ps8_soln - Physics 105 Problem Set 8 Solutions Problem 1 (3...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 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 by the volume V of a sphere with radius x . = M e 4 3 R 3 e V = 4 3 x 3 (2) M = V = 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 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 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 = 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 = radicalbigg k m = radicalBigg GM e R 3 e (6) Thus for the period of oscillation T we find, T = 2 = 5058 s = 1 . 4 h (7) 1 R r e 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 theres gravity. Define 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 = vector r vector F pointing into the page for small displacements have: = l 2 mg sin lk x t cos l 2 k x m cos (8) I = l 2 mg sin l 2 k sin cos l 2 4 k sin cos (9) For a rod pivoting about an end point the moment of inertia I = 1 3 ml 2 . For small angles sin and cos 1. Substituting this in and simplifying we find, + parenleftBig 15 k 4 m 3 g 2 l parenrightBig = 0 (10) In this form we recognize the coefficient of as 2 , thus: = radicalbigg 15 k 4 m 3 g 2 l (11) Problem 3...
View Full Document

Page1 / 8

ps8_soln - Physics 105 Problem Set 8 Solutions Problem 1 (3...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online