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Unformatted text preview: Physics 105 Problem Set 8 Solutions Problem 1 (3 pts) a) Dig a narrow tunnel straight through the center of the Earth. What is the force F ( x ) as a function of distance x from the center of the Earth? With gravity, a spherical shell produces no force on a mass inside of it. Hence only the mass of the Earth with a radius less than x contributes to the force. For a gravitational force we have: vector F ( x ) = − GMm x 2 ˆ x (1) However now M does not equal M e the mass of the Earth. We can find M by multiplying the density of Earth ρ by the volume V of a sphere with radius x . ρ = M e 4 3 πR 3 e V = 4 3 πx 3 (2) M = ρV = M e x 3 R 3 e (3) Inserting this into our equation for force we get: vector F ( x ) = − GM e mx R 3 e ˆ x (4) b) What is the component F x ( x ) of the force along the 1000 km tunnel as a function of position x from the center of the tunnel? Here we follow the same prescription as above, except now we take the component of force along the direction of the tunnel. Let F ( r ) be the force from part a) and let θ be the angle between the tunnel and a radial vector to position x . Here r is the distance between the center of earth and position x . See Figure 1. F x ( x ) = F ( r ) cos θ = F ( r ) x r = − GM e mx R 3 e (5) c) What is the period of the oscillation in each tunnel? Notice that the force has the form F = − kx , the same as a spring force. The frequency of oscillation for a spring is ω = radicalbigg k m = radicalBigg GM e R 3 e (6) Thus for the period of oscillation T we find, T = 2 π ω = 5058 s = 1 . 4 h (7) 1 R r e θ d Figure 1: Tunnel Coordinates Problem 2 (3 pts) Here we have an inverted pendulum of mass m , and length l , with springs of spring constant k attached to its midpoint and top end. And there’s gravity. Define θ to be the angle from vertical in the clockwise direction and let ∆ x t , ∆ x m be the linear displacements of the top and middle of the rod respectively. We will make the further assumption that the equlibrium length of the spring is much greater than l the length of the rod and thus sthe springs remain essentially horizontal for small displacements. Taking vertical to be the equilibrium position and looking at the torque vector τ = vector r × vector F pointing into the page for small displacements have: τ = l 2 mg sin θ − lk ∆ x t cos θ − l 2 k ∆ x m cos θ (8) I ¨ θ = l 2 mg sin θ − l 2 k sin θ cos θ − l 2 4 k sin θ cos θ (9) For a rod pivoting about an end point the moment of inertia I = 1 3 ml 2 . For small angles sin θ ≈ θ and cos θ ≈ 1. Substituting this in and simplifying we find, ¨ θ + parenleftBig 15 k 4 m − 3 g 2 l parenrightBig θ = 0 (10) In this form we recognize the coefficient of θ as ω 2 , thus: ω = radicalbigg 15 k 4 m − 3 g 2 l (11) Problem 3...
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 '08
 LYMANA.PAGE
 mechanics, Force, Gravity, Complex number, fo, ωo, Rω, sthe springs

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