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 860 times greater than Earths. What would be the gravitational 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 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 ω0 =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 = ω0t + 1/2αt ∆θ − ω 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 = ωr.

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 is tilted away from the center of curvature of the road. If the coefficient of static 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 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

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