We will see more of this kind of thing in section 106

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Unformatted text preview: r path of radius r with constant angular velocity ω . Suppose that at time t, the object has swept out an angle measuring θ radians. If we assume that the object is at the point (r, 0) when t = 0, the angle θ is in standard position. By definition, ω = θ which we rewrite as θ = ωt. According t to Theorem 10.3, the location of the object Q(x, y ) on the circle is found using the equations x = r cos(θ) = r cos(ωt) and y = r sin(θ) = r sin(ωt). Hence, at time t, the object is at the point (r cos(ωt), r sin(ωt)).11 y r Q (x, y ) = (r cos(ωt), r sin(ωt)) 1 θ = ωt 1 rx Equations for Circular Motion Example 10.2.7. Suppose we are in the situation of Example 10.1.5. Find the equations of motion of Lakeland Community College as the earth rotates. π Solution. From Example 10.1.5, we take r = 2960 miles and and ω = 12 hours . Hence, the π π equations of motion are x = r cos(ωt) = 2960 cos 12 t and y = r sin(ωt) = 2960 sin 12 t , where x and y are measured in miles and t is measured in hours. In addition to circular motion, Theorem 10.3 is also the key to developing what is usually called ‘right triangle’ trigonometry.12 As we...
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