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Assignment 8

# Assignment 8 - PHYS 230 Assignment 8 Fall 2007 1 French...

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PHYS 230 Fall 2007 Assignment 8 1. French, problem 2.10, extended. 10 points. A particle moves along the curve y = Ax 2 such that its x position is given by x = Bt . (a) Express the vector position of the particle in the form r ( t ) = x i + y j . Since y = Ax 2 = AB 2 t 2 , we get r = Bt ˆ i + AB 2 t 2 ˆ j . (b) Calculate the speed v (= ds/dt ) of the particle along this path at an arbitrary instant. Differentiate x and y , separately, with respect to t . We get v = B ˆ i + 2 AB 2 t ˆ j . Therefore, the speed v is v = B 2 + 4 A 2 B 4 t 2 . (c) Show that the angular velocity ω = ( dθ/dt ) at an arbitrary instant is AB/ (1 + A 2 B 2 t 2 ). The component of the velocity v along ˆ θ is just rdθ/dt = r ˙ θ = - v x sin θ + v y cos θ . But sin θ = y/r and cos θ = x/r , so: r ˙ θ = - AB 3 t 2 (1 - 2) /r ˙ θ = + AB 3 t 2 / ( B 2 t 2 + A 2 B 4 t 4 ) = AB/ (1 + A 2 B 2 t 2 ) 2. 10 points. A penny is rolling upright in a straight line along a horizontal surface. Its speed is constant. A red spot is painted at one point on its circumference. Plot a graph of the position of this spot as a function of time as seen by a stationary observer. Show that once every revolution the velocity of the red spot is exactly zero. &% ’\$ - v ¡ ¡ s θ - ω The velocity of the red spot has two pieces. They are a constant velocity v = v ˆ x , and an angular velocity about the centre of the penny. Note, however, that when the rolling motion is to the right (positive x ), the rolling motion is clockwise, so that ω is negative, ω = - v/R . (This is the condition for rolling . . . ) In the frame moving with velocity v ˆ x , the position of the red spot is ´ x = R cos θ and ´ y = R + R sin θ , where we choose θ = - ωt - π/ 2 , so that x = y = 0 at t = 0 . In the frame of the surface, at time t , the centre of the penny is at ( vt, R ) . The red spot is therefore at ( vt + R cos θ, R + R sin θ ) . When you draw this, you get a so-called cycloid. The graph has R = 1 , ω = 2 π . 1

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0 5 10 15 20 25 0 1 2 3 4 The velocity is just d dt v . We find this by differentiating each component independently . . . v x = v + sin - ωt - π/ 2 = v (1 - sin ωt + π/ 2) v y = - cos - ωt - π/ 2 = - v cos ωt + π/ 2 By inspection, whenever t = n 2 π/ω ; n = 0 , 1 , 2 . . . , we get both v x and v y = 0 . This is whenever the red spot is on the ground: y = 0 . 3. French, problem 2.16 (part (a) only). Slightly modified, for clarity. 10 points. This problem is about the relationship between polar and Cartesian coordinates. Research Project: complete the questionnaire for this question. The orbital radius of Venus is 0.72 times the radius of Earth’s orbit. Its period is about 0.62 times Earth’s year. Using these data, find how the apparent angular positions of Venus changes with time as seen from Earth, assuming that the orbits of both planets lie in the same plane. Present your results as a plot of angular position versus time, and compare them semi-quantitatively with Figure 2-20 (see next page).
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