 13:5m SOLUTIONS 3 7. (10 points) An amusement park ride consists of a rotating circular platform d=11.4 m in diameter from which 10.0 kg seats are sus- pended at the end of l=1.18 m massless chains. When the system rotates, the chains makes an angle of θ = 43:7◦ with the vertical. What is the speed of each seat? Solution: The radius of the circular motion is given by R = l sin θ + d=2 R = 6:52m From the conditions of the force equilibrium in the horizontal and vertical direction we obtain: v2 T sin θ = m R T cos θ = mg tan θ = v2=gR v = pgR tan θ v = 7:82m=s SOLUTIONS 4 8. (10 points) A curve of radius 80 m is banked at θ = 45◦. Suppose that an ice storm hits, and the curve is effectively frictionless. What is the safe speed with which to take the curve without either sliding up or down? Solution: The simplest approach to this problem is if we consider the force equilibrium in the horizontal and vertical direc- tions. v2 N sin θ = m R N cos θ = mg v = pgR tan θ v = 28m=s SOLUTIONS 5 9. (10 points) A car travels along a reversed banked turn. That is the road is tilted away from the center of curvature of the road. If the coefficient of static friction is µs = 0:5, the radius of the curvature is R = 15 m, and the banking angle is θ = 10◦, what is the maximum speed at which a car can safely navigate such a turn? Solution: The normal reaction force N is due to the sum of the components of the gravity and the centripetal force in the normal to the surface direction. v2 N = (mg cos θ − m sin θ) R v2 µN = mg sin θ + m cos θ R Solving for v we get v2 (µ sin θ + cos θ) = g(µ cos θ − sin θ) R µ cos θ − sin θ v2 = Rg µ sin θ + cos θ v = 6:61m/s v2 Where we used the expressions for the centripetal and friction force Fc = m R and f = µN respectively. SOLUTIONS 6.